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Long back, I posted a question with title Is the vacuum of a local ${\rm U(1)}$ 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

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    $\begingroup$ 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, 2020 at 7:35

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This is indeed a very confusing question and I have spent a lot of time parsing the literature looking for an answer. The references I found most useful are section 3 of the paper by Harlow and Ooguri, a set of lectures notes, and a study of SSB in superconductors(1) by van Wezel and van den Brink. Below, I summarize my current understanding; any comments and corrections from gauge theory experts are welcome. Apologies for the long answer, because of all the confusion, I thought it's useful to be careful and detailed. The upshot is that the precise answer depends on a choice of boundary conditions, but I think there is a way in which you can consistently say that there is spontaneous symmetry breaking (SSB) in the gauge theory.

Global vs local symmetries

Firstly, as pointed out in Ruben's answer, a breaking of local gauge symmetry is ruled out both by Elitzur's theorem and by the fact that these are un-physical, 'do-nothing' transformations (we require states in the physical Hilbert space do be invariant under them). The symmetry that is supposed to be broken in the ground state is the global U(1) symmetry. It's important to clarify the relationship between these. Local gauge transformations are defined by the requirement that they tend to the identity at (spatial) infinity. The simplest examples are those that act only on some finite region. The global symmetry operations are those that do not vanish at infinity. More precisely, the global symmetry group is the factor of the group of all gauge transformation by the local ones: $G_\text{global} = G_\text{gauge} / G_\text{local}$. Note that by this definition, $G_\text{global}$ is not a subgroup of $G_\text{local}$

The global symmetry transformations thus defined act entirely on the boundary at spatial infinity (for this reason, they are also called 'asymptotic symmetries'). There is a good physical reason for that. We expect the global symmetry to be the one generated by the total electric charge of the system, $Q_\text{tot}$. But by Gauss's law, $Q_\text{tot}$ can be measured by looking at the outgoing electric flux at infinity. Another visualization is that charged objects always exist at the endpoints of Wilson lines; if $Q_\text{tot} \neq 0$ then the Wilson lines of 'unpaired' charges extend all the way to infinity and you can evaluate the total charge by counting them.

Choices of boundary conditions

Is $G_\text{global}$ a physical symmetry that can be spontaneously broken? In some sense, the answer is a matter of choice. As Harlow and Ooguri discuss (p. 58), when varying the EM action, one picks up a boundary term which is a product of the variation of the gauge field $A$ and the electric flux leaving the system(2). There are therefore two ways of ensuring gauge invariance:

  1. You can require that there is no outgoing flux, in which case you can freely vary the gauge field. This is equivalent to enforcing that $Q_\text{tot} = 0$ for all physical states.
  2. Alternatively, if you want to allow configurations with finite charge in your theory, then you have to restrict variations of $A$ to those that vanish on the boundary.

Both choices are still consistent with requiring physical states to be invariant under $G_\text{local}$.

The picture becomes clearer in lattice gauge theory (p. 66-67 in Harlow and Ooguri). One has to make a choice of boundary conditions, whether to enforce the Gauss constraint on the 'boundary sites', i.e. endpoints of edges 'sticking out' of the lattice (see their Fig. 9 for illustration). Doing so is equivalent to option 1 above, because in this case, Wilson lines cannot end at the boundary so $Q_\text{tot} = 0$ by construction. But you can also choose not to enforce the constraint on these boundary sites, in which case you keep a larger Hilbert space that has physical states with $Q_\text{tot} \neq 0$. From these, you could in principle construct the states with fixed phase that would describe spontaneous symmetry breaking. States with different phases would be physically distinct states in the Hilbert space, leading to a degenerate vacuum in the thermodynamic limit just like in a theory with only global $U(1)$ symmetry.

Dirac order parameter

As noted before, Elitzur's theorem precludes any truly local order parameter, since if one existed, it would also be non-invariant under $G_\text{local}$ and hence non-physical. Instead, we have the Dirac order parameter (Eq. 7.5 in the lectures notes)

$\Psi_D(\mathbf{r}) \equiv \Psi(\mathbf{r}) e^{i \int d^d\mathbf{r'} \mathbf{E}_\text{cl}(\mathbf{r}-\mathbf{r'})A(\mathbf{r'})}$

Here $\Psi(\mathbf{r})$ is the local matter (e. g. Cooper pair) field while $\mathbf{E}_\text{cl}(\mathbf{r})$ is a classical function corresponding to the electrostatic field of a point charge, i.e. it satisfies the equation $\nabla \cdot \mathbf{E}(\mathbf{r}) = \delta(\mathbf{r})$. This is clearly non-local; however, there is a gauge choice (Coulomb gauge, $\nabla \cdot \mathbf{A} = 0$) where the expression in the exponential vanishes and $\Psi_D$ just looks like the local fermion field.

Note that $\Psi_D$ is invariant under $G_\text{local}$ but not under $G_\text{global}$ (if you don't require $\mathbf{A}$ to vanish at infinity, you pick up a boundary term in the exponent under gauge transformation), so it is a good order parameter for the putative phase where this latter symmetry is spontaneously broken. Indeed, given the 'asymptotic' nature of this symmetry, it's unsurprising that its order parameter has to be non-local in this way.

A closer look at SSB

This might still seem a bit unsatisfying. There are at least two possible complaints:

  • I could have defined my theory on a manifold without a boundary (e.g. by taking periodic boundary conditions), in which case the choice above does not arise and I'm forced to have $Q_\text{tot}=0$.
  • Even if I take the case where all charges are allowed, there is a theorem in QFT showing that no physical (=local and gauge invariant) operator can change $Q_\text{tot}$(3). This implies that there is a superselection rule: superpositions of states with different charges are physically meaningless.

