Chiral symmetry and quark condensate

1. Dec 11, 2012

Einj

I'm studying chiral symmetry in QCD. I understand that in order for a spontaneous symmetry breaking to occur, there must be some state with a vacuum expectation value different from zero. My question is: can someone prove that is the chiral symmetry is an exact symmetry of the QCD then necessarily:

$$\langle 0 | \bar{\psi}\psi |0\rangle = 0$$

In understand that this has to be derived from the invariance of the vacuum but I can't prove it explicitely.

Thanks

2. Dec 11, 2012

fzero

It's actually easy to see that $\langle 0 | \bar{\psi}\psi |0\rangle$ is not invariant under chiral symmetry. For QCD with one flavor,

$$\psi = \begin{pmatrix} u \\ d \end{pmatrix},$$

where we can consider the up and down quarks $u,d$ as Dirac spinors. Then we can write the chiral symmetry as

$$\psi \rightarrow \exp\left[i\gamma^5 \left( \vec{\theta}\cdot\vec{\tau}\right) \right] \psi,$$

where the $\vec{\tau}$ are the generators of $SU(2)$ flavor transformations.

Some algebra will show that the kinetic term $\bar{\psi} \gamma^\mu\partial_\mu \psi$ is chiral invariant, but $\bar{\psi} \psi$ is not, because the $\gamma^5$ in the chiral transformation anticommutes with the factor of $\gamma^0$ in the Dirac conjugate.

3. Dec 12, 2012

Einj

Ok, I knew that. Actually my question was a little bit different (or maybe it's the same thing but I can't see it :tongue2:). I was asking: IF the theory is invariant under chiral symmetry (i.e. the vacuum state is invariant) how can I show that necessarily $\langle \bar{\psi}\psi\rangle=0$??

4. Dec 12, 2012

fzero

Zero is the only value that is invariant under the chiral symmetry. You can be as explicit as you like by picking a one-parameter transformation and using the $u,d$ parameterization. Show that, if $\rho = \langle \bar{\psi}\psi\rangle_0$, then $\delta\rho \neq 0$ unless $\rho =0$.