# Spontaneously broken gauge symmetry

1. Jul 20, 2012

### atyy

I have read 2 arguments that a gauge symmetry cannot be spontaneously broken.

1. Wen's textbook says a gauge symmetry is a by definition a "do nothing" transformation, so it cannot be broken.

2. Elitzur's theorem, eg.http://arxiv.org/abs/hep-ph/9810302v1

The first argument seems sound and simple, while Elitzur's theorem needs some calculation. Is the notion of gauge symmetry the same in both arguments, or is Elitzur's theorem more powerful, covering cases where there is a local symmetry without a gauge redundancy?

2. Jul 23, 2012

### DrDu

Gauge symmetry breaking can also be formulated in terms of gauge invariant properties only, e.g., as off diagonal long range order (ODLRO).
As the article you cite shows, Elitzurs theorem only refers to local observables, but not, e.g. to Wilson loops.

3. Jul 26, 2012

### atyy

Could it be clearer to replace "gauge symmetry" with "local symmetry", since a gauge symmetry by definition cannot be spontaneously broken?

4. Jul 26, 2012

### DrDu

Maybe it is kind of a misnomer, but the current definition of a broken symmetry is that the generator of the symmetry operation cannot be represented as an operator in the Hilbert space. When you decide to work in a Hilbert space which supports the action of non-gauge invariant operators (like creation and anihilation operators for charged particles) the generator of gauge transformations is indeed not included in this space once the symmetry is broken.

5. Jul 26, 2012

### atyy

So under this definition, Wen's textbook would be wrong that a gauge symmetry can never be spontaneously broken?

I guess he's using a different definition of gauge symmetry?

6. Jul 27, 2012

### DrDu

I must say that I was thinking in global gauge symmetries. I don't know the situation for local gauge symmetries very well.

7. Jul 27, 2012

### DrDu

8. Jul 28, 2012

### atyy

I copied the discussion from https://www.physicsforums.com/showthread.php?t=623786.

In Friedrich's example of spontaneous breaking of a global gauge symmetry using Bose-Einstein condensation of a non-relativistic free Bose gas at zero temperature, is the ground state degenerate? He says that there are an infinite number of pure ground states, and that they are physically equivalent. I guess the ground state should be degenerate, since I naively think of spontaneous symmetry breaking as requiring degenerate ground states. But if the ground state is degenerate, how can the different ground states be physically equivalent?

Is Greiter's claim that the ground state of a superconductor is degenerate right or wrong?

Last edited: Jul 28, 2012
9. Jul 29, 2012

### DrDu

Every ground state of an infinite system with finite particle density is degenerate, whether superconducting or not. We can add a finite number of particles to the system without changing the overall particle and energy density. The point is that we have to take this degeneracy into consideration in a superconductor, while it has no consequence in a normal metal.
This was brought out most clearly in the analysis by Haag:

10. Jul 31, 2012

### atyy

@DrDu, in the paper by Struyve you mentioned, there are two definitions of gauge symmetry. In his second definition (section 7), the gauge symmetry leads to non-unique time evolution. Is the global gauge symmetry that you or Haag are talking about not a gauge symmetry in the sense of Struyve's second definition?

11. Jul 31, 2012

### DrDu

Haag exclusively talks about global gauge transformations, so the EOM are deterministic.
I was speculating how his conclusions change when local transformations are taken into account.
It may be that this induces some indeterminism. I would have to read Struyves article in depth.

12. Jul 31, 2012

### atyy

In Haag's section 2, "spontaneously broken global gauge symmetry" is due to degenerate ground states labelled by $\alpha$, while Greiter says that there is "spontaneously broken global phase symmetry" due to degenerate ground states labelled by $\lambda$ (Eq 98) or $\phi$ (Eq 106).

Are Haag's $\alpha$-labelled states and Greiter's $\phi$-labelled states the same?

If they are the same, is the following an acceptable explanation for why Haag calls these states 'gauge' while Greiter doesn't: Haag excludes observing the states with a second superconductor via the Josephson effect, so that all the $\alpha$-labelled states are observationally indistinguishable, while Greiter includes observing the states with a second superconductor so that the different $\phi$-labelled states are observationally distinguishable?

