Long back, I posted a question with title "Is the vacuum of a local U(1) gauge theory unique?", which, as the title suggests asked whether the vacuum of a "spontaneously broken" the gauge theory is unique.

If I understand it correctly, the answer by @ACuriousMind was that the vacuum is degenerate even in presence of a gauge field. However, here is a lecture by Prof. Gerard 't Hooft which seems to disagree. Between $10$-$11$ minutes he emphasizes that the vacuum is actually gauge-invariant and non-degenerate or unique!

Can someone explain what makes the vacuum of a gauge theory unique, if really so? I failed to understand Prof. 't Hooft's line of reasoning. Below are two screenshots from this lecture. From the slide below, the comment about the vacuum is important enter image description here

Again, the comment below regarding the vacuum of gauge theory is important enter image description here

  • $\begingroup$ While I'm quite happy that you didn't just accept my wrong answer, you shouldn't have asked the question again, but instead offered this bounty (with the same reason!) on your original question. In order not to deprive current answerers of the bounty, I've left the bounty here and now closed the older question as a duplicate of this one, but please do not re-ask questions if you are not satisfied with the answers. (Also, since I have deleted my wrong answer, you might want to edit your post) $\endgroup$ – ACuriousMind Apr 22 at 7:35

Unsurprisingly, 't Hooft's claim is correct: the ground state of the Higgs phase is unique. More generally, in any gauge theory, the ground state will be invariant(1) under all gauge symmetries (including the global subgroup).

The easiest way of seeing this, is by noting that a gauge symmetry is unphysical whereas a (ground) state is physical---and hence the former cannot affect the latter. However, it seems this argument leads to confusion, especially when it is sometimes claimed that the global symmetry is not a 'gauge' symmetry and hence can be broken. This is wrong. Fortunately, to clear up this perennial confusion, we don't even need to presume that what we are talking about a gauge symmetry. Instead, one can make the argument for general local symmetries, independent of whether or not they are 'gauge' or otherwise(2). So, for the rest of this post, I will refrain from using the words 'gauge' and rather use the words 'local' (and 'global'). More precisely, I will explain:

A theory with local symmetries cannot spontaneously break the local symmetries or the global symmetry related to it.

The first part of this claim is that we cannot spontaneously break the local symmetries. This is the content of Elitzur's theorem, and there is already a beautiful post by Dominic Else on this.

The second part of the claim is that the corresponding global symmetry is also unbroken. This is in fact an easy consequence from the previous point since the global symmetry is a special case of the local symmetry. For instance, suppose the ground state is invariant under a local gauge transformation of the particular form $$ \psi(x) \to e^{i\lambda_n(x)} \psi(x) \qquad \textrm{ where } \lambda_n(x) = \left\{ \begin{array}{lll} \theta & & \textrm{if } x \in [n,n+1), \\ 0 & & \textrm{otherwise}. \end{array} \right. $$ Then it will clearly also be invariant under doing this many times: $$ \psi(x) \to e^{i \sum_n \lambda_n(x)} \psi(x) = e^{i\theta} \psi(x), $$ which is simply a global transformation!

To say it more mathematically: the group of global transformations is a subgroup of the local transformations, and hence if we preserve the latter, we preserve the former. Any claim to the contrary is mathematically inconsistent.

It is sometimes pointed out that the subgroup of global transformations does not affect the gauge field. This is true. But that in no way implies that one can spontaneously break it (and indeed, the previous paragraph shows that it is in fact impossible).

There is another way of seeing this impossibilty (which is equivalent but it might help to drive the point home): if one really claims that the global symmetry is spontaneously broken, then what is the order parameter? Any local order parameter that one would write down that is charged under the global symmetry, will automatically also be charged under the local symmetry. Hence, if such an order parameter has a nonzero expectation value, we would violate Elitzur's theorem.

(1) Note that this does not exclude degeneracies for other reasons such as topological order. However, this does not happen in the Higgs phase

(2) Another common misconception is that all local symmetries are gauge symmetries. This is wrong, except if one defines the term 'gauge symmetry' as a synonym for 'local symmetry'. A gauge symmetry is by definition a do-nothing symmetry; in the context of quantum mechanics, acting with a gauge symmetry does nothing to any physical state. A local symmetry is, as the name suggests, a symmetry which only acts non-trivially in a finite region of space. For an example of a system with a local symmetry that is not a gauge symmetry: consider a paramagnet $H = \sum_n S^z_n$, which has a local symmetry at every site generated by $S^z_n$. (The silliness of this example is meant to demonstrate the silliness of claiming that a local symmetry = gauge symmetry.)

| cite | improve this answer | |
  • 1
    $\begingroup$ @ChiralAnomaly Well as you say, the gauge group is a group. If we leave out the global ones, it would no longer be a group (the global symmetry action is generated by the local ones). $\endgroup$ – Ruben Verresen Apr 22 at 1:16
  • 1
    $\begingroup$ @ChiralAnomaly My answer was based on the idea that the global transformations are not gauge transformations (they are not do-nothing). This answer makes explicit that the idea of uniqueness here is not dependent on any of the transformations being gauge. But I find your line of thought also interesting. $\endgroup$ – ACuriousMind Apr 22 at 7:53
  • $\begingroup$ Could you maybe be a little bit mir explicit for the funny example in (2). I somehow fail to see what you mean. I would like to understand your local not necessarily gauge statement better. $\endgroup$ – ungerade Apr 22 at 9:52
  • $\begingroup$ @ungerade Using a gauge group that acts trivially asymptotically (or on the "boundary") seems to be a common theme (see arxiv.org/abs/1703.05448). If $H$ denotes the group of transformations with $\theta(x)\to 0$ as $|x|\to \infty$, and if $G$ denotes the larger group with no restrictions on the asymptotic behavior of $\theta(x)$, then we could choose the group $H$ to be do-nothing transformations without requiring that $G/H$ (that is, $G$ modulo transformations in $H$) be do-nothing transformations. However, now I see that this is irrelevant (see my next comment)... $\endgroup$ – Chiral Anomaly Apr 22 at 17:44
  • $\begingroup$ @ACuriousMind Okay, now I see what you're saying. This answer uses a theorem that local symmetries can't be spontaneously broken, and that theorem doesn't care whether or not they're do-nothing transformations. Thank you for clarifying this. I'm deleting my earlier comments where I missed that important point. $\endgroup$ – Chiral Anomaly Apr 22 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.