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:(adsbygoogle = window.adsbygoogle || []).push({});

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

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

Thanks

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# Chiral symmetry and quark condensate

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