How can the Cauchy integral and Fourier integral produce the same result?

[Avodyne]

It doesn't matter what kind of physical device will be actually used for this measurement. It is only important that this is a position-measuring device ...

The issue (as I see it) is whether we can actually build such a device.

In the "source model" of measurement, described above, I can show that, as I conjectured, there is not a disturbance outside the lightcone. If our ability to manipulate particles corresponds to having sources that we control, then I think this result shows that the fact that the wave function does not vanish outside the lightcone cannot be measured.

Here's a sketch of the proof. In the interaction picture, the time evolution operator in the source model is
U(t) = \exp\!\!\left[-{i\over\hbar}\int_0^t dt&#039;\int_{-L}^{+L}dx&#039;<br /> \,J(x&#039;,t&#039;)\varphi(x&#039;,t&#039;)\right].
Usually, the right-hand side must be time-ordered; however, because the commutator of two fields is a c-number, we can drop the time-ordering, up to an overall c-number phase factor. (See, e.g., Merzbacher, 3rd edition, p.339, for a similar analysis of a harmonic oscillator.) So the state at time t is |\psi(t)\rangle =U(t)|0\rangle, and the expectation value of the field at time t is
\langle\psi(t)|\varphi(x,t)|\psi(t)\rangle = \langle 0|U^\dagger(t)\varphi(x,t)U(t)|0\rangle,
where \varphi(x,t) is the free field. Now we use
e^A B e^{-A}=B + [A,B] + {\textstyle{1\over2}}[A,[A,B]]+\ldots\;.
Because the commutator of two fields is a c-number, this expansion terminates with the second term, and we get
\langle\psi(t)|\varphi(x,t)|\psi(t)\rangle = \langle 0|\varphi(x,t)|0\rangle<br /> +{i\over\hbar}\int_0^t dt&#039;\int_{-L}^{+L}dx&#039;\,J(x&#039;,t&#039;)[\varphi(x&#039;,t&#039;),\varphi(x,t)].
Of course \langle 0|\varphi(x,t)|0\rangle is zero; the interesting part is the second term. But we know the commutator vanishes outside the lightcone, so the expectation value of the field at a particular point in spacetime will not register the disturbance by the source until a speed-of-light signal can get there. The same will be true of any operator built out of fields at a particular spacetime point, because we will always end up with a commutator of fields somewhere in every term.

 

[meopemuk]

The issue (as I see it) is whether we can actually build such a device.

What about a ruler?

In the "source model" of measurement, described above, I can show that, as I conjectured, there is not a disturbance outside the lightcone. If our ability to manipulate particles corresponds to having sources that we control, then I think this result shows that the fact that the wave function does not vanish outside the lightcone cannot be measured.

What is the relevance of the scalar field \phi(x,t) to measurements of position? In my opinion, it is not relevant at all.

The position observable of a single particle is represented by the corresponding Newton-Wigner position operator \hat{\mathbf{r}}. Eigenvalues of this operator are found in the arguments of the particle wave function \psi(\mathbf{r},t). The square of the modulus of this wave function is the probability density for measuring position. So, it contains all information that one would need to interpret position measurements.

Eugene.

 

[Avodyne]

What about a ruler?

A ruler is made out of local excitations of quantum fields. The source, a classical field, is a simple model of these. (It may, of course, be too simple.)

What is the relevance of the scalar field \phi(x,t) to measurements of position?

In QFT, position is defined as the argument of the quantum fields, just like time.

The position observable of a single particle is represented by the corresponding Newton-Wigner position operator

The Newton-Wigner operator is a global object that cannot be written as the integral of a local density (in this respect, it is unlike momentum or energy). So it's not at all obvious to me how you would build a device that measures it.

 

[Haelfix]

Here we go with the Newton-Wigner position operator schtick again.

The literature on that particular is vast, and needless to say controversial and murky to the nth degree. It is completely undefined for interacting field theories, and likely the operator itself is a nogo by various powerful theorems.

Likewise, going back a few posts the physicist will simply object to the propagator argument by Avodyne b/c its a one particle free field approximation. The real physical situation requires a hard cut off on E(k) to make sense at all, and to turn on interactions.

So even though I completely agree the path integral shouldn't be thought off as saying something like Feynman originally thought (eg amplitudes between space like separated paths), he is perfectly justified in pointing out that upon a measurement, you will have pair creation ambiguities. No amount of free field handwaving gets around this.

 

[strangerep]

In Avodyne's earlier calculation of the propagator,

[a(k),a^\dagger(k&#039;)]=f(k)\delta(k-k&#039;),

Those k's are 3-vectors, right? If we were talking about 4-vectors,
we'd need in general an extra factor like \delta^{(4)}(k^2 - m^2).
The latter is what normally causes a 1/E_k to appear when a
4-momentum integral is reduced to a 3-momentum integral (and the latter
depends in turn on a choice of contour).

Now we need to decide what a position eigenstate is. Certainly two
eigenstates at different positions should be orthogonal, so we have
\langle x&#039;|x\rangle = h(x)\delta(x&#039;-x),

I think there's an error in the above. If your x's are 4-vectors, then the
simplistic expression above is not justified. But if your x's are 3-vectors, then
the way you act on them with e^{-iHt} below is incorrect. (You
can't move between equal-time points via a pure time-translation.)

Now let's compute the propagation amplitude. This is given by
\langle x&#039;|e^{-iHt}|x\rangle<br /> =\int {dk\over f(k)}\langle x&#039;|e^{-iHt}|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}e^{-iE(k)t}\langle x&#039;|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x&#039;-x)}<br /> =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x&#039;-x)}.
This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk.

It would all need to be re-done with 4-momentum integrals, and 4-positions, using
the \delta^{(4)}(k^2 - m^2) in appropriate places. Note also that
introducing such a constraint in 4-momentum space would also imply some kind
of admissability constraint on the set of 4-position eigenstates.

Also, this does not vanish outside the lightcone. [...]

It's well-known that the Feynman propagator does not have causal support,
unlike the Pauli-Jordan function. But all that matters is that the matrix elements
of all states in our Hilbert space (in position basis) reflect only causal relationships.
I think you'd find that the matrix element between two arbitrary position states which
are mutually outside each other's light cones are either zero, or the two states are not
in the same Hilbert space.

 

[strangerep]

Here we go with the Newton-Wigner position operator schtick again.

The literature on that particular is vast, and needless to say controversial
and murky to the nth degree. It is completely undefined for
interacting field theories,

Is that just because 4D interacting theories are themselves undefined, or for some
other reason?

and likely the operator itself is a nogo by various powerful theorems.

Which theorems?

 

[Avodyne]

Those k's are 3-vectors, right?

Yes. This is standard; see any QFT book.

I think there's an error in the above. If your x's are 4-vectors, then the simplistic expression above is not justified.

They're 3-vectors.

But if your x's are 3-vectors, then the way you act on them with e^{-iHt} below is incorrect. (You can't move between equal-time points via a pure time-translation.)

The time evolution operator can be applied to any state. I have defined a perfectly sensible linear combination of single-particle momentum eigenstates. I want to call it a position eigenstate; but doing so does not mean that I am suddenly not allowed to time-evolve it in the way that any state is time-evolved.

If we don't call |x\rangle a position eigenstate, then everything I did is straight out of the textbooks. So the only possible controversy is whether |x\rangle "really" represents a particle localized exactly at the point x.

 

[meopemuk]

In QFT, position is defined as the argument of the quantum fields, just like time.

Is there any evidence that parameter x of quantum fields is related to particle positions? I don't know of any experimentally verifiable prediction of QFT that would suggest that x can be measured as the position. All experimental predictions of QFT are contained in the S-matrix. When S-matrix is calculated the parameter x gets integrated out. So, in my opinion, x is simply an integration variable in QFT.

Moreover, in quantum theory (including QFT) each observable should have a corresponding Hermitian operator. If we assume that x are eigenvalues of some Hermitian operator X, then what is X?

The Newton-Wigner operator is a global object that cannot be written as the integral of a local density (in this respect, it is unlike momentum or energy). So it's not at all obvious to me how you would build a device that measures it.

Why do you think that operators of observables should be written as integrals of a local density?

Eugene.

 

[meopemuk]

It would all need to be re-done with 4-momentum integrals, and 4-positions, using
the \delta^{(4)}(k^2 - m^2) in appropriate places.

Why? Obeying the Poincare invariance (which is the true mathematical expression of the principle of relativity) does not require you to perform all calculations in a 4-dimensional notation. A theory is relativistically invariant if and only if it has a representation of the Poincare group of inertial transformations. In most cases the 3D notation is entirely appropriate for relativistic theories and their comparison with experiment. The 4D notation is often (e.g., in this case) redundant and confusing.

Eugene.

 

[meopemuk]

So even though I completely agree the path integral shouldn't be thought off as saying something like Feynman originally thought (eg amplitudes between space like separated paths), he is perfectly justified in pointing out that upon a measurement, you will have pair creation ambiguities. No amount of free field handwaving gets around this.

I read this argument (particle localization = pair creation) very often, but I've never seen a detailed explanation or proof. Could you please clarify?

Eugene.

 

[Avodyne]

I don't know of any experimentally verifiable prediction of QFT that would suggest that x can be measured as the position.

What about measuring electric and magnetic field strengths? These are given by Maxwell's equations as functions of x and t, and of course are well verified. Classical electric and magnetic fields have to be understood as coherent states of photons.

Moreover, in quantum theory (including QFT) each observable should have a corresponding Hermitian operator.

There is no time operator. In QFT, position is treated as an external parameter, like time, that labels operators. (Time labels operators in the Heisenberg picture.)

Why do you think that operators of observables should be written as integrals of a local density?

Because measurements are done locally. We never measure the total energy of the universe, only the energy of some local pieces. If energy were not the integral of a local density, this would not be possible.

 

[Avodyne]

I read this argument (particle localization = pair creation) very often, but I've never seen a detailed explanation or proof. Could you please clarify?

There is no proof because it's wrong. The construction of a single-particle position eigenstate works just as well in an interacting theory (with massive particles only). Even in an interacting theory, the one-particle states are well defined, because they form an isolated hyperboloid in energy-momentum space, given by E^2-\vec p^{\,2} = m^2. (All multiparticle states have E^2-\vec p^{\,2} &gt; m^2, and the ground state has E=0.) Once we have well-defined single-particle momentum eigenstates, we can take a Fourier superposition to get something that has all the properties we would want a position eigenstate to have.

 

[meopemuk]

What about measuring electric and magnetic field strengths? These are given by Maxwell's equations as functions of x and t, and of course are well verified. Classical electric and magnetic fields have to be understood as coherent states of photons.

But it is impossible to detect even a single particle (photon) in a static electric or magnetic field even if sensitive photomultipliers or photograhic plates are used. So, the relationship between fields and particles looks rather dubious in this case.

There is no time operator. In QFT, position is treated as an external parameter, like time, that labels operators. (Time labels operators in the Heisenberg picture.)

