- #1

- 36

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Rocky Raccoon
- Start date

- #1

- 36

- 0

- #2

- 220

- 7

Electromagnetism: U(1) gauge symmetry -> conservation of charge ?

- #3

tom.stoer

Science Advisor

- 5,778

- 170

The meaning of a gauge symmetry, how it is treated in quantization, etc. is partially different from global symmetries, but Noether's theorem works as expected.

That applies to non-abelian gauge symmetries like SU(3) in QCD as well.

- #4

- 36

- 0

- #5

- 368

- 12

- #6

tom.stoer

Science Advisor

- 5,778

- 170

Zero! The charge is zero, but this is more complicated than it seems.

Of course there is no charge associated to photons b/c they are charge-neutral - and there is no charge associated to electrons as you decided to eliminate them :-) Therefore there is no conserved current and no conserved charge. But there is still a Gauss law which - when promoted to an operator during quantization - still acts as a generator of the local gauge symmetry.

But this is basically due to the fact that in QED w/o electrons and positrons the photon is free field. If you do the same calculation in QCD you get a conserved current which is purely gluonic.

The reason why the gauge symmetry in the photon sector is trivial in QED is that the photon field strength [tex]F_{\mu\nu}[/tex] is itself invariant w.r.t. to the gauge symmetry [tex]A_{\mu} \to A_{\mu} - \partial_{\mu}\chi[/tex]. This is a very special case. In QCD the gluon field strength is NOT invariant w.r.t. the gauge symmetry but transforms in the adjoint rep. Only the quadratic term in the Lagrangian is invariant.

@Chopin: of course you can study QED w/o electrons - it's a free field theory.

Of course there is no charge associated to photons b/c they are charge-neutral - and there is no charge associated to electrons as you decided to eliminate them :-) Therefore there is no conserved current and no conserved charge. But there is still a Gauss law which - when promoted to an operator during quantization - still acts as a generator of the local gauge symmetry.

But this is basically due to the fact that in QED w/o electrons and positrons the photon is free field. If you do the same calculation in QCD you get a conserved current which is purely gluonic.

The reason why the gauge symmetry in the photon sector is trivial in QED is that the photon field strength [tex]F_{\mu\nu}[/tex] is itself invariant w.r.t. to the gauge symmetry [tex]A_{\mu} \to A_{\mu} - \partial_{\mu}\chi[/tex]. This is a very special case. In QCD the gluon field strength is NOT invariant w.r.t. the gauge symmetry but transforms in the adjoint rep. Only the quadratic term in the Lagrangian is invariant.

@Chopin: of course you can study QED w/o electrons - it's a free field theory.

Last edited:

- #7

- 36

- 0

A free photon field exist even in the classical limit - it's the usual EM waves.

- #8

- 36

- 0

So, the conserved quantity of free EM field is zero?

- #9

- 525

- 7

The conserved quantity is electric charge. So yes, for a free EM field it is zero.

- #10

tom.stoer

Science Advisor

- 5,778

- 170

But for local gauge symmetries there is a quantity that is conserved locally, namely the Gauss law. In case of QED w/o electrons this is (promoted to a constraint equation acting on physical states)

[tex]\nabla E|\psi\rangle = 0[/tex]

The conservation law is just

[tex][H, \nabla E] = 0[/tex]

i.e. the Gauss law operator commutes with the Hamiltonian. In that sense the globally defined Noether charge is identically zero but the gauge symmetry is still present due to the Gauss law. This is different for global symmetries as they do not have something like Gauss law.

- #11

- 368

- 12

@Chopin: of course you can study QED w/o electrons - it's a free field theory.

Sure. I guess what I meant by that (but put into words poorly) is that if you're dealing only with the EM field, the gauge transformation itself becomes uninteresting, since the covariant derivative term is gone. So I guess that basically leads to what you said--the Noether current of the gauge symmetry of a free photon field results in the conservation of 0. :)

- #12

tom.stoer

Science Advisor

- 5,778

- 170

yes - it's d0/dt = 0

study QCD w/o quarks - it's fun

study QCD w/o quarks - it's fun

- #13

RUTA

Science Advisor

- 1,366

- 401

So, the conserved quantity of free EM field is zero?

The stress-energy tensor (SET) for the free EM field is divergence-free (as are all physically realizable SETs), so momentum and energy are also conserved for this field. I realize this is a bit off topic, but just to be precise

- #14

tom.stoer

Science Advisor

- 5,778

- 170

correct, but the stress-energy tensor is not related to the gauge symmetry

- #15

- 980

- 2

But for local gauge symmetries there is a quantity that is conserved locally, namely the Gauss law. In case of QED w/o electrons this is (promoted to a constraint equation acting on physical states)

[tex]\nabla E|\psi\rangle = 0[/tex]

The conservation law is just

[tex][H, \nabla E] = 0[/tex]

i.e. the Gauss law operator commutes with the Hamiltonian. In that sense the globally defined Noether charge is identically zero but the gauge symmetry is still present due to the Gauss law. This is different for global symmetries as they do not have something like Gauss law.

Might be worth mentioning that these equations hold in the "pre"-Hilbert space before gauge fixing. In the "physical" space the conservation law will be the identity operator.

