I was reading through the first chapter of Polyakov's book "Gauge-fields and Strings" and couldn't understand a hand-wavy argument he makes to explain why in systems with discrete gauge-symmetry only gauge-invariant quantities can have finite expectation value. This is known as Elitzur's theorem (which holds for continuous gauge-symmetry).

Polyakov says : [...] there could be no order parameter in such systems (in discrete gauge-invariant system) [...], only gauge invariant quantities are nonzero. This follows from the fact, that by fixing the values of $\sigma_{\mathbf{x},\mathbf{\alpha}}$ at the boundary of our system we do not spoil the gauge invariance inside it.

Here $\sigma_{x,\alpha}$ are the "spin" variables that decorate the links of a $\mathbb{Z}_2$ lattice gauge theory. I would like to understand the last sentence of this statement. Could anyone clarify what he means and why this implies no gauge-symmetry breaking ?


2 Answers 2


1) Gauge theory is a theory where we use more than one label to label the same quantum state.

2) Gauge “symmetry” is not a symmetry and can never be broken.

This notion of gauge theory is quite unconventional, but true.

When two different quantum states $|a\rangle$ and $|b\rangle$ (i.e. $\langle a|b\rangle=0$) have the same properties, we say that there is a symmetry between $|a\rangle$ and $|b\rangle$. If we use two different labels “$a$” and “$b$” to label the same state, $|a\rangle=|b\rangle$, then $|a\rangle$ and $|b\rangle$ obviously have (or has) the same properties. In this case, we say that there is a gauge “symmetry” between $|a\rangle$ and $|b\rangle$, and the theory about $|a\rangle$ and $|b\rangle$ is a gauge theory (at least formally). As $|a\rangle$ and $|b\rangle$, being the same state, always have (or has) the same properties, the gauge “symmetry”, by definition, can never be broken.

Usually, when the same “thing” has the same properties, we do not say that there is a symmetry. Thus, the terms “gauge symmetry” and “gauge symmetry breaking” are two of the most misleading terms in theoretical physics. Ideally, we should not use the above two confusing terms. We should say that there is a gauge structure (instead of a gauge “symmetry”) when we use many labels to label the same state. When we change our labeling scheme, we should say that there is a change of gauge structure (instead of “gauge symmetry breaking”).

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    $\begingroup$ Thank you for your comments Prof. Wen. So if there is no such thing as a gauge "symmetry", then asking wether you can have gauge-symmetry breaking does not make any sense. However, I think the question I'm asking, i.e. "How can you understand Elitzur's theorem from Polyakov's simple argument ?" is still legitimate. In other words, why Polyakov's argument implies that you cannot have a finite expectation value of operators that are not gauge invariant ? $\endgroup$ Mar 28, 2014 at 18:41
  • $\begingroup$ @VanillaSpinIce Because the expectation values of the operator that is not gauge invariant would correspond to the "name" that we use to label the state. As emphasized by Prof. Wen, the name is not a physical observable, all names are equivalent in the gauge theory. Any expectation value of such operator will single out one name from the others, which is illegal. $\endgroup$ Apr 2, 2014 at 19:57
  • $\begingroup$ @ Everett You: I see. So following 1) and 2) statements given above, this makes sense. Ok, but why are these statement true ? Is this in fact a consequence of Elitzur's theorem ? There's a gap in my understanding (from what I understand, it seems there is something circular in this argument). $\endgroup$ Apr 3, 2014 at 14:51
  • $\begingroup$ @ Everett You: I'm also still not sure why Polyakov's argument (and only Polyakov argument) implies no finite expectation of gauge dependent quantities for the lattice $\mathbb{Z}_2$ gauge theory. $\endgroup$ Apr 3, 2014 at 14:54
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    $\begingroup$ @EverettYou It seems extremely obvious to me that gauge symmetries can't be spontaneously broken (indeed, I don't even know what it would mean for a gauge symmetry to be "broken"). So Elitzur's theorem seems totally trivial and can be proven in one sentence with no math, as Prof. Wen did above. But I assume that to get a "theorem" named after him, Elitzur did a little more work than that. Exactly what nontrivial result did Elitzur prove? If he actually did some math, then he must have been addressing a slightly more nontrivial problem. $\endgroup$
    – tparker
    May 24, 2016 at 0:47

Say differently than this other answer somewhere else on this page, an expectation value should be given by some averaging process $\left\langle O\right\rangle $ of the observable $O$. Now if you want to calculate the expectation value of a gauge-covariant quantity, you must average over all the gauge redundancy, something like $\left\langle O\right\rangle \sim\left\langle \emptyset\right|\int dR\left[ROR^{\dagger}\right]\left|\emptyset\right\rangle $ say, with $\left|\emptyset\right\rangle$ the vacuum state (the ground state if you prefer) and $R$ the gauge transform. It corresponds to a sum over all the internal (redundant) degrees of freedom which does not change the outcome of an experiment (so a gauge transformation spans a set of states which can not be distinguished from each other). The gauge transformation can be seen as a rotation in the parameter space onto which $R$ applies too, and thus the average gives always zero.

So a gauge covariant quantity can not be an observable, henceforth it can not be an order parameter.

As for the Polyakov's argument explicitly (what is his $\sigma_{x,\alpha}$ ?), I cannot say, since I never open his book.


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