Yes, there is no time operator in QM and QFT simply because time is not an observable. Measured values of observables (position, energy, momentum, spin) normally depend on the physical system being measured and on its state. The value of time can be read off the clock (which is a part of the laboratory or reference frame rather than observed physical system) regardless on what system is observed or even if the system is not present at all. There is a fundamental difference between time (a numerical parameter) and position (a true observable represented by a Hermitian operator) which is correctly reflected in the formalism of quantum mechanics.

I already mentioned that the argument x in the quantum field \phi(x,t) cannot be regarded as measured position of anything. It is also misleading to interpret parameter t as time of any event, because in QFT when S-matrix elements are calculated this parameter serves as an integration variable, so it is not present in any final result for scattering cross-sections. I haven't seen any QFT predictions of time-dependent processes that can be reliably compared with experiment.

Because measurements are done locally. We never measure the total energy of the universe, only the energy of some local pieces. If energy were not the integral of a local density, this would not be possible.

Is local energy density also measurable? What is the energy density of a single electron? or single photon?

Eugene.

 

[meopemuk]

There is no proof because it's wrong. The construction of a single-particle position eigenstate works just as well in an interacting theory (with massive particles only). Even in an interacting theory, the one-particle states are well defined, because they form an isolated hyperboloid in energy-momentum space, given by E^2-\vec p^{\,2} = m^2. (All multiparticle states have E^2-\vec p^{\,2} &gt; m^2, and the ground state has E=0.) Once we have well-defined single-particle momentum eigenstates, we can take a Fourier superposition to get something that has all the properties we would want a position eigenstate to have.

I am happy to tell that I agree with everything you wrote here. However, with one caveat. This doesn't apply to "bare" particles whose creation and annihilation operators are normally used in QFT. In most quantum field theories single "bare" particles are not eigenstates of the total Hamiltonian, so they even don't have well-defined energies. Your description is perfectly valid for "dressed" particles.

Eugene.

 

[strangerep]

Even in an interacting theory, the one-particle states are well defined,

No, they're not. Exact expressions for single-particle states are only known for a few
very simple examples of interacting QFT (e.g: the Lee model, or certain non-4D theories).
There are no known existence proofs for any physically-interesting 4D interacting QFTs.

 

[Micha]

I am still trying to understand, if there could be in principle measurable superluminal quantum fluctuations, which could not be immediately used to transmit information with superluminal speed. How could such a purely statistical effect look like? Naively I would say, this is not possible.

Edit: If this is true, we know, that leaking out of the lightcone is something for virtual particles only. And we know, that they can do all kinds of weird stuff.

 

[meopemuk]

I am still trying to understand, if there could be in principle measurable superluminal quantum fluctuations, which could not be immediately used to transmit information with superluminal speed. How could such a purely statistical effect look like? Naively I would say, this is not possible.

Edit: If this is true, we know, that leaking out of the lightcone is something for virtual particles only. And we know, that they can do all kinds of weird stuff.

I don't think there exists an easy solution. I consider it well established that wave functions of real (not virtual) particles do "leak out of the lightcone". This leakage can be used to transmit information superluminally. However, I am not convinced that these facts can be used to build a machine that sends signals back to the past. If the possibility of such a machine can be proved, then we are in a deep trouble. Until then there is no need to worry.

Eugene.

 

[Avodyne]

I am still trying to understand, if there could be in principle measurable superluminal quantum fluctuations, which could not be immediately used to transmit information with superluminal speed.

If we could prepare a position eigenstate |x\rangle (as defined earlier) at a definite time t, and if we had a detector that gave a signal if and only if the particle appeared at point x&#039; at time t&#039;, and if we could arrange things so that no particle other than the one we prepared could possibly ever show up in the detector, then we could do this experiment repeatedly, and we would find that, in a fraction roughly \exp(-ms) of the time, where s=(|x-x&#039;|^2-(t-t&#039;)^2)^{1/2} is the invariant separation between the points, the particle would appear at x&#039; before a speed-of-light signal could get there.

All big ifs, though.

 

[meopemuk]

If we could prepare a position eigenstate |x\rangle (as defined earlier) at a definite time t, and if we had a detector that gave a signal if and only if the particle appeared at point x&#039; at time t&#039;, and if we could arrange things so that no particle other than the one we prepared could possibly ever show up in the detector, then we could do this experiment repeatedly, and we would find that, in a fraction roughly \exp(-ms) of the time, where s=(|x-x&#039;|^2-(t-t&#039;)^2)^{1/2} is the invariant separation between the points, the particle would appear at x&#039; before a speed-of-light signal could get there.

All big ifs, though.

If instead of releasing just one particle we release a large number of them from point x simultaneously, then the probability of registering the superluminal signal by the detector at x' will be much greater than \exp(-ms). By doing this we can guarantee that the superluminal signal arrives to the detector with almost 100% certainty.

Eugene.

 

[Micha]

One thing, which makes me skeptical about the whole effect being real is, that obviously the leaking gets bigger, the lighter the rest mass of the particle is. But then for a particle with zero rest mass the effect disappears. Shouldn't there be a continuous change of the effect between zero restmass and very light restmass?

 

[Micha]

If instead of releasing just one particle we release a large number of them from point x simultaneously, then the probability of registering the superluminal signal by the detector at x' will be much greater than \exp(-ms). By doing this we can guarantee that the superluminal signal arrives to the detector with almost 100% certainty.

Eugene.

This was my thought as well. We also do not need to worry about, whether it is the same particle. A statistical effect is good enough. The concept of the identity of a particle does not make sense anyway in QM.

