Quark condensate and spontaneous symmetry breaking?

Question

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?

Answer 1

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.