A hint that these objections are not fatal is given by the fact that the latter one arises also for superfluids, where there is no gauge redundancy, and in quantum optics, where the photon number is not even conserved.

To resolve the issues, one has to consider more carefully what we mean by spontaneous symmetry breaking. This is done in the latter two references cited at the beginning(4). The main point is that, whenever we say that a system spontaneously breaks a symmetry, we need to have some other, external reference system in mind to which we can compare it. This is true of any symmetry breaking, independently of whether we consider a gauge theory or not. So both for a superconductor and a superfluid, when we assign a phase to it, what we really mean by it is the relative phase with some previously chosen reference system(5). Another way of saying this is that while two decoupled superfluids/superconductors are in fact symmetric under changing their phases independently, an infinitesimal coupling(6) between them is enough to reduce this to the case when the ground state is only invariant under transforming both at the same time.

This gets rid of both complaints above. The more natural thing to consider for SSB is a system that does have a boundary, with the reference system being on the 'outside'; and taking the reference system into account removes the superselection rule. These considerations are the same whether there are gauge fields involved or not. So I think it is fair to say that the $U(1)$ gauge theory has SSB just as much as system with only global $U(1)$ symmetry does (and indeed, in their paper, van Wezel and van den Brink explicitly check that there is a tower of (physical) states that becomes a degeneracy in the thermodynamic limit).


(1) For those less comfortable with condensed matter jargon, feel free to replace 'superconductor' with '$U(1)$ gauge theory' everywhere. In fact, I'm cheating a bit: I have in mind a system where the matter fields that condense carry charge $e$, instead of $2e$ as they wound in an actual superconductor. Near the end, I also mention superfluids for comparison, which you should think of as 'some theory with spontaneously broken global $U(1)$ symmetry (and no gauge fields)'.

(2) Formally, it reads $\delta S_\text{boundary} \propto \int_{\partial M} \delta A \wedge \star F$, where $\partial M$ is the boundary of the manifold $M$ on which the theory is defined, and $\star F$ is the Hodge dual of the EM field tensor $F$. This latter object is the electric flux, as can be shown using Maxwell's equations.

(3) The intuition for this should be clear from the discussion above: to change $Q_\text{tot}$, you need to bring in a charge from infinity.

(4) In particular, there is a nice discussion of this in Sec. 1.5.3 of the lecture notes on SSB.

(5) This relative phase is clearly physical, since it leads to a measurable Josephson current. As pointed out in the lecture notes, this current is proportional to the correlation between the Dirac order parameter, $\langle \Psi_D(\mathbf{r})\Psi_D^\dagger(\mathbf{r'})\rangle$, in the case when the two points, $\mathbf{r}, \mathbf{r'}$, are located in the two different superconductors.

(6) By 'infinitesimal' I mean the same thing as in the definition of SSB by adding an infinitesimal symmetry-breaking field. So the precise statement is that if we first take the size of both superconductors to infinity and then take the coupling between them to zero, the remaining symmetry is still only $U(1)$ and not $U(1) \times U(1)$.

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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 impossibility (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.)

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    $\begingroup$ @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). $\endgroup$ Apr 22, 2020 at 1:16
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    $\begingroup$ @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, 2020 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, 2020 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)... $\endgroup$ Apr 22, 2020 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. $\endgroup$ Apr 22, 2020 at 17:44
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I want to present a "counter"example to Ruben's answer --- a state that preserves local gauge symmetry but breaks the global symmetry.

The issue is rooted in the thermodynamic limit. For simplicity, let us imagine $\mathbb{Z}_2$ gauge field on a one-dimensional non-compact lattice. The gauge symmetry is generated by $\sigma_{x,i\in\mathbb{Z}}$, while the global symmetry $\mathbb{Z}_2$ is generated by $\prod_{i\in\mathbb{Z}}\sigma_{x,i}$. We consider the following state (up to normalization): \begin{eqnarray} |\Psi\rangle=\sum_{\mathcal{C}:\text{ all finite subsets of $\mathbb{Z}$}}\left\{\prod_{i\in\mathcal{C}}\sigma_{x,i}|\text{all up}\rangle\right\}, \end{eqnarray} which respects $\sigma_{x,j}$ for any $j\in\mathbb{Z}$ thereby gauge invariant. However, it is explicitly breaking the global $\mathbb{Z}_2$.

Thus, I don't think Elitzur's theorem could exclude the SSB of the global "part" of the gauge symmetry, if we are careful enough about the thermodynamic limits through the proof of Elitzur's theorem.

Unfortunately, gauged $\mathbb{Z}_2$-SSB Ising chain has a unique gapped ground state (obvious from Kramers-Wannier duality) and it is also the same situation for the two-dimensional generalization. Thus I guess that it is due to the finiteness of $\mathbb{Z}_2$, and if we have chosen U(1), the situation may be different, like the superconductor \begin{eqnarray} \prod_k(u_k+v_kc^\dagger_{k\uparrow}c^\dagger_{-k\downarrow})|\text{empty}\rangle,\alpha\in\text{U}(1)/\mathbb{Z}_2, \end{eqnarray} has a huge ground state degeneracy parametrized by $\alpha$, due to the SSB of the global part, but I don't know to what extend we could trust it.

If the superconductor does not have degenerate vacua, how to explain Josephson effect? Namely, if the ground state is unique, two superconductors touching together shouldn't have any observable effect.

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