- #1
izh-21251
- 32
- 1
First of all, let me remind about an older thread on this topic:
https://www.physicsforums.com/showthread.php?t=330517
Here I'd like to thank again to everybody, who participated in that discussion.
However, I still find myself at a deadlock with some questions about Gauge Invariance (GI) principle. Thus, I decided to start a new thread about GI.
Before I come up to my problems, let me start with refreshing what I derived from books and what has been discussed previously.
So, the principle of GI arise from the fact, that a 4-vector quantum field cannot be built from
creation/annihilation opeartors of spin 1 massless particles (say, photons) (refer to S. Weinberg, "The Quantum Theory of Fields", Cambridge University Press, 1995, vol.1, p.246-250)
Namely, the coefficient functions of a desired field appear to transform not like 4-vectors under LORENTZ transformations! Instead,
D^{\mu}_{\nu}(W(\Theta,\alpha,\beta))e^{\nu}(k,\pm1)=e^{\pm i\Theta}\left\{e^{\mu}(k,\pm1)+\frac{\alpha\pm i\beta}{\sqrt{2}k}k^{\mu}\right\}
where e^{\mu}(k,\pm1) are the coefficient functions of vector field for helicities +1 and -1, \alpha, \beta, \Theta - arbitrary parameters of Lorentz-transformation.
We see - an additional term appeared, proportional to particle momentum k^{\mu}. This means, that photon field is not a 4-vector field.
However, the way out to build Lorentz-invariant S-matrix was suggested. It consists in coupling "vector field" to conserved currents only, so that additional term k^{\mu} does not affect Lorentz-invariance. Different additional terms k^{\mu} correspond to different gauges. As long as "vector potential" is coupled with conserved currents, one is free to choose any gauge, and it will not affect the observable results.
But - here begins my misunderstanding - what happens, when we build Poincare generators?
The Hamiltonian operator does not remain unchanged, if we added a term k^{\mu} to a photonic coefficient function.
In other words, if we apply a LORENTZ transformation to Hamiltonian it will vary due to the fact, that coefficients e^{\mu}(k,\pm1) will not transform as 4-vectors, but acquire a k^{\mu}-term. Stress, Hamiltonians of different gauges are connected with each other by LORENTZ transformations!
So far, my first question is - AM I RIGHT with my logic up to this point?
Ok, if I am right, Hamiltonian is not a Lorentz-scalar any more... However, the S-matrix still seems to be the same for all gauges, because, as I understand, all these different Hamiltonians are scattering-equivalent... (see Ekstein H. Equivalent Hamiltonians in scattering theory // Phys. Rev.,1960. Vol. 117. P. 1590 – 1595).
What about other Poincare generators? To ensure relativistic invariance of the theory, they must obey the set of commutation relations - the Lee algebra. For example - in instant form of relativistic dynamics, another generator, that carries interaction is the boost operator. Thus, the interaction in boost-operator will vary when one moves to another Lorentz-frame - in the same way as interaction in Hamiltonian did. No doubts - the whole set of Poincare-Lee commutation relations must remain unchanged.
So, we've come to a point, where in each Lorentz-frame we have unique set of Poincare generators.
My second question again - am I right here? And does it have any consequences for the theory?
Especially, if anybody is familiar with the so-called "clothing-transformation" method in QFT, what all this will mean for this approach??
Thanks.
Ivan.
https://www.physicsforums.com/showthread.php?t=330517
Here I'd like to thank again to everybody, who participated in that discussion.
However, I still find myself at a deadlock with some questions about Gauge Invariance (GI) principle. Thus, I decided to start a new thread about GI.
Before I come up to my problems, let me start with refreshing what I derived from books and what has been discussed previously.
So, the principle of GI arise from the fact, that a 4-vector quantum field cannot be built from
creation/annihilation opeartors of spin 1 massless particles (say, photons) (refer to S. Weinberg, "The Quantum Theory of Fields", Cambridge University Press, 1995, vol.1, p.246-250)
Namely, the coefficient functions of a desired field appear to transform not like 4-vectors under LORENTZ transformations! Instead,
D^{\mu}_{\nu}(W(\Theta,\alpha,\beta))e^{\nu}(k,\pm1)=e^{\pm i\Theta}\left\{e^{\mu}(k,\pm1)+\frac{\alpha\pm i\beta}{\sqrt{2}k}k^{\mu}\right\}
where e^{\mu}(k,\pm1) are the coefficient functions of vector field for helicities +1 and -1, \alpha, \beta, \Theta - arbitrary parameters of Lorentz-transformation.
We see - an additional term appeared, proportional to particle momentum k^{\mu}. This means, that photon field is not a 4-vector field.
However, the way out to build Lorentz-invariant S-matrix was suggested. It consists in coupling "vector field" to conserved currents only, so that additional term k^{\mu} does not affect Lorentz-invariance. Different additional terms k^{\mu} correspond to different gauges. As long as "vector potential" is coupled with conserved currents, one is free to choose any gauge, and it will not affect the observable results.
But - here begins my misunderstanding - what happens, when we build Poincare generators?
The Hamiltonian operator does not remain unchanged, if we added a term k^{\mu} to a photonic coefficient function.
In other words, if we apply a LORENTZ transformation to Hamiltonian it will vary due to the fact, that coefficients e^{\mu}(k,\pm1) will not transform as 4-vectors, but acquire a k^{\mu}-term. Stress, Hamiltonians of different gauges are connected with each other by LORENTZ transformations!
So far, my first question is - AM I RIGHT with my logic up to this point?
Ok, if I am right, Hamiltonian is not a Lorentz-scalar any more... However, the S-matrix still seems to be the same for all gauges, because, as I understand, all these different Hamiltonians are scattering-equivalent... (see Ekstein H. Equivalent Hamiltonians in scattering theory // Phys. Rev.,1960. Vol. 117. P. 1590 – 1595).
What about other Poincare generators? To ensure relativistic invariance of the theory, they must obey the set of commutation relations - the Lee algebra. For example - in instant form of relativistic dynamics, another generator, that carries interaction is the boost operator. Thus, the interaction in boost-operator will vary when one moves to another Lorentz-frame - in the same way as interaction in Hamiltonian did. No doubts - the whole set of Poincare-Lee commutation relations must remain unchanged.
So, we've come to a point, where in each Lorentz-frame we have unique set of Poincare generators.
My second question again - am I right here? And does it have any consequences for the theory?
Especially, if anybody is familiar with the so-called "clothing-transformation" method in QFT, what all this will mean for this approach??
Thanks.
Ivan.
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