# Goldstone boson without symmetry?

1. Jul 1, 2010

### QuantumCosmo

Hi,

I was wondering about the U(1)_A problem. The Lagrangian exhibits a (in the limit of vanishing quark masses) U(1)_A symmetry but due to the chiral anomaly, the current $$J_5^{\mu}$$ is not conserved:

$$\partial_{\mu}J_5^{\mu} = G\tilde{G} + 2i\bar{u}\gamma_5 u +...$$

The $$G\tilde{G}$$ term is itself the divergence of the (not gauge invariant) current $$K^{\mu}$$.
(I have left out constant factors etc)

So in the limit of vanishing quark masses, the current $$\tilde{J}_5^{\mu} = J_5^{\mu} - K^{\mu}$$ is conserved and so is the charge

$$\tilde{Q}_5 = \int \tilde{J}_5^{\mu} \d^3$$

Now it seems that although there actually isn't a U(1)_A symmetry in my theory, I still get a Goldstone boson because $$\tilde{Q}_5$$ is conserved.

But I thought Goldstone bosons occured because of spontenously broken continous symmetries and not because of conserved charges?

Can anyone help me with that?

Thank you very much,
Quantum

2. Jul 1, 2010

### blechman

AH! But this charge is not gauge invariant!! Goldstone's Theorem as you normally see it requires gauge invariance.

In fact, you DO get poles in NON-Gauge invariant correlation functions, which Coleman refers to as "Goldstone Dipoles" in his "Aspects of Symmetry". But since these aren't gauge invariant, they don't contribute to physics (which is only described by gauge-invariant quantities).

3. Jul 1, 2010

### blechman

I just noticed this last sentence!

Of course, Goldstone bosons only come from spontaneously broken symmetries. So if you don't have that, then there are no Goldstone bosons at all.

This stuff about the Goldstone Dipole I mentioned earlier is relevant for the chiral Lagrangian, for example, to explain why there is no singlet after chiral symmetry breaking.