I've always wondered at the back of my mind what the precise relation is between the corresponding structures in quantum theory and classical theory. I know it involves Dirac constraints, but I confess to never having fully internalised how that's supposed to work Would it be fair to say that the whole tortured thing is a result of topological obstructions to constructing a decent coordinate system on the physical symplectic manifold? Such that we find it easier notationally to talk about an enlarged system and then impose the "symmetry"?

- #16

tom.stoer

Science Advisor

- 5,778

- 170

Exactly; thanks for adding this comment.Might be worth mentioning that these equations hold in the "pre"-Hilbert space before gauge fixing. In the "physical" space the conservation law will be the identity operator.

Yes; you should read the "old" Dirac papers regarding constraint quantization. It is essentially a qm framework to implement a foliation of the gauge fiber bundle.I've always wondered at the back of my mind what the precise relation is between the corresponding structures in quantum theory and classical theory. I know it involves Dirac constraints, but I confess to never having fully internalised how that's supposed to work Would it be fair to say that the whole tortured thing is a result of topological obstructions to constructing a decent coordinate system on the physical symplectic manifold? Such that we find it easier notationally to talk about an enlarged system and then impose the "symmetry"?

- #17

- 980

- 2

Yes; you should read the "old" Dirac papers regarding constraint quantization. It is essentially a qm framework to implement a foliation of the gauge fiber bundle.

Yes, I've tried ;-) The thing which I guess I'm not clever enough to infer, is actually the other way round. It's not so hard (conceptually) to understand the quantisation procedure and its motivations, but I wonder what it looks like going the other way: what can be said about the classical structures if you "ungauge fix" a quantum theory? Now, clearly (?) we need to be specific about what kinds of overlabelling is allowed, but a good starting point would be the usual Yang-Mills type thing. Are we always going to end up with a proper Hamiltonian formalism in the classical limit? What does the lack of a proper inner product in the pre-Hilbert space correspond to?

I suspect that it's all there already in Dirac's work, but I guess (since it's not my actual research) I'd like someone to just tell me :-)

- #18

tom.stoer

Science Advisor

- 5,778

- 170

First of all these issues can be treated in different formalisms - which should be identical w.r.t. to physical results :-)

So essentially one MUST gauge fix - depending on the formalism in different ways and due to different reasons. In the canonical formalism gauge fixing is required in order to have a physical Hilbert space w/o negative norm states (one can try to work with negative norm states as well - Gupta-Bleuler-formalism - but in the end again the unphys. d.o.f. must cancel; in addition this formalism does not work for non-abelian gauge theories). In the PI formalism one must gauge fix as otherwise one integrates not only over physically different paths but over gauge-copies which leads to a divergent PI b/c each physical paths appears in infinite number of times due to gauge-copies.

Gauge fixing in the canonical formalism is essentially described by Dirac; gauge fixing in the PI formalism uses Fadeev-Popov ghosts (killng unphysical d.o.f.) or the more sophisticated (but related) BRST invariance. In the very end the gauge d.o.f. are removed / cancelled / ... w/o breaking gauge invariance

There is a simple analogy for a two-particle system with translational invariance and interaction V(|x-y|). Going to the c.o.m frame means to use p (rel. mom.) and P (c.o.m. mom.). "Gauge fixing" now means setting P=0 (or in QM P|phys>=0) i.e. looking at the rel. momemtum p in the c.o.m. frame. Of course this does not break translational invariance but it reduces the physical space to states with vanishing P (all other states are still there in the larger Hilbert space but we are clever enough to project them away).

The difference between gauge fixing classically and quantum mechanically is not that we do not gauge-fix but that the details of the implementationa re different. This is most clearly seen in the canonical approach as one can relate both the classical and the qm case

{.,.} => [,.]; Poisson bracket => commutator

H => H; Hamiltonian (function) => Hamiltonian (operator)

C=0 => C~0; C|phys>=0; constraint => constraint acting an states; kernel = phys states

K => U; canonical transformation => unitary gauge fixing transformation

(in the exmaple above this transforms from the two coordinates describing the two particles to the c.o.m. frame)

Your sentence "Are we always going to end up with a proper Hamiltonian formalism in the classical limit?" is misleading. The proper Hamiltonian is constructed both in the classicasl and in the qm case and has nothing to do with the classical limit.

Of course one can work in an unphysical Hilbert space as well, but one has to use other mechanisms to cancel the unphysical d.o.f. The Fadeev-Popov ghosts do something like that; they cancel certain contributions coming from unphysical gluons.

There is one major benefit in fxing the gauge (such that it becomes the identity in the physical subspace): After gauge fixing you can use any approx. you like w/o ever violating gauge symm. Dealing with a larger Hilbert space is risky as an approx. my break gauge inv. which typically kills the consistency!

- #19

- 120

- 0

The conserved charge in local gauge theories is not truly conserved. But local gauge symmetry is unphysical and has to be removed (gauge fixed), unlike global gauge symmetry. This "fixes" the local gauge charge into the a conserved charge (similar to global gauge charge). I have also be told that Noether's second theorem is related to all this - but I never got to read and understand it.

- #20

tom.stoer

Science Advisor

- 5,778

- 170

nice paper; but note that it uses BRST and not Dirac

Share: