I am aware that when some operator $\phi_n(x)$ which transforms nontrivially with respect to some symmetry group acquires a VEV, it signals the spontaneous breakdown of a particular symmetry, since there must be different vacua to accomodate for the different values of $\langle\phi_n(x)\rangle$ under the symmetry transformation.

My question is, why doesn't the same logic apply to two-point functions like $\langle T\phi_n(x)\phi_m(y)\rangle$? These are expected to be nonzero, and they transform nontrivially. Somehow we don't make the conclusion that a nonzero two-point function implies spontaneous symmetry breaking.

What I am thinking:

I can make an analogy to a different situation that might help the discussion. If we consider Yang-Mills theory, there are 1-form global symmetries, which act on Wilson and 't Hooft line operators. This 1-form symmetry can be spontaneously broken, and the criterion is as follows. If a Wilson loop $\langle W(L)\rangle$ follows an area law for large loops the symmetry is unbroken, while if it follows a circumference law the symmetry is broken. The logic for this is that a circumference law can be canceled by some local counterterm, and so the Wilson loop won't go to zero for large loops.

The situation may be similar for the two-point case, where it can either have an exponential decay (which signals a mass gap) or some slower decay.

However, if this is the case it seems weird to me that we don't care what happens when $x\to y$ (which probes the UV), we only care about about the IR behavior of the two point function.

Response to the singlet answers

The field strength two point correlation function is

$$\langle T F_{\mu\nu}(x)F_{\lambda\sigma}(0)\rangle=\frac{4}{(x^2)^2}\Big(\eta_{\mu\lambda}\eta_{\nu\sigma}-2\eta_{\nu\sigma}\frac{x_{\mu}x_{\lambda}}{x^2}-2\eta_{\mu\lambda}\frac{x_{\nu}x_{\sigma}}{x^2}-(\mu\leftrightarrow\nu)\Big)$$

Looking at the tensor structure of this, the first term with just $\eta$'s is certainly a singlet of the Lorentz group, but I don't see how the other parts are singlets. This suggests that Lorentz symmetry is spontaneously broken based on what you are saying.

Response to MannyC's answer

While I agree that what you did shows no contradiction with the fact that $Q|\Omega\rangle=0$ and the correlator is non-zero, there does appear to be a contradiction by considering the quantity

$$0\neq\langle\Omega|[Q,\phi(x_1)O_1(x_2)\cdots O_{n-1}(x_n)]|\Omega\rangle=0$$

This is nonzero by the assumption that the operator transforms non-trivially, but it is zero by the assumption that $Q|\Omega\rangle=0$. So actually, if $Q|\Omega\rangle=0$ then the correlator must be in the singlet representation. This observation doesn't bode well for the two point correlator of field strengths however.

  • $\begingroup$ I think the answer is much simpler than you suspect: If the product of the two reps in the correlator conatins a singlet, then a non-zero correlation fct does not signal symmetry breaking. Or, if the two reps refer to a space-time symmetry then the correlator can be proportional to suitable tensors constructed from the coordinate vectors. $\endgroup$
    – Thomas
    Mar 4, 2020 at 19:25
  • $\begingroup$ So are you saying that if the symmetry is not broken, then the correlation function will always be a singlet? $\endgroup$
    – fewfew4
    Mar 4, 2020 at 20:01
  • $\begingroup$ Unless I'm missing something, in the solid state two point correlators can indeed signal symmetry breaking and long range order. For example the liquid-to-solid phase transition. $\endgroup$
    – KF Gauss
    Mar 4, 2020 at 23:20
  • 1
    $\begingroup$ @KFGauss Isn't it true that in those examples they are really composite operators? That is, operators of the form $\phi^2(x)$. These are different right? $\endgroup$
    – fewfew4
    Mar 5, 2020 at 3:08
  • $\begingroup$ @LucashWindowWasher, that I'm not sure about. The key point is that you get nonzero values for $\phi_i(x)\phi_j(y)$ for $x\neq y$, and multiple components $i$ and $j$ that could be distinct atoms or electrons of different "flavors". Would that classify as a composite operator? $\endgroup$
    – KF Gauss
    Mar 5, 2020 at 12:44

2 Answers 2


I have two different ways to look at it.

Charge acting on the vacuum

Take a symmetry group $G$ and a field $\phi$ which is not a singlet. There is a Cartan generator $Q$ in $G$ that is non vanishing on $\phi$, namely $$ [Q, \phi ] = q_\phi \phi\,, $$ for some non zero number $q_\phi$. Let me denote the vacuum as $|\Omega\rangle$. If $\phi$ has a non zero one-point function then $$ 0 \neq q_\phi\langle \Omega |\phi| \Omega\rangle = \langle \Omega| [Q,\phi]|\Omega\rangle\,. $$ If this has to be non-zero, then necessarily $Q |\Omega\rangle \neq 0$, thus the vacuum is not invariant. On the other hand, any other $n$-point funtion will be $$ q_\phi \langle \Omega|\phi(x_1)\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle = \langle \Omega| [Q,\phi(x_1)]\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle\,. $$ Even if $\langle \Omega | Q = 0$ there will be a piece like $$ \langle \Omega| \phi(x_1)\,Q\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle\,, $$ which doesn't have to vanish. So a non vanishing $(n>1)$-point function is consistent with a $G$ invariant vacuum.

