[SOLVED] Gauge Transformations in Momentum Space?

[mikem@despammed.com]

Most textbook treatments of gauge transformations do it in
position space. So far, I haven't found any that discuss
in detail what they look like in momentum space, and what
issues arise in the QFT Fock space.

For example, consider the U(1) group for electromagnetism
acting on Dirac electrons in QED. Textbooks write it as
something like:

Psi(x) -> exp(i theta(x)) Psi(x)

where theta(x) is a real scalar function of x (in 3+1
spacetime of course).

To pass to momentum space, we need to assume that
exp(i theta(x)) has a reasonable Fourier transform,
such that the transformation can be represented
in momentum space as an integral operator whose
kernel is a distribution.

Take a simple case: theta(x) = wt, where 'w' is a real constant.
I.e: in position space we have

Psi(x) -> exp(iwt) Psi(x)

In momentum space, this just shifts the energy by an amount 'w'
I.e: E -> E - w.

So old positive-energy modes in the energy range 0 to w get
transformed into negative-energy modes. In the 2nd-quantized Fock
space this means we're mixing some of the annihilation and creation
operators - because they were defined in terms of the original +ve
and -ve energy modes. Such mixing usually means that we're mapping
between unitarily inequivalent representations, i.e: between
orthogonal Fock spaces.

I'm interested in finding explicit operators which are form-invariant
in both representations. I tried Google-Scholar but didn't have much
success.

So my question is:

Do any textbooks or review papers discuss this stuff at length?
(I don't mean just the usual Bogoliubov transformations from
condensed matter physics which map between inequivalent reps,
but specifically for standard model gauge transformations
in momentum space, and hence Fock space(s).)

TIA.

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[Igor Khavkine]

On 2005-09-13, mikem@despammed.com <mikem@despammed.com> wrote:
> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
> For example, consider the U(1) group for electromagnetism
> acting on Dirac electrons in QED. Textbooks write it as
> something like:
>
> Psi(x) -> exp(i theta(x)) Psi(x)
>
> where theta(x) is a real scalar function of x (in 3+1
> spacetime of course).
>
> To pass to momentum space, we need to assume that
> exp(i theta(x)) has a reasonable Fourier transform,
> such that the transformation can be represented
> in momentum space as an integral operator whose
> kernel is a distribution.

That's pretty much all there is to it.

> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> I.e: in position space we have
>
> Psi(x) -> exp(iwt) Psi(x)
>
> In momentum space, this just shifts the energy by an amount 'w'
> I.e: E -> E - w.

You forget that a gauge transformation acts either on a pure E&M system,
or on an E&M system minimally coupled to some matter field Psi(x). In
this case, you are subtracting grad phi(x) to the E&M vector potential
A(x), which comes out just subtracting w from the time component of
A(x). Because the E&M field is minimally coupled, this constant shift by
w enters into the action/Hamiltonian in the same way as i d/dt. In
other words, the entire spectrum of the matter states gets shifted by
the same amount, including the energy of the ground state. Positive and
"negative" energy states are measured relative to the ground state
energy. Thus, when the whole energy spectrum gets shifted by the same
constant, no mixing between positive and "negative" energy states
occurs. So gauge transformations don't do much, just as they should.

Hope this helps.

Igor

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[thomas_larsson_01@hotmail.com]

mikem@despammed.com skrev:

> Most textbook treatments of gauge transformations do it in
> position space. So far, I haven't found any that discuss
> in detail what they look like in momentum space, and what
> issues arise in the QFT Fock space.
>
[...]
>
> So old positive-energy modes in the energy range 0 to w get
> transformed into negative-energy modes. In the 2nd-quantized Fock
> space this means we're mixing some of the annihilation and creation
> operators - because they were defined in terms of the original +ve
> and -ve energy modes. Such mixing usually means that we're mapping
> between unitarily inequivalent representations, i.e: between
> orthogonal Fock spaces.
>

Your statement is incorrect. Negative-energy gauge transformations,
like all negative-energy operators, annihilate the vacuum, and
thus we stay within the Hilbert space. However, this is closely
related to an interesting and difficult set of questions, which I
have spent quite a lot of time thinking about.

Let us first consider the free Maxwell field. The kinematical
Hilbert space is spanned by canonical variables E_u(x) and A_v(x),
where x is a point in 3-space, Greek indices u,v = 0,1,2,3 and
Latin indices i,j = 1,2,3. The constraints are

E_0(x) = 0, G(x) = d_i E_i(x) = 0.

The latter is Gauss' law, which is the generator of gauge
transformations. The Maxwell field is just a bunch of harmonic
oscillators a_u(k), a*_u(k), where (up to numerical factors)

A_u(x) = int dk ( a_u(k) exp(ikx) + a*_u(k) exp(-ikx) )

E_u(x) = int dk ( a_u(k) exp(ikx) - a*_u(k) exp(-ikx) )

The oscillators are labelled by 3-momenta k, and their energy is
w_k = |k| and -|k|, respectively. The normal-ordered Hamiltonian
is

H = int dk w_k a^u(k) a*_u(k),

and the Fock vacuum is annilated by the annihition operators
a_u(k),

a_u(k)|vac> = 0.

