Why cannot fermions have non-zero vacuum expectation value?

Question

In quantum field theory, scalar can take non-zero vacuum expectation value (vev). And this way they break symmetry of the Lagrangian. Now my question is what will happen if the fermions in the theory take non-zero vacuum expectation value? What forbids fermions to take vevs?

Answer 1

Why can't fermions have a non-zero vacuum expectation value (VEV)? Lorentz invariance.

If anything other than a Lorentz scalar has a non-zero VEV, Lorentz invariance would be spontaneously broken.

For example, suppose we have a Lorentz invariant term in a Lagrangian for a vector $$ \mathcal{L} \supset m^2 A_\mu A^\mu. $$ Now suppose the vector obtains a VEV, $A_\mu \to v + A_\mu$, $$ m^2 A_\mu A^\mu \to m^2 v A^\mu + m^2 vA_\mu + m^2v^2 + m^2 A_\mu A^\mu. $$ The first two are clearly not Lorentz invariant. One can construct idential arguments for any non-scalar field term. If $\psi\to v+\psi$, the VEV, $v$, won't have the same Lorentz transformation properties as the field, $\psi$ unless $\psi$ is a scalar.

Answer 2

I think it is a general fact about grassmannian field, and this has nothing to do with Lorentz invariance or other symmetries (you can invent a lot of QFTs without this kind of symmetry, but the VEV of a fermionic operator will be always zero (in the absence of sources)).

In a functional integral formulation, the VEV of a grassmannian field $\psi$ is written as $$ \langle \psi \rangle= \int D\psi D\bar\psi\, \psi \,e^{-S},$$ where the action S is bosonic (involves products even products of $\psi$ and $\bar\psi$). Therefore, unless there are source terms of the form $\bar\eta\psi$ in the action, the integral over the $\psi e^{-S}$ will give zero, since we are integrating over an odd number of grassmannian fields (when the exponential is expanded).

Answer 3

Can't we just evaluate the partition function and then find the expectation value? $Z=\int \mathcal{D}\phi \mathcal{D}\phi ^* \exp \left [ \int \left ( -\phi^* G^{-1}\phi +\phi j^*+j \phi ^* \right ) \right ]=\det\left(G^{\mp 1}\right) \exp \left ( \int j^* G j \right )$,

where $\mp$ corresponds to bosons and fermions respectively. Summing over the Matsubara frequencies of $j^* G j$,

$\sum _{\omega } j^*(\omega ,k) G(\omega ,k) j(\omega ,k) =\frac{1}{(2 \pi )^3} \sum _{\omega } \frac{j(\omega ,k) j^*(\omega ,k)}{\frac{k^2}{2 m}-\mu -i \omega } = \pm j^*(k) \frac{1}{(2 \pi )^3} \left(\frac{1}{\exp \left(\beta \left(\frac{k^2}{2 m}-\mu \right)\right)\mp 1}\pm \frac{1}{2}\right) j(k)$.

$\Rightarrow \langle \phi(k) \rangle= \pm \left[\frac{\delta \log Z}{\delta j^*(k)}\right]_{j^*,j=0} = \left[ \frac{1}{(2 \pi )^3} \left(\frac{1}{\exp \left(\beta \left(\frac{k^2}{2 m}-\mu \right)\right)\mp 1}\pm \frac{1}{2}\right) j(k)\right]_{j^*,j=0}$.

The Bose function has singularity at $\frac{k^2}{2m}=\mu$, and we encounter $\frac{0}{0}$. The Fermi function has no singuarity, and we get the expectation value strictly zero.