Chiral symmetry and quark condensate

[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

 

[fzero]

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

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.

 

[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 ). 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??

 

[fzero]

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 ). 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??

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.