Does spontanous symmetry breaking affect the existence of a conserved charge?

And how does depend on whether we look at a classical or a quantum field theory (e.g. the weak interacting theory)?

(In the quantum case, if we don't want to speak of the Noether theorem, the question can be worded as how does the field breaking the symmetry affect the identities resulting from the Lagrangian symmetry.)

The question came up as I wondered if you can make a gauge theory out of every "kernel" of the process in which you compute obervables. In the sense that if

$$\langle A \rangle_\psi=\int (\psi^*A\psi)\text d V,$$

the transformation $T:\phi\rightarrow \text e^{i\alpha}\phi$ is in the kernel of $\langle - \rangle_\psi$.

  • $\begingroup$ See e.g. the Wikipedia page for Goldstone's Theorem. $\endgroup$
    – Qmechanic
    Aug 10, 2012 at 17:21
  • $\begingroup$ SSB only has consequences in quantum theories with infinite degree os freedom, i.e., in QFT. The consequences depend on which kind of symmetry is spontaneously broken. If it is a rigid (=no gauge) symmetry you have the goldstone theorem. However, if you are spontaneously breaking the global part of a gauge symmetry you have the Higgs mechanism. $\endgroup$ Aug 10, 2012 at 17:39
  • $\begingroup$ Do these two comments already imply a statement about the existence of the charge associated with the symmetry, which is broken? I mean like if we compare the theory with the conserved charges and the theory with an additional term for symmetry breaking. If yes, then from $\frac{\text d}{\text d t}Q\ne 0$ in the quantum case, where Q is supposed to act on the states, I would say the answer to my question is that there is no conserved charge left. $\endgroup$
    – Nikolaj-K
    Aug 10, 2012 at 17:51
  • 1
    $\begingroup$ At the quantum level, asking whether $\frac{d\hat{Q}}{dt}$ vanishes or not is not a very physical question because it could depend on the picture (Heisenberg or Schrödinger) , for instance. What makes more sense is asking about the expectation value of $\hat{Q}$. But if a rigid symmetry is SB then $Q$ is not an operator of the Hilbert space: it brings states from one Hilbert to other Hilbert space. These different Hilbert spaces do not talk each other (they are superselector sectors) and the only trace of the classical symmetry are the goldstone bosons. $\endgroup$ Aug 10, 2012 at 18:29

1 Answer 1


When you have a symmetry, we can find the conserved current $J$ and the cut-off charge $Q_f = \int_\Sigma f J$, where $\Sigma$ is a spatial hypersurface and $f:\Sigma\rightarrow\mathbb{R}$ is a Schwarz function. Note that we cannot use $f=1$ in infinite volume, which is the case we must consider since there is no spontaneous symmetry breaking in finite volume.

If the symmetry is spontaneously broken, then there is a realisation of the theory above some vacuum state $|0\rangle$ and an n-point operator $\phi_n(x_1,...,x_n)$ which is not preserved by the symmetry. Using the OPE, we can argue that we can also find a 1-point operator $\phi(x)$ which is not preserved, ie.

$$\langle 0|[Q_f,\phi(0)]|0\rangle\neq0.$$

for a suitable $f$. This implies

$$\langle 0| [J(x),\phi(0)] |0\rangle\neq 0$$

for some $x$. Now look at $M_+(x) = \langle 0| J(x)\phi(0) |0\rangle$, $M_-(x) = \langle 0| \phi(0)J(x) |0\rangle$. By what I just said, $M_+ \neq M_-$.

Using the spectral decomposition of the Hilbert space of the theory we can write

$$M_{\pm}=\int_{V_+}K_\pm(x,p)dp ,$$

the integral being over the forward light cone in momentum space. Lorentz invariance requires these guys are of the form

$$K_\pm = p e^{\pm ipx}\rho_\pm (-p^2).$$

We can write

$$M_\pm (x) = i \frac{d}{dx} \int_0^\infty \rho_\pm (m^2) W_m (\pm x)dm^2$$

where $W_m$ is the Klein-Gordon propagator


the integral being over the positive mass shell with mass $m$.

Now Let $M=M_+ -M_-$. For spacelike $x$, $W_m(x)=W_m(-x)$, so we have

$$M(x) = i \frac{d}{dx} \int_0^\infty (\rho_+ (m^2) - \rho_- (m^2))W_m(x)dm^2$$

By Lorentz invariance, for spacelike $x$, $\phi(0)$ and $J(x)$ commute, so $M(x)=0$. The formula above then implies $\rho_+ = \rho_- := \rho$. In general then,

$$M(x) = i \frac{d}{dx} \int_0^\infty \rho(m^2)(W_m(x)-W_m(-x))dm^2.$$

Now we take the derivative of both sides and use the Klein-Gordon equation $\frac{d}{dx}^2 W_m = -m^2W_m $ :

$$\frac{d}{dx}M(x) = \int_0^\infty m^2 \rho(m^2)(W_m(x)-W_m(-x))dm^2.$$

This should be zero since the current is conserved. This then implies (take the Fourier transform)

$$p^2 \rho(-p^2)=0.$$

The only possibility is $\rho(m^2)=c \delta(m^2)$, where c is some (necessarily nonzero, since the commutator above must be nonzero) constant. We have shown that all the contributions to the 2-point function $\langle 0| J(x)\phi(0)|0\rangle$ come from intermediate states of zero mass. Thus, $L^2(O_0^+)$ is contained in the discrete spectrum of the Hilbert space as a subrepresentation of the Poincare group. in other words, the spectrum includes a massless scalar field : it's the Goldstone boson!

In conclusion, the Noether charge of a spontaneously broken symmetry (not a gauge symmetry, for which the argument does not apply) generates the Goldstone boson when acting on the noninvariant ground state.

Reference : IAS course on fields and strings vol 2, Witten's lecture 1


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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