 

[Micha]

I don't think there exists an easy solution. I consider it well established that wave functions of real (not virtual) particles do "leak out of the lightcone". This leakage can be used to transmit information superluminally. However, I am not convinced that these facts can be used to build a machine that sends signals back to the past. If the possibility of such a machine can be proved, then we are in a deep trouble. Until then there is no need to worry.

Eugene.

How does this go together with the axiom (or theorem?) of QFT, that two quantum operators separated by a spacelike distance commute?

 

[meopemuk]

One thing, which makes me skeptical about the whole effect being real is, that obviously the leaking gets bigger, the lighter the rest mass of the particle is. But then for a particle with zero rest mass the effect disappears. Shouldn't there be a continuous change of the effect between zero restmass and very light restmass?

Are you sure that the effect is absent for zero mass particles? As far as I know, the position operator and localized states have not been successfully defined for such particles yet.

Eugene.

 

[meopemuk]

How does this go together with the axiom (or theorem?) of QFT, that two quantum operators separated by a spacelike distance commute?

Yes, people keep writing these things in textbooks, but I couldn't understand their logic.

Yes, there are quantum fields \phi(x,t) in QFT and they (anti)commute at space-like separations. As shown in Weinberg's vol. 1 this property is important to ensure that interaction operators constructed as polynomials of quantum fields are relativistically invariant. But quantum fields do not correspond to any physical observable measured in experiment.

It is true that we can build some operators of total observables (total momentum, total energy, etc.) as integrals of certain field products. Then x is an integration variable, and the resulting operators have no dependence on x. So, it doesn't make sense to say "operators of total energy separated by a spacelike interval".

There are also operators of observables (position, momentum, energy, etc.) of individual particles. (For example in the 2-particle sector of the Fock space there are two distinct operators of particle momentum). These particle observables cannot be represented as functions of quantum fields. So, the idea of "spacelike separation" is not applicable to them as well.

Eugene.

 

[Micha]

Are you sure that the effect is absent for zero mass particles? As far as I know, the position operator and localized states have not been successfully defined for such particles yet.

Eugene.

I thought, we would only need the photon propagator presented by Hans earlier.
Also with m=0, hbar drops out of the KG equation.

Sorry if this is too simple minded.

EDIT: I remember, you did a different calculation with position Eigenstates with a slightly different result.

 

[meopemuk]

I thought, we would only need the photon propagator presented by Hans earlier.
Also with m=0, hbar drops out of the KG equation.

Sorry if this is too simple minded.

EDIT: I remember, you did a different calculation with position Eigenstates with a slightly different result.

Propagators are great tools for calculating scattering cross-sections via Feynman diagrams, but they (propagators) are not designed to describe the evolution of one-particle wave functions in space and time. Avodyne wrote a good post #188 about that.

Eugene.

 

[Micha]

Propagators are great tools for calculating scattering cross-sections via Feynman diagrams, but they (propagators) are not designed to describe the evolution of one-particle wave functions in space and time. Avodyne wrote a good post #188 about that.

Eugene.

ok, but what would we put into the exponential?

 

[meopemuk]

ok, but what would we put into the exponential?

Do you mean the exponential decay of photon's wave function outside the light cone?
I don't know about that. As I mentioned, the definition of the position operator and localized states for massless particles is a controversial issue. I have references to (at least) a dozen papers discussing this situation, but I don't think they found an acceptable solution yet.

Eugene.

 

[strangerep]

It would all need to be re-done with 4-momentum integrals, and
4-positions, using \delta^{(4)}(k^2 - m^2)

Why? Obeying the Poincare invariance (which is the true mathematical expression of
the principle of relativity) does not require you to perform all calculations in a
4-dimensional notation. A theory is relativistically invariant if and only if it has a
representation of the Poincare group of inertial transformations. In most cases the
3D notation is entirely appropriate for relativistic theories and their comparison
with experiment. The 4D notation is often (e.g., in this case) redundant and confusing.

Construct a (single-particle) vector space of states indexed by all the possible tuples
of 4-momentum. At this point we don't yet have a Hilbert space because we haven't
specified an inner product. It also does not yet carry a Poincare representation, and
hence doesn't represent the states of a physical particle type.

Now consider a spin-0 particle type of mass m. To turn the 4-momentum vector space into
a (single-particle) Hilbert space for this type of particle, we restrict to the subset consisting
of 4-momentum vectors which satisfy the usual relativistic mass formula. That's done via
multiplying by the familiar \delta^{(4)}(k^2 - m^2) expression. Now (and only
now) we can turn our (restricted) vector space into a Hilbert space by specifying the inner
products between all states such that we only mention the 3-momentum:
&lt;k&#039;|k&gt; = ... \delta^{(3)}(k&#039;-k), (and taking the completion, which I won't worry
about here).

But now, if we try to change from momentum basis to coordinate basis, extra care is
needed. If we start from the original 4-vector momentum space (non-Hilbert) we
can do the usual Fourier transformations and pass to a 4-vector coordinate space (like
Minkowski space), but this is not a Hilbert space of states. To get a Hilbert space
in a coordinate basis, we must start from the Hilbert space in momentum basis,
with the restriction \delta^{(4)}(k^2 - m^2), and then do the Fourier-transform
stuff. Multiplication by \delta^{(4)}(k^2 - m^2) in momentum space
corresponds to convolution with F_x(\delta^{(4)}(k^2 - m^2)) in coordinate space,
where F_x denotes Fourier transform k\to x, hence the allowable
4-vectors from the full 4D coordinate space are restricted to those which are obtained
via this complicated convolution. Only those states are in our Hilbert space, so talking
about position states indexed by arbitrary 4-position coordinates is incorrect in
general, because they might not be in the Hilbert space of the relativistic particle.

