Why a field theory containing only fermions does not show spontaneous symmetry breaking?

Question

For a real scalar field $\phi$, a theory as simple as the $\phi^4$ theory, can exhibit the phenomenon of Spontaneous Symmetry Breaking (SSB). For a complex scalar field $\phi$, a theory as simple as the one described by a potential of the form $V(\phi^*\phi)$ is capable of showing SSB. Why is it that a simple quantum field theory containing only a fermion field $\Psi$ incapable of showing this feature of SSB?

Added after @Dwagg's answer I was a bit sloppy in saying that theories with only fermions do not admit SSB. In QCD, there is chiral symmetry breaking which is related to a nonzero VEV of the operator $\bar{q}q$ for a quark field $q$. But the route is not straightforward.

In classical field theories, the standard way to check whether SSB has occurred is the following. One has a Lagrangian with a potential term $V(\phi)$. When minimized, the minimum of $V(\phi)$ occurs at a nonzero value of $\phi$. In QFT, one says that $\phi$ has got a VEV.

I have not seen such a straightforward route to understand QCD chiral symmetry breaking i.e., no such potential in the QCD Lagrangian. So the question is

Is there a similar way of showing that a nonzero VEV of $\bar{q}q$ is possible by minimizing some "effective potential" $V(\bar{q}q)$ of QCD.?

For the Gross-Neveu model in Dwagg's answer there is already a quartic term in the Lagrangian, and minimizing the potential w.r.t $\bar{\psi}_a\psi_a$ can give rise to SSB. Such a term is not there in QCD Lagrangian.

Answer 1

First, as already discussed, there are several (very similar) purely fermionic model field theories that exhibit spontaneous symmetry breaking, the models of Gross-Neveu, Thirring, and Nambu-Jona-Lasionio. These theories contain a fundamental four-fermion interaction ${\cal L}\sim G(\bar\psi\psi)^2$.

Second, there are interesting and important condensed matter systems that exhibit spontaneous symmetry breaking with a fermion-bilinear order parameter, BCS superconductivity in metals, neutron matter, liquid helium 3, and atomic gases. Fundamentally, these are not purely fermionic theories, but they can often be reduced to effective four-fermion theories. This is already a hint that a fundamental four-fermion interaction is not essential; such an interaction can always appear from integrating out other degrees of freedom.

In QCD we can define an effective potential for $\langle\bar\psi\psi\rangle$ in the usual way. Couple an external field to the order parameter (QCD already has such a term, the mass term) and compute the partition function. Then Legendre transform to get the effective action as a function of the order parameter. The static part is the effective potential.

Chiral symmetry breaking takes place at strong coupling, so we cannot compute the effective potential reliably (except in certain limiting cases), but we can identify the diagrams that contribute to it. The simplest is a fermion loop with a gluon going across. If the gluon was heavy (which it is not), we would be able to contract the gluon propagator to a point, and this diagram would be the same four-fermion closed-off-by-two-loops diagram that appears in the Gross-Neveu model (which is why, historically, Gross and Neveu studied it).

Answer 2

The typical method for SSB for fermions is the formation of bosonic bound states, such as Cooper pairs, which can then have an effective potential which looks like Landau-Ginzburg. There is no problem with, eg. $\langle \bar \psi(x) \psi(x) \rangle \neq 0$ while $\langle \psi(x) \rangle = 0$. There are also interesting systems in condensed matter physics such as superfluid He-3 which have order parameters with angular momentum such as $\langle \bar \psi \gamma^\mu \psi \rangle$.

Answer 3

Check out the Gross-Neveu model, which involves only fermions. The chiral symmetry is spontaneously broken by a composite field of fermions.