13. Aug 1, 2012

### DrDu

As I said already I am convinced that Greiters argumentation is flawed from the very beginning.
Therefore I do not bother too much about how his ground states are related to other ones.

14. Aug 1, 2012

### atyy

Well, let me ask the question directly then - in what sense are the transformations in Haag 'global gauge symmetries', instead of being simply 'global symmetries'? Is Haag using Struyve's definition (p4, just after Eq 8) that "both the global and local symmetries can be regarded as gauge symmetries. They both connect observationally indistinguishable solutions"?

Last edited: Aug 1, 2012
15. Aug 1, 2012

### TrickyDicky

Even though I don't have access to Haag's article I have often had that doubt , and haven't found a clear answer, hopefully some expert can clarify what distinguishes a global gauge from a global symmetry, is it just the possibility of restricting it to a local gauge?

16. Aug 1, 2012

### DrDu

After Eq. 8, Struyve writes:
"In the above examples, both the global and local symmetries can be regarded as
gauge symmetries. They both connect observationally indistinguishable solutions (at
least in a hypothetical world, where those classical field equations would hold). But
while observational indistinguishability is necessary to label a symmetry as gauge, it is
not sufficient. One could still regard observationally indistinguishable states or solutions
as physically distinct (see [19] for a detailed discussion of such matters). This is the case,
for the alternative notion of gauge symmetry which relates gauge symmetry to a failure
of determinism, and which will be discussed in detail in sections 6 and 7. In the following
sections, we will regard the global and local symmetries as gauge symmetries."

So Struyve also calls global U(1) transformations gauge transformations. "Operational indistinguishability" is also what Haag had in mind. He talks explicitly of gauge transformations of the 1st kind (cf. eq. 21) which is synonymous to global gauge transformations.

17. Aug 1, 2012

### DrDu

A global gauge transformation allows us to implement charge superselection although it acts only nontrivially on operators which are not observables (in contrast to "real" symmetry operations). That was the central message of Doplicher Haag and Roberts.

18. Aug 1, 2012

### atyy

Struyve discusses that a global symmetry can be considered operationally distinguishable or not depending on whether the system is viewed as being the whole universe or a subsystem (p15):

"The other partial transformations do not correspond to a local transformation and can hence be viewed as changing the physical state of the subsystem. However, the observational difference only arises when comparing the fields of the subsystem with those of the environment and not from within the subsystem itself. This is analogous to the familiar view one can take on, for example, translation and rotation symmetry in classical mechanics. While translations or rotations of the whole universe are unobservable, and hence may be regarded as gauge transformations, translations or rotations of (approximately isolated) subsystems relative to their environment will yield (in principle) observable differences."

Presumably Haag must be taking the superconductor to be the whole universe so that the ground states are operationally indistinguishable? OTOH, since the global symmetry is not a gauge symmetry in the sense of producing nondeterministic EOM, if the superconductor is considered a subsytem, then the ground states are operationally distinguishable, say with respect to a second superconductor via the Josephson effect?

19. Aug 2, 2012

### atyy

I believe Greiter's Eq 11-13 are correct. The are the same definition of gauge invariance as Scholarpedia's Eq 17 & 18. There isn't any conflict with Haag's analysis, because Greiter's gauge invariance is not the same as Haag's gauge invariance. Rather Haag's gauge invariance is Greiter's global invariance (Eq 98), which is not the same as Greiter's gauge invariance (Eq 11-13).

20. Aug 3, 2012

### DrDu

You (and Greiter) are mixing up state space and position representation. A state $|\phi>$ transforms into $U|\phi>$. To get the transformation in the position representation, you have to project onto $<x|$, ie $\phi(x)=<x|\phi>$ gets transformed into $\phi'(x)=<x|U|\phi>$. Greiter claims that $U|\phi>=|\phi>$, so that $\phi(x)=\phi'(x)$ (note that the function $\phi$ in eq. 13 is, in contrast to the claim by Greiter, not the wavefunction in position representation) which is at variance with Scholarpedia.

Last edited: Aug 3, 2012