In particular, the negative-energy part of the constraints
annihilate the vacuum. So the answer to your question is negative:
the gauge generators do not create states of negative energy,
because such states are annilated by the vacuum.

Physical quantitities should be gauge invariant, so we want to mod
out the constraints. The physical Hilbert space is spanned by
states satisfying

E_0(x)|phys> = 0, G(x)|phys> = 0.

The crucial property that makes this possible is that the
kinematical Hilbert space carries a quantum representation of the
constraint algebra. The representation is of lowest-energy type
(any operator O with negative H eigenvalue annihilates the
vacuum), and the Gauss law algebra holds without anomalies,
[G(x), G(y)] = 0. The physical Hilbert space has a positive-definite
inner product, so time evolution is unitary.

Interacting theories such as non-abelian Yang-Mills or QED with
matter are more subtle. E.g., in Yang-Mills the canonical
variables are E^a_u(x) and A^b_v(x), and the constraints read

E^a_0(x) = 0,

G^a(x) = d_i E^a_i(x) + f^abc A^b_i(x) E^c_i(x) = 0,

where f^abc are the structure constants. The last term is not
linear, which makes a huge difference. In order to obtain a
well-defined action on the Fock vacuum, we must normal order the
bilinear term. If we try to calculate the bracket of two
normal-ordered Gauss constraints, we will find something like

[G^a(x), G^b(y)] = f^abc G^c(x) delta(x-y)

+ f^acd f^bcd delta(x-y) delta(y-x).

The second term is proportional to delta(0)! The same kind of
extension arises if we consider QED with matter. In general the
extension is proportional to the value of the second Casimir
operator. It vanishes only in the free Maxwell theory, because the
adjoint rep of U(1) is trivial.

The appearence of the infinities in interacting field theories is
hardly surprising. We can render some observables finite by
renormalization, but I think that in this process G^a(x) ceases to
be a well-defined operator. The renormalized operators presumably
violate associativity, e.g.

(G^a(x))_ren (G^b(x))_ren != (G^a(x) G^b(x))_ren,

which means that they are not operators at all. AFAIU, every
renormalization scheme more complicated that normal ordering will
have such problems. If the renormalized generators were operators,
they would obey some extension of the Gauss law algebra, and the
kinematical Hilbert space would carry a lowest-energy
representation of the extended gauge algebra. The rep would not
need to be unitary, but the Hamiltonian must be bounded from below
by construction, i.e. the rep must be of lowest-energy type. It is
known how to construct lowest-energy reps of anomalous gauge
algebras, and subtracting off counterterms is not how you do it.

To understand what is going on, it is useful to consider a simpler
model, which is not quite as trivial as the free Maxwell field,
but where constraint generators remain operators after
renormalization: the free bosonic string in D dimensions. String
theory is a complex subject, but my point only requires that we
know that it has a gauge symmetry with generators L_m, m in Z, and
brackets

[L_m, L_n] = (n-m) L_m+n.

It is important that the gauge generators carry energy, because
the Hamiltonian H = L_0 does not commute with them;
[H, L_m] = m L_m. Actually, we have two commuting sets of
gauge generators and the Hamiltonian is the sum of two terms,
but this is irrelevant here.

When we quantize the theory, we demand that the kinematical
Hilbert space should have a lowest energy state |vac>, and thus

L_m |vac> = 0, for all m < 0.

So negative energy states are not allowed to arise even before
gauges are modded out! Next we want to pass to the physical
Hilbert space, by restriction to physical states satisfying

L_m |phys> = 0, for all m, positive or negative.

However, it is well known that there is an obstruction to
implementing this condition - the constraint algebra acquires a
central extension, the conformal anomaly c:

[L_m, L_n] = (n-m) L_m+n - (c/12) (m^3 - m) delta_m+n,

where delta_m is the Kronecker delta. Only if c = 0, which happens
if D = 26, can we impose the physical state condition. If we try
to impose it for non-zero c, we find that

[L_-m, L_m] |phys> = (c/12) (m^3 - m) |phys> = 0,

so the physical Hilbert space would be completely empty - not a
very interesting theory. This is the origin of the special role of
D = 26 in bosonic string theory. However, the free string defines
a consistent theory even if D < 26. This is because the inner
product in the kinematical Hilbert space is already
positive-definite (or at least can be made so by without factoring
out much less than all constraints).

Everything up til now is, I believe, completely undisputable.
However, the remainder of this post is not. I believe that it will
be possible to construct a formulation where the constraints are
well-defined operators on the kinematical Hilbert space, perhaps
not for all non-abelian gauge theories but at least for the
standard model. The Gauss law constraints will then be anomalous,
but the full algebra must be a well-defined Lie algebra extension.
This is a very powerful condition. In terms of the smeared gauge
generators

G(X) = \int X^a(x) G^a(x),

the only extension for which representations of lowest-energy type
are known is given by

[G(X), G(Y)] = G([X,Y]) +
+ k \int dt dq^i/dt <d_i X(q(t)),Y(q(t))>,

where <.,.> denotes the Killing form. Here q^i(t) is a privileged
curve in space.

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[mikem@despammed.com]

I wrote:

>> Take a simple case: theta(x) = wt, where 'w' is a real constant.
>> I.e: in position space we have
>>
>> Psi(x) -> exp(iwt) Psi(x)
>>
>> In momentum space, this just shifts the energy by an
>> amount 'w'. I.e: E -> E - w.

Igor Khavkine wrote:

> [...] when the whole energy spectrum gets shifted by the same
> constant, no mixing between positive and "negative" energy states
> occurs. So gauge transformations don't do much, just as they should.

Hmmm. I now see that my simple example of "theta(x) = wt" was too
simple to illustrate my question adequately. Let me try again...

I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field. This is because I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

I'll change my earlier notation slightly and use lower case
"psi(x)" for the position-space field operator, and upper case
"Psi(p)" for the corresponding momentum-space field operator. With
this notation, a U(1) transformation acting on psi(x) is written:

psi(x) -> psi'(x) = exp(i theta(x)) psi(x)

Now consider the not-so-simple example: theta(x) = st^2
where s is a real constant.

Suppressing the 3-space coordinates for brevity, we have:

psi'(t) = exp(ist^2) psi(t)

Taking Fourier transforms...

Psi'(p) = F(exp(ist^2)) * Psi(p), where "*" is convolution.

= Int dq sqrt(2 pi i/s) exp(-iq^2/4s) Psi(p-q)

For psi(t) = exp(iEt), it Fourier transform is a delta fn, so:

Psi'(E) = sqrt(2 pi i/s) exp(-iE^2/4s)

which has support from E = -inf to +inf.

I.e: this gauge transformation maps a +E energy delta fn
delta(E) into a function with support from E = -inf to +inf. The
new field is thus an infinite superposition of both +ve and -ve
frequencies.

In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation *and* creation
operators. That's usually the signal that we're dealing with a
Bogoliubov transformation between disjoint Fock spaces, and that's
what I'm trying to prove (or disprove) definitively, for arbitrary
theta(x).

Any further thoughts would be most welcome...

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[Igor Khavkine]

mikem@despammed.com wrote:
> I wrote:
>
> >> Take a simple case: theta(x) = wt, where 'w' is a real constant.
> >> I.e: in position space we have
> >>
> >> Psi(x) -> exp(iwt) Psi(x)
> >>
> >> In momentum space, this just shifts the energy by an
> >> amount 'w'. I.e: E -> E - w.
>
> Igor Khavkine wrote:
>
> > [...] when the whole energy spectrum gets shifted by the same
> > constant, no mixing between positive and "negative" energy states
> > occurs. So gauge transformations don't do much, just as they should.
>
> Hmmm. I now see that my simple example of "theta(x) = wt" was too
> simple to illustrate my question adequately. Let me try again...
>
> I know that we need the EM potential to get a gauge-covariant
> derivative, and all that. But I want to focus just on U(1)
> acting on the fermion field. This is because I want to know
> whether or not the gauge-transformed fermion field lives in
> the same Fock space as the original untransformed field.

I think you still get a null effect, at least in the way that I
understand what you are doing. What you are doing is a simple field
redefinition:

chi(x) = exp(i theta(x)) psi(x).

If you want to keep the same dynamics then you'll have to perform the
same transformation in the Lagrangian. The gradient terms in the
Lagrangian will force extra theta(x) dependent terms into the
Lagrangian. This will change the spectrum of the the solutions of the
linear part of the equations of motion for chi(x). When theta(x) was
linear in t, this only gave a constant energy shift. Now the
rearrangement of the spectrum might be more complicated, but the result
is the same. When you quantize, there will be a map between the
creation/annihilation operators for the chi(x) and psi(x) fields.

Or, perhaps, I am still misunderstanding what you are trying to do.

Igor

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).

[mikem@despammed.com]

In response to an earlier post of mine,
Thomas Larsson wrote:

> [...] Negative-energy gauge transformations, like all
> negative-energy operators, annihilate the vacuum,
> and thus we stay within the Hilbert space. However,
> this is closely related to an interesting and
> difficult set of questions, which I have spent quite
> a lot of time thinking about.
>
> Let us first consider the free Maxwell field. [...]

Er, thanks for responding to my post, but I'm actually more interested
in the details of what U(1) gauge transformations do to the electron
field. See also my Sep-18 followup to Igor Khavkine's Sep-16 response,
where I talk about a more challenging example: Psi -> exp(ist^2) Psi.

Regarding the rest of your post, I'm a bit lost about what point you
were trying to make. (I sense you composed it in haste, as I spotted as
least one typo - in the normal-ordered H).