- #36
reilly
Science Advisor
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Sorry, I haven't a clue to do anything but directly cite post 32 for the algebra that I can't get to show up here. Sorry 'bout that.
Regards,
Reilly Atkinson
Regards,
Reilly Atkinson
reilly said:One small quibble, however.
Normally the unitary transformation operator, U, will act on the "a" operators and the vacuum state. U will commute with p and the volume measure in p-space.
I've assumed all along that we're talking about a relativistic quantum theory. That meansjostpuur said:I realized that if the Hamiltonian is known, the state [tex]|\psi\rangle[/tex] immediately fixes the time evolution [tex]|\psi(t)\rangle[/tex] for all [tex]t\in\mathbb{R}[/tex]. Consequently the boost [tex]|\psi\rangle\mapsto \Lambda|\psi\rangle[/tex] should exists. Could it be, that you are implicitly using the H in those transformations?
If that's all you've got, then you don't have a relativistic quantum theory.Surely we all agree, that if the wave function is given only on a fixed time in some fixed frame, [tex]\psi(x)[/tex], there is no chance of [...]
strangerep said:I've assumed all along that we're talking about a relativistic quantum theory. That means
we must have already constructed a Hilbert space carrying a unitary representation of the
Poincare group. So indeed, all the generators [itex]J_{\mu\nu}, P_i, H[/itex] of the Poincare group
are represented there as operators on the Hilbert space. In particular, if [itex]|\psi\rangle[/itex]
is a state vector in the Hilbert space, then so is [itex]|\phi\rangle := e^{i\tau H} |\psi\rangle[/itex] .
Haelfix said:I thought i'd add a classic thought experiment about localization often taught in early QFT courses (eg Coleman's lectures). This very much relates to the nonvanishing of the propagator outside lightcone and things like that, as well as the early tensions between field theory and relativistic QM.
If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.
Enter field theory, which is a theory with infinite degrees of freedom, and the (somewhat sad) requirement to only use momentum operators.
You are free to try to sidestep this somehow (see Newton Wigner) but it starts becoming more and more intractable as you add interaction terms, and the technical and conceptual difficulties start increasing (the photon for instance becomes the source of much debate).
I guess I don't the question, then. The Lorentz generators (in isolation) are defined byjostpuur said:This doesn't yet immediately imply that the Hamiltonian is used in the definition of the generators of the Lorentz subgroup (or in the definition of boosts) which is what I asked about.
koolmodee said:(Haelfix):
If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.
I subscribe to this view, which I believe is the standard one.
you are forced to no longer deal with 1 particle, but rather many.
strangerep said:But we know that position and momentum do not commute, so it should
be no surprise that position and particle-number don't commute either.
By restricting to NW operators for single particles, there's an implicit projectionmeopemuk said:This doesn't look convincing. It is known that Newton-Wigner position operators for individual particles do commute with particle number operators. So, it should be possible to measure particle positions as precisely as one wants without disturbing the number of particles in the system.But we know that position and momentum do not commute, so it should
be no surprise that position and particle-number don't commute either.
strangerep said:By restricting to NW operators for single particles, there's an implicit projection
operator to the 1-particle sector floating around, in whose presence I'm not sure that
your larger conclusion necessarily follows.
jostpuur said:At any instant in some particular frame the spatial and momentum space representations are related by equations
[tex]
\psi(t,\boldsymbol{x}) = \int\frac{d^3p}{(2\pi)^3}\hat{\psi}(t,\boldsymbol{p})e^{i\boldsymbol{x}\cdot\boldsymbol{p}},\quad\quad \hat{\psi}(t,\boldsymbol{p})=\int d^3x\;\psi(t,\boldsymbol{x})e^{-i\boldsymbol{x}\cdot\boldsymbol{p}},
[/tex]
and the spatial representations of the state, in the spacetime, between two different frames are related by equation
[tex]
\psi'(t',\boldsymbol{x}') = \psi(t,\boldsymbol{x}) = \psi\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}}, (x')^1, (x')^2,\frac{(x')^3 + ut'}{\sqrt{1-u^2}}\big) = \psi((\Lambda^{-1})^0{}_{\mu} (x')^{\mu},\ldots, (\Lambda^{-1})^3{}_{\mu} (x')^{\mu}).
[/tex] (1)
In any particular frame, with initial configuration [tex]\psi(0,\boldsymbol{x})[/tex], [tex]\hat{\psi}(0,\boldsymbol{p})[/tex], the time evolution is fixed with (by definition equivalent) equations
[tex]
\psi(t,\boldsymbol{x}) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(0,\boldsymbol{x}),\quad\quad \hat{\psi}(t,\boldsymbol{p})=e^{-it\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}).
[/tex]
These are my assumptions.
meopemuk said:your "manifestly covariant" transformation in the (x,t) space (I marked it by (1)) is perfectly OK if [tex]\psi(t,\boldsymbol{x}) [/tex] denotes quantum field. However, I would disagree if you applied it to particle wavefunctions in the position representation.
jostpuur said:Do you also believe that there is something wrong in defining momentum eigenstates [tex]|p\rangle[/tex], which can be acted on by Lorentz transformation operators? I believe I just showed how this approach can be made equivalent with the approach where the Lorentz transformation is postulated with [tex]\psi'(t',x')=\psi(t,x)[/tex]. Logically, you should either believe that the (t,x) transforming approach is right, or that the |p> transforming approach is wrong, or then that there is something wrong with my calculation.
Those equations above, and your later one:jostpuur said:[...]
In any particular frame, with initial configuration [tex]\psi(0,\boldsymbol{x})[/tex], [tex]\hat{\psi}(0,\boldsymbol{p})[/tex], the time evolution is fixed with (by definition
equivalent) equations
[tex]
\psi(t,\boldsymbol{x}) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(0,\boldsymbol{x}),\quad\quad \hat{\psi}(t,\boldsymbol{p})=e^{-it\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}).
[/tex]
These are my assumptions. [...]
are just a representation of the Poincare Casimir operator[tex]i\partial_t\psi = \sqrt{-\nabla^2 + m^2}\psi[/tex]
Haelfix said:This debate keeps popping up on these boards, and I'll side with Strangereps point of view on the subject.
I thought i'd add a classic thought experiment about localization often taught in early QFT courses (eg Coleman's lectures). This very much relates to the nonvanishing of the propagator outside lightcone and things like that, as well as the early tensions between field theory and relativistic QM.
If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.