# why cannot fermions have non-zero vacuum expectation value?

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?

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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.

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Hi @innisfree, thanks for your answer. But, it will be very helpful, if you could explain it a bit more mathematically? – Paul Apr 12 '14 at 18:11
Take the fermion field to have spin 1/2. A spin 1/2 defines a direction in space, the direction such that the spin 1/2 is "up" with respect to that direction. If this field interacts with other fields, this breaks rotational symmetry. The only spin that does not define a preferred direction in this way is spin 0, that is, a scalar. – Robin Ekman Apr 12 '14 at 18:14
@Paul, is it clearer now? – innisfree Apr 12 '14 at 18:20
Having said that, I suppose there could be very special theories, in which e.g. vectors appeared always as $\partial_\mu A^\mu$, allowing a VEV. – innisfree Apr 12 '14 at 18:24
@Paul: VEV = Vacuum expectation value i.e. a property of the vacuum. So if any object in a non-trivial representation of the Poincare algebra picks up a VEV, then some of the spacetime symmetries will be spontaneously broken by the vacuum (state). One can expect the same to apply to fluctuations around the vacuum. That would mean that the corresponding conserved quantities are not really conserved. And as far as we can see, conservation of energy-momentum and angular momentum apply quite perfectly to our universe. – Siva Apr 12 '14 at 18:43

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).

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Can't one make similar arguments for a scalar vev $<\phi>=0$? It ought to be zero, which is why we expand about the homogeneous nonzero part in Higgs mechanisn – innisfree Apr 12 '14 at 20:42
Any field linear in creation and annihilation operators will give a zero vev – innisfree Apr 12 '14 at 20:44
@innisfree: Not necessarily. If the action is not symmetric (say to $\phi\to-\phi$, where $\phi$ is bosonic), then you can have $\langle\phi\rangle\neq 0$. For creation operator, this is not the case, if you include a source $J$, compute $\langle\hat a^\dagger\rangle$ at finite source, and the let $J\to0$ (and if there is a degeneracy between two states with a difference of one particle). For fermions, you will always find that as $\eta\to0$, then $\langle\psi\rangle\to0$. – Adam Apr 12 '14 at 22:46
@Adam I think your argument is incorrect. Think of the bosonic example that innisfree has mentioned with the $Z_2$ symmetry $\phi\rightarrow-\phi$, with your argument you would conclude that $\langle\phi\rangle=0$ which is wrong as spontaneous symmetry breaking of $Z_2$ may in fact happen. Your error is simply setting the sources J for the fermion to zero, but when there is spontaneous symm breaking the result depends on the way you take the limit as the action is non-analytic in J. Moreover, in my answer I have provided an explicit weakly coupled example that does break Lorentz – TwoBs Apr 24 at 7:13
@TwoBs: SSB won't change anything, since already for explicit symmetry breaking (for a condensate $\bar\psi\psi$) won't give you a finite $\langle\psi\rangle$, because Grassman variables are completely different from bosonic variables. Stated in a different way: I challenge you to find a reference talking about $\langle\psi\rangle\neq 0$ (without linear sources). – Adam Apr 24 at 8:15

Well, they could develop a vev, depending of the interactions you consider. Take e.g. this term in the lagrangian $L=\frac{1}{\Lambda^2}[\bar{\psi}\psi- v^3]^2$. Of course this would imply that you also break Lorentz, which you may want to avoid.

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Won't this give a fermion condensate? That won't break Lorentz! – innisfree Apr 22 at 13:22
@innisfree let me borrow an example from the Higgs boson. Doesn't $L=\lambda(|H|^2-v^2)^2$ break the EW sym. even though $|H|^2$ is a singlet? Only singlets can appear in $L$ but a non vanishing $|H|^2$ implies a non-vanishing doublet $H$ which breaks the sym. Back to the $\psi$: a vev for $\bar{\psi}\psi$ implies a non-vanishing vev for $\psi$, which breaks Lorentz. There is an exception, that is when the theory for $\psi$ is strongly coupled at the scale $v$: in this case they form a strongly coupled bound state and it is thus meaningless to talk of $\psi$ and $\bar{\psi}$ separately. – TwoBs Apr 23 at 16:59
Non-zero $<x^2>$ doesn't imply non-zero $<x>$. – innisfree Apr 23 at 17:04
@innisfree As I have argued above with a well known example, in weakly coupled models it does imply so. (Otherwise, we have a problem with the Higgs sector of the SM). – TwoBs Apr 23 at 17:06
Hmm I think I agree that this won't give a condensate unless strongly coupled. But still, nonzero $<x^2>$ never implies non-zero $<x>$ – innisfree Apr 23 at 17:17