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?
[Physics] Why cannot fermions have non-zero vacuum expectation value
fermionsgrassmann-numbersquantum-field-theorysymmetry-breakingvacuum
Related Solutions
We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation:
$$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$
$$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$
Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be considered as a visible "response" of the system on the source and it usually denoted as a new variable, called the "classical field". We would like to find it when there are no external sources: $J=0$.
For that, one then does the Legendre transform trick, arriving at the effective action:
$$\Gamma[\phi_{cl}] = - E - J\phi_{cl}\quad\quad\frac{\delta\,\Gamma}{\delta \phi_{cl}} = - J$$
Remembering our goal to find $\phi_{cl}$ at $J=0$, we arrive at the equation.
$$\frac{\delta\,\Gamma}{\delta \phi_{cl}} = 0$$
Adding an extra assumption that $\phi_{cl}$ is space and time independent: $\phi_{cl}(x) = v$, the effective action functional $\Gamma[\phi_{cl}]$ is then reduced to effective potential $V_{eff}(v)$ and the equation becomes.
$$\frac{dV_{eff}}{dv} = 0$$
Now, as David Vercauteren correctly pointed out, $V_{eff}(v)$ is not the same function as $V(\phi)$. But usually it is a good first approximation, because we usually consider systems where the "real" quantum field fluctuates weakly around its vacuum: $\phi(x)=v+\eta(x)$ with $\eta$ being small.
A simple (and quite accurate) answer is that quantum fluctuations are the fluctuations that exist at zero temperature. What it means is that even at zero temperature, there might be fluctuations in the measurements of observables, which does not happen for classical systems at zero temperature, due to the non-commutativity of the dynamical and potential parts of the Hamiltonian. (Although frustrated classical systems and quantum systems with Hamiltonians described by only commutating operators slightly complicate this picture.)
Of course, at finite temperature, there will be both types of fluctuations, and how to take them apart (that is, know which part of the fluctuations are quantum or thermal) is still an active subject of research.
In a thermal QFT, one is dealing with fields $\phi(\tau,x)$ with $\tau$ the imaginary time that serves to encode the quantumness of the system (when constructing the path integral). In particular, one easily sees that if $\phi(\tau,x)$ is time-independent (but still depends on $x$), the field theory looks like a classical statistical field theory, and that's why people sometime says that this time-independent field (or the zero Matsubara frequency field) is the classical field. This, however, is not telling us which fluctuation is thermal or quantum. I don't think that is easy to tell (see above).
Concerning the phase diagram: at zero temperature, the transition is obviously driven by quantum fluctuations. A way to see that is that the frequencies are continuous (instead of discrete at finite temperature), so that there are always a large (in fact infinite) number of quantum modes that participate to the low energy physics, even when the low energy scale (say the gap) vanishes as one gets closer and closer to the transition.
For a finite temperature phase transition, there is always a value of the gap, smaller than the temperature, and therefore the quantum modes (with non-zero Matsubara frequencies) are gapped out by the temperature, and they cannot participate to the critical physics of the transition. They can be integrated out, and one is left with a classical (thermal) field theory with renormalized parameters.
Best Answer
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.