Wigner-Eckart-like argument

Another point of view, which came up in the comments, is to think of it in the same way as the Wigner-Eckart theorem. Take two operators $O_1$, $O_2$ in the representations $\rho_1$ and $\rho_2$ of some group $G$. If the vacuum is $G$ invariant (i.e. no spontaneous symmetry breaking), the expectation value $$ \langle \Omega |O_1(x_1) O_2(x_2) | \Omega\rangle\,, $$ will be proportional to some matrix element $$ \langle \Omega | T_{12} |\Omega \rangle \,, $$ where $T_{12}$ is kind of a "placeholder" operator that transforms under the representations in the tensor product $\rho_1 \otimes \rho_2$. If such a tensor product contains a singlet, the above matrix element does not need to vanish. Clearly even when $\rho_1$ and $\rho_2$ are non trivial one can find singlets in $\rho_1\otimes \rho_2$.

  • $\begingroup$ I made an edit to my question which is a response to your answer. $\endgroup$
    – fewfew4
    Mar 5, 2020 at 3:39
  • $\begingroup$ You misunderstood the implication of my statement. The fact that the singlets contribute to the three-point function doesn't mean that the tensor structure itself will be a singlet. It's just a counting statement: the number of independent tensor structures is the same as the number of singlets. You can see for yourself that the product of two antisymmetric tensors in 4d contains two singlets ($(1,0)+(0,1) = 2(0,0) + \cdots$). $\endgroup$
    – MannyC
    Mar 5, 2020 at 9:20
  • $\begingroup$ the Wigner-Eckart theorem says that if the product of the two representations contain a singlet, then the answer is not necessarily zero. But I don't believe it says anything about what representation it transforms with respect to. Your first argument appears to address this. But I'm having trouble seeing why this first argument works, since you need to consider the commutator of the full product of operators, not just with $\phi(x_1)$ right? If you consider the full commutator it seems like you arrive at the same conclusion as with the single operator. $\endgroup$
    – fewfew4
    Mar 5, 2020 at 20:25
  • $\begingroup$ "But I don't believe it says anything about what representation it transforms with respect to" sorry I'm being a bit slow, what does the it refer to there? As for the second part of your comment. If you refer to the third equation in my post, that's just an equality. Due to the first equation you can trade $q\phi$ for $[Q,\phi]$, without worrying about what's around it. $\endgroup$
    – MannyC
    Mar 5, 2020 at 21:01
  • $\begingroup$ Sorry, by it I meant the correlator. The point I am making is that if you are considering how the n-point correlator changes under the symmetry, you should take the commutator of the full thing $[Q,\phi(x_1)O_1(x_2)\cdots O_{n-1}(x_n)]$. Hopefully that clarifies things. $\endgroup$
    – fewfew4
    Mar 5, 2020 at 22:33

In a field theory with O(N) symmetry, spontaneous symmetry breaking implies that $\varphi_n\equiv\langle \phi_n\rangle\neq0$. The two-point correlators can then be written in all generality as $$ \langle\phi_n(x)\phi_m(y)\rangle=\delta_{nm}G_1(x,y)+\varphi_n\varphi_m G_2(x,y), $$ where $G_1$ and $G_2$ are invariant functions (which will in general depend on $\sum_n\varphi_n^2$. Thus, a non-trivial tensor structure of the correlation function implies spontaneous symmetry breaking.

On the other hand, the absence of spontaneous symmetry breaking just implies that $\varphi_n=0$ and thus $\langle\phi_n(x)\phi_m(y)\rangle=\delta_{nm}G_1(x,y)$, which transforms trivially under $O(N)$.

  • $\begingroup$ You can't say the same thing for space-time indices though. Consider the two point function between field strengths $\langle F_{\mu\nu}(x)F_{\rho\sigma}(y)\rangle$. The result involves a term with $(x-y)_{\mu}(x-y)_{\nu}(x-y)_{\rho}(x-y)_{\sigma}$. Does this imply that Lorentz symmetry is spontaneously broken? $\endgroup$
    – fewfew4
    Mar 4, 2020 at 21:34
  • $\begingroup$ @LucashWindowWasher It doesn't have to, it can also involve the metric $\eta_{\mu\nu}$. $\endgroup$
    – Adam
    Mar 5, 2020 at 13:38
  • $\begingroup$ I wrote the explicit form of the two point function I mentioned in my question. It doesn't have the term I said but it has a different term which isn't I singlet. $\endgroup$
    – fewfew4
    Mar 5, 2020 at 20:34

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