So 3D notation is reasonably ok in momentum basis, because the relativistic constraint
is easy to keep in mind (and it's easy enough to convert a 4D integral containing a
\delta^{(4)}(k^2 - m^2) into a 3D integral containing a 1/E_k).
The same cannot be said for the coordinate basis.

 

[meopemuk]

Construct a (single-particle) vector space of states indexed by all the possible tuples
of 4-momentum. At this point we don't yet have a Hilbert space because we haven't
specified an inner product. It also does not yet carry a Poincare representation, and
hence doesn't represent the states of a physical particle type.

Yes, this 4D momentum-energy space is not suitable for describing single particle states. Then why do you begin your construction with this irrelevant space? From Wigner's theory of irreducible representations of the Poincare group we know that elementary particles are characterized by a single value of the Casimir invariant m = + c^{-2}\sqrt{H^2 - c^2 \mathbf{P}^2}. Therefore the complete set of mutually commuting operators can be chosen as (P_x, P_y, P_z). The fourth operator H is redundant, because it is a function of the three components of momentum and the constant m. Thus the single particle Hilbert space is a span of common eigenvectors of (P_x, P_y, P_z), i.e., the set of functions \psi( \mathbf{p}) in 3 dimensions.

Now consider a spin-0 particle type of mass m. To turn the 4-momentum vector space into
a (single-particle) Hilbert space for this type of particle, we restrict to the subset consisting
of 4-momentum vectors which satisfy the usual relativistic mass formula. That's done via
multiplying by the familiar \delta^{(4)}(k^2 - m^2) expression. Now (and only
now) we can turn our (restricted) vector space into a Hilbert space by specifying the inner
products between all states such that we only mention the 3-momentum:
&lt;k&#039;|k&gt; = ... \delta^{(3)}(k&#039;-k), (and taking the completion, which I won't worry about here).

Yes, if you like, you can build the 1-particle Hilbert space as a set of functions confined to the mass hyperboloid.

But now, if we try to change from momentum basis to coordinate basis, extra care is
needed. If we start from the original 4-vector momentum space (non-Hilbert) we
can do the usual Fourier transformations and pass to a 4-vector coordinate space (like
Minkowski space), but this is not a Hilbert space of states.

To get a Hilbert space
in a coordinate basis, we must start from the Hilbert space in momentum basis,
with the restriction \delta^{(4)}(k^2 - m^2), and then do the Fourier-transform
stuff. Multiplication by \delta^{(4)}(k^2 - m^2) in momentum space
corresponds to convolution with F_x(\delta^{(4)}(k^2 - m^2)) in coordinate space,
where F_x denotes Fourier transform k\to x, hence the allowable
4-vectors from the full 4D coordinate space are restricted to those which are obtained
via this complicated convolution. Only those states are in our Hilbert space, so talking
about position states indexed by arbitrary 4-position coordinates is incorrect in
general, because they might not be in the Hilbert space of the relativistic particle.

Yes, this would be completely unphysical, because this would lead to "wave functions" localized in space and time like \delta^{(4)}(x-x&#039;) = \delta^{(3)}(\mathbf{r-r&#039;})\delta(t-t&#039;). Such "wave functions" would describe particles that exist for an infinitesimally small period of time unlike any familiar particle in nature.

So 3D notation is reasonably ok in momentum basis, because the relativistic constraint
is easy to keep in mind (and it's easy enough to convert a 4D integral containing a
\delta^{(4)}(k^2 - m^2) into a 3D integral containing a 1/E_k).
The same cannot be said for the coordinate basis.

To me this indicates that the 4D formalism is not helpful at all. If we want to construct the position representation we can find the Newton-Wigner operator \mathbf{R} as a function of 10 generators of the Poincare group. Then we can find the action of \mathbf{R} on 3D momentum-space wave functions \psi( \mathbf{p}), find eigenvalues and eigenvectors of \mathbf{R} and thus define the position-space basis.

Eugene.

 

[strangerep]

Yes, this 4D momentum-energy space is not suitable for describing single particle states.
Then why do you begin your construction with this irrelevant space?

Because I was trying to illustrate the non-obvious pitfalls that occur when one naively
tries to mix position-momentum Fourier transformation with relativistic QM.
But I ended up in essentially the same place as you.

[...] If we want to construct the position representation we can find the Newton-Wigner
operator \mathbf{R} as a function of 10 generators of the Poincare group. [...]

Actually, I happen to share Haelfix's opinions about the N-W operator. I don't buy it.
I think one should start with the larger Heisenberg-Poincare algebra (15 generators)
rather than merely Poincare(10 generators), and try to construct a QFT from its
unitary irreps. But this is an unsolved task, and the H-P algebra also has the
possible problem of not being stable under small deformations. (I mention this only
as a possible research direction, not as established mainstream.)

 

[meopemuk]

Actually, I happen to share Haelfix's opinions about the N-W operator. I don't buy it.
I think one should start with the larger Heisenberg-Poincare algebra (15 generators)
rather than merely Poincare(10 generators), and try to construct a QFT from its
unitary irreps. But this is an unsolved task, and the H-P algebra also has the
possible problem of not being stable under small deformations. (I mention this only
as a possible research direction, not as established mainstream.)

The Poincare group has the clear advantage that it is composed of real physical transformations that we can easily observe in everyday life. Of course, one can assume that there is also some hidden reality and try to enlarge this group. Then one would enter the shaky ground of mathematical speculations.

