# Quark condensate and spontaneous symmetry breaking?

It is known the quark condensate $$<\bar{\psi}^{i}_L\psi^j_R>=\sigma \delta^{ij}$$($$i,j$$ are flavour indices ) breaks the symmetry group $$SU(N_f)_L\times SU(N_f)_R$$. Because it is only invariant under diagonal subgroup of $$SU(N_f)_L\times SU(N_f)_R$$. This kind of breaking is generally identified as spontaneous symmetry breaking(SSB).

Here is my confusion: Accroding to some QFT textbooks, SSB in a theory is caused by non-vannishing vacuum expectation value(VEV) of a field $$<\Phi>\neq 0$$. That is, if we want to investigate SSB, we need to consider VEV of just one field, and its non-vanshing value will give SSB. Now if we look at $$<\bar{\psi}^{i}_L\psi^j_R>$$, there are two fields inside ket and bra. So it seems the condition $$<\bar{\psi}^{i}_L\psi^j_R>=\sigma \delta^{ij}$$ does not meet the requirment of SSB? Why is it still called SSB?

• $\psi_i\psi_j$ is two fields, but it is also one field. It is a composite, sure, but still one field. – AccidentalFourierTransform Oct 18 '20 at 22:38

## 1 Answer

Define $$\Psi_{pair}^i=\bar{\psi}_L^i \psi_R^i$$.

Quarks are fermions, so unable to form a BEC. Unless they pair up into bosonic quasiparticles, that is. A similar story is present in superconductors or fermionic atomic condensates, where cooper pairing must occur before condensation.

$$\Psi^i_{pair}$$ is, unlike $$\psi_{L/R}^i$$ an appropriate order parameter.