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.

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    $\begingroup$ What makes you think that fermions can’t cause SSB? $\endgroup$
    – knzhou
    Commented Mar 31, 2019 at 14:49
  • $\begingroup$ @knzhou I have not seen any textbook example where a theory consisting of fermions only lead to SSB. In QCD, there is chiral symmetry breaking which is related to a nonzero VEV of the operator $\bar{q}q$ for a quarks field $q$. But the route is not straightforward. $\endgroup$
    – SRS
    Commented Mar 31, 2019 at 14:52
  • $\begingroup$ @SRS You've provided an example of SSB with a theory with only fermions, so what's your question? $\endgroup$ Commented Mar 31, 2019 at 17:24
  • $\begingroup$ @SRS As soon as you don't want to break Lorentz, a spinor cannot take vacuum expectation value. I think this is an underlying requirement in your case. $\endgroup$
    – apt45
    Commented Mar 31, 2019 at 21:16
  • $\begingroup$ @Runlikehell Please see the newly added part of the question. $\endgroup$
    – SRS
    Commented Apr 1, 2019 at 6:03

3 Answers 3


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).


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 $\langle \bar \psi \gamma^\mu \psi \rangle$.

  • $\begingroup$ My concern is how would such a term be induced in some effective QCD potential. @RyanThorngren $\endgroup$
    – SRS
    Commented Apr 3, 2019 at 8:09
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    $\begingroup$ @SRS I dunno, but in general I expect that any term allowed by symmetry can and does appear. $\endgroup$ Commented Apr 3, 2019 at 8:37
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    $\begingroup$ Nice point, I would also mention the topcolor models modeling the Higgs as $\left<\bar{t} t\right>$. $\endgroup$ Commented Apr 4, 2019 at 1:36

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

  • $\begingroup$ Thanks for your answer. Can you illuminate more about the newly added part of the question? @Dwagg $\endgroup$
    – SRS
    Commented Apr 1, 2019 at 6:06

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