I agree that Newton-Wigner position has some unusual properties, like boost non-invariance of localization and superluminal spreading. However, it hasn't been demonstrated that these properties are in contradiction with hard experimental data. So, I don't see any compelling reason to dismiss the Newton-Wigner position operator.

Eugene.

 

[Haelfix]

Hi Strangerep, I was thinking about the Reeh-Schlieder theorems. Actually the N-W operator was modernized in hopes of bypassing this theorem, but people still argue about that today. Its a tremendously thorny subject, and its also largely irrelevant, b/c no one hsa the faintest clue how to make a falsifiable experiment that could detect such a thing.

Anyway I view the N-W operator as a mathematical construct that probably has limited applicability. If there is such a thing as a position operator in qft then it would be the NW operator but alas its probably at best only well defined in simple free field theories with massive particles. (This leads into the tired debates about whether a photon really exists or not, etc)

On philosophical grounds I also somewhat object to these sort of inbetween meta constructs. You either introduce QFT with full mathematical rigor (alla Aqft or other axiomitized attempts) or you should be satisfied with sloppy physicist traditional field theory from textbooks that is largely experimentally and simplicity motivated.

 

[Micha]

Do you mean the exponential decay of photon's wave function outside the light cone?
I don't know about that. As I mentioned, the definition of the position operator and localized states for massless particles is a controversial issue. I have references to (at least) a dozen papers discussing this situation, but I don't think they found an acceptable solution yet.

Eugene.

Yes, that's what I mean.

I mean, just from dimensional analysis, not from a full blown calculation.
First of all, the exponent as a whole must be dimensionless. Obviously a spatial coordinate should appear. Then I think, there should be an hbar in it. And then you need something with dimensions energy. At the moment, I can not think of anything, that makes sense here.

 

[Hans de Vries]

Anyway I view the N-W operator as a mathematical construct that probably has limited applicability.

In my opinion, historically, a number of interpretation difficulties have
unfortunately led people to the NW position operator. Most of all there
is the velocity operator which seems wrong when it produces +c/-c.
But it's not wrong, it's right. the interpretation is wrong.

It gives -c for the left chiral and +c for the right chiral component, this
is just a one step further subdivision of the physical process as the series:

\frac{1}{p^2-m^2}\ =\ \frac{1}{p^2}+\frac{m^2}{p^4}+\frac{m^4}{p^6}+\frac{m^6}{p^8}+...

Where all the right hand terms propagate on the light cone. To see this
we can write down the linearized chiral Klein Gordon equation:<br /> \left[\ \left( \begin{array}{ll} 0 &amp; 1 \\ 1 &amp; 0 \end{array} \right) \frac{\partial }{\partial t} + c \left( \begin{array}{cc} 0 &amp; 1\ \\ -1\ &amp; 0 \end{array} \right) \frac{\partial }{\partial r}\<br /> \right] \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right) = \frac{mc^2}{i\hbar} \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right)<br />

The operator between the brackets becomes the d'Alembertian when squared.
Put sigma's in the second matrix and you've got the Dirac equation. So the
velocity operator gives us:

\vec{v}\ =\ \frac{\partial\vec{r}}{\partial t}\ =\ \frac{i}{\hbar}[\ H,\vec{r}\ ]\ =\ -c\ \mbox{for}\ \psi_L,\ \ \ +c\ \mbox{for}\ \psi_R

The role of m in the equation is that it reflects the two counter propagating
components back and forward into each other. The average speed is v.
This whole process is actually pretty classical. It occurs in a simple mass/
spring transmission line described by the Klein Gordon equation like here:Klein Gordon mass/spring system

Where the 'm' masses cause the reflections of the left and right propagating
waves into each other. I've been doing some lattice simulations lately of the
above linear chiral Klein Gordon equation and get very nice results, see:

Propagating particle

Well, that's pretty much unitary isn't it... and guess, no propagation
outside the lightcone. Regards, Hans

 

[Hans de Vries]

So, the correct relativistic velocity-density operator is:

\vec{v}\ \ =\ \ \bar{\psi}\ \frac{i}{\hbar}[\ H,\vec{r}\ ]\ \psiWith the plane wave solution,


\left( \begin{array}{c} \sqrt{E-p} \\ \sqrt{E+p} \end{array} \right)we get for the velocity-density v=p, which, when integrated over the entire
(Lorentz contracted) wave function gives v for the speed of the particle.
Edit(2x): The formula holds as well for the full Dirac equation, since:

\frac{i}{\hbar}\ [\ H,x^i\ ]\ = \ c\gamma^iRegards, Hans

 

[Avodyne]

It's easy to do my calculation of the propagation of a localized particle (in post #188) for the massless case, ignoring the potential complications of what it means to have a single massless particle, etc.

In d space dimensions, we have

\langle \vec x{\,}&#039;|e^{-iHt}|\vec x\,\rangle<br /> =\int {d^d k\over(2\pi)^d}\,e^{-ikt}e^{i\vec k\cdot(\vec x{\,}&#039;-\vec x{\,})}<br /> \propto {it\over(r^2-t^2)^{(d+1)/2}}

where r=|\vec x{\,}&#039;-\vec x{\,}|. (To get this I did the integral in euclidean time, and then continued back.) This formula shows no exponential suppression outside the lightcone.

 

[strangerep]

The Poincare group has the clear advantage that it is composed of real physical transformations that we can easily observe in everyday life. Of course, one can assume that there is also some hidden reality and try to enlarge this group. Then one would enter the shaky ground of mathematical speculations.

Those are exactly the reasons for my interest in the Heisenberg-Poincare algebra (or
some stabilized deformation thereof).

The basic H-P algebra is of the following form (I won't write out all the indices):

[ J , J] = ...terms in J only...

[ J , P] = ...terms in P only...

[ J , X] = ...terms in X only...

[P, X] = ...E...

Here, J is the 6 Lorentz generators, P is the 4 momentum generators,
so that part is just the Poincare algebra. X is the 4 position generators,
and E is a central element, so the last equation above is just the usual
Heisenberg commutation relation. So there's 15 generators in total.

The X generators have perfectly reasonable physical foundations.
The central element E is just from elementary QM. So there's nothing
physically wrong with this as a starting point. No "hidden reality"
is involved.

However, it turns out that the algebra is not stable under small deformations,
and this leads one to consider stabilized variants of the algebra, but I won't
delve into that here (because I don't yet understand it very well).

 

[meopemuk]

The X generators have perfectly reasonable physical foundations.

What are these foundations? Other generators are related to well-known inertial transformations of reference frames (space and time translations, rotations and boosts). What finite transformations are generated by X?

Eugene.

 

[Hans de Vries]

Those are exactly the reasons for my interest in the Heisenberg-Poincare algebra (or
some stabilized deformation thereof).

The basic H-P algebra is of the following form (I won't write out all the indices):

[ J , J] = ...terms in J only...

[ J , P] = ...terms in P only...

[ J , X] = ...terms in X only...

[P, X] = ...E...

Interesting, I always felt that the x,p relations were somehow
missing from the commutation prescription.

Regards, Hans

 

[Hans de Vries]

To recapitulate why there is nothing wrong with the normal
relativistic position and velocity operators: If we have the
Dirac equation:

\gamma^0\ \partial_t\ \psi\ \ =\ \ -\left[\ c\ \gamma^i\ \partial_{x^i} + i\frac{mc^2}{\hbar}\ \right]\psi \qquad\qquad\qquad\qquad\qquad (1)

and we define:

\gamma^0\ H\ \ =\ \ -\gamma^0\ i\hbar\ \partial_t\ \ =\ \ i\hbar\ \left[\ c\ \gamma^i\ \partial_{x^i} + i\frac{mc^2}{\hbar}\ \right]\psi \qquad\qquad (2)

Then we find the usual result for the velocity:

v\ =\ \frac{i}{\hbar}\ [\ H,x^i\ ]\ = \ c\ \gamma^i\ /\ \gamma^0\ \ =\ \ c\ \alpha^i \qquad\qquad\qquad\qquad\qquad (3)

Then, instead of concluding that something must be terrible wrong
with the relativistic position and velocity operators we simple ask
ourself what we have to do to go from the H as defined in (2) to
get {\cal E}, the energy-density of the wave-function, and we find:

{\cal E}\ \ =\ \ \frac{1}{2E}\left(\ \bar{\psi}\ \gamma^0\ H\ \psi\ \right)

Thus to get the velocity-density we need to evaluate

v^i\ \ =\ \ \frac{1}{2E}\left(\ \bar{\psi}\ \gamma^i\ \psi\ \right) \ \ = \ \ \frac{cp^i}{E}\ \ =\ \ v^i

So, there is nothing wrong after all. Instead we could interpret
(3) as giving us some interesting information about the internal
components of the spinor wave function.Regards, Hans

 

[meopemuk]

Interesting, I always felt that the x,p relations were somehow
missing from the commutation prescription.

The Newton-Wigner position operator x is a function of Poincare generators x(P, J) and it has correct (i.e., physically acceptable) commutators with all these generators, including usual Heisenberg commutators with momentum p. So, I don't see any reason to invent an artificial substitute for x.

Eugene.

 

[meopemuk]

Hi Hans,

QM has a strictly defined formalism in which observables are represented by Hermitian operators and states are represented by vectors in the Hilbert space. The energy-densities and velocity-densities you are talking about are rather unusual notions for quantum mechanics. Are they measurable? Shall we also consider the position-density?
Why not?

Eugene.

 

[Hans de Vries]

The Newton-Wigner position operator x is a function of Poincare generators x(P, J) and it has correct (i.e., physically acceptable) commutators with all these generators, including usual Heisenberg commutators with momentum p. So, I don't see any reason to invent an artificial substitute for x.

Eugene.

Because, as you claim a few post back, the NW-position operator violates
Special Relativity and allows for faster than light signal communication, and,
yes, not really everybody finds that "physically acceptable" ...

Therefor I would prefer to use the standard operators used in classical
physics and QM also in relativistic quantum mechanics.Regards, Hans

 

[strangerep]

The X generators have perfectly reasonable physical foundations.

What are these foundations?

Er,... observations of position? ;-)

I.e., the inclusion of position observables X in the H-P algebra
just represents an attempt to have a single Lie algebra of all
the observables which are obviously relevant in a relativistic
quantum theory.

Like Hans, I too often wondered why the x,p relations are somehow
missing from the usual commutation prescription. In fact, the idea of
the H-P algebra goes right back to Heisenberg himself. But I guess it
wasn't until much later, after (axiomatic/algebraic) QFT based on
operator fields over Minkowski space was looking decidedly sick, and
alternatives like Newton-Wigner fields found to be problematic, that
people started revisiting the H-P idea.

Other generators are related to well-known inertial transformations of
reference frames (space and time translations, rotations and boosts). What finite
transformations are generated by X?

The H-P algebra should be viewed as a Lie algebra of observables.
X corresponds to the position observable. Exp(X) transformations correspond
to translations in momentum space, and this does indeed have certain
difficulties of its own (so I read).

The Newton-Wigner position operator x is a function of Poincare
generators x(P, J) and it has correct (i.e., physically acceptable)
commutators with all these generators, including usual Heisenberg
commutators with momentum p. So, I don't see any reason to invent an
artificial substitute for x.

Hmm. I certainly would never have described the X's in the H-P algebra
as artificial. Quite the opposite.

The N-W operator seems far more artificial to me (IMHO). The explicit
expression for the N-W position operator "R" involves dividing by the mass
and hamiltonian operators. That means it's not in the Lie algebra, nor even
in the (polynomial) enveloping algebra. One must extend the algebra to
some kind of division algebra, and show that it is (mathematically and
physically) sensibly well-defined on all of one's Hilbert space before such
a construction can be considered satisfactory. That's exactly where the
N-W operator runs into trouble.

But I guess it's a subjective judgement whether something is/isn't "artificial".

 

[meopemuk]

X corresponds to the position observable. Exp(X) transformations correspond
to translations in momentum space, and this does indeed have certain
difficulties of its own (so I read).

That's what I am having problems with. For Poincare group generators H, P, J, K, the corresponding finite transformations Exp(H), Exp(P), Exp(J), Exp(K) have a well-defined and easily observable physical meaning as transformations of inertial reference frames. On the other hand, momentum-space translations Exp(X) are rather abstract things. That's why I don't feel like treating X on the same footing as other generators.

But I guess it's a subjective judgement whether something is/isn't "artificial".

Here I should agree with you. Relativistic quantum theory is so weird. It seems that no matter which approach we take, we face artificial or counterintuitive things of one kind or another. Most often, these things cannot be directly measured. So, which of them are less "artificial" remains largely a matter of taste. Apparently, we have different tastes for these things. I don't think it's bad.

Eugene.

 

[OOO]

This leakage can be used to transmit information superluminally. However, I am not convinced that these facts can be used to build a machine that sends signals back to the past. If the possibility of such a machine can be proved, then we are in a deep trouble.

Hi meopemuk,

I haven't followed the thread for the last few days. But what struck me was your statement above about signals being sent back to the past. I feel I have to object to this.

First, I think, the term "past" is defined by the backward lightcone and not just by some event that has t<0 in some frame. I guess the region outside the lightcone is commonly called present (even if t<0 in some frame).

Second, even if it was possible to "send signals back to the past" (the backward lightcone), this would not be interpreted by the observer as such. Think of the advanced propagator in classical ED: what we would see in such a case was a spherical wave that runs towards a specific spacetime point, and when it reaches it, it vanishes and creates a delta-source (the reverse of what the retarded propagator does: create a delta source in connection with an outgoing spherical wave). So I guess the term "sending signals back to the past" is a bit misleading because it implies that the past can be changed, which it can't. Rather signals seemingly sent to the past correspond to very unlikely course of events (in some sense similar to the broken roof tile that heals magically and flies back to the roof...).

Do you agree with that ?

 

[meopemuk]

So I guess the term "sending signals back to the past" is a bit misleading because it implies that the past can be changed, which it can't.

I agree with you here. I think it is impossible to send signals back to the past even if it is quite possible to have particle wavefunctions that propagate superluminally or action-at-a-distance interactions. You can find in the literature designs of "time machines" that are based on superluminal signals. These "time machines" should supposedly allow you to make terrible things, like killing your grandfather before you were born, etc.
I think that these designs involve subtle flaws that wouldn't allow them to work as intended.

Eugene.

 

[strangerep]

X corresponds to the position observable. Exp(X) transformations correspond
to translations in momentum space, and this does indeed have certain
difficulties of its own (so I read).

That's what I am having problems with. For Poincare group generators H, P, J, K, the
corresponding finite transformations Exp(H), Exp(P), Exp(J), Exp(K) have a well-defined
and easily observable physical meaning as transformations of inertial reference frames.
On the other hand, momentum-space translations Exp(X) are rather abstract things.
That's why I don't feel like treating X on the same footing as other generators.

If we write R (the N-W position operator) as (for example):

R = -c^2H^{-1}K - \frac{i \hbar c^2 P}{2H^2} - \frac{cP \times W}{MH(Mc^2 + H)}<br />

then one could also ask what exp(R) means. Since it satisfies (at least formally)
the Heisenberg CRs, I think you'd reach an identical conclusion that R generates
translations in momentum space.

The important thing is the basic (linear) Lie algebra of quantities corresponding to observable
properties of physical systems. Then the task is to go further and construct a 4D
infinite-dimensional interacting quantum theory rigorously from the algebra, which of
course no one has done.

 

[meopemuk]

If we write R (the N-W position operator) as (for example):

R = -c^2H^{-1}K - \frac{i \hbar c^2 P}{2H^2} - \frac{cP \times W}{MH(Mc^2 + H)}<br />

then one could also ask what exp(R) means. Since it satisfies (at least formally)
the Heisenberg CRs, I think you'd reach an identical conclusion that R generates
translations in momentum space.

Yes, it is true that \exp (R) performs translations in momentum space. But it is important to realize that there is no inertial transformations of reference frames, which correspond to such a translation. (Boost transformations come close, but their action on definite momentum states is more complicated than simple translation) So, \exp (R) is not a representative of any Poincare group element. This is the reason why R does not belong to the Poincare Lie algebra and should be constructed as a (rather complicated) function of proper Lie algebra elements that you correctly reproduced.

Eugene.