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?


3 Answers 3


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.

  • $\begingroup$ Hi @innisfree, thanks for your answer. But, it will be very helpful, if you could explain it a bit more mathematically? $\endgroup$
    – Paul
    Apr 12, 2014 at 18:11
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    $\begingroup$ 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. $\endgroup$ Apr 12, 2014 at 18:14
  • $\begingroup$ @Paul, is it clearer now? $\endgroup$
    – innisfree
    Apr 12, 2014 at 18:20
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    $\begingroup$ @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. $\endgroup$
    – Siva
    Apr 12, 2014 at 18:43
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    $\begingroup$ @innisfree Shouldn't we take $\langle\Omega|A_{\mu}|\Omega\rangle=v_{\mu}$? It can not be just $v$. The Lagrangian obviously cannot contain a term like $m^2 vA_{\mu}$, it is a four-vector. However, if you consider $v_{\mu}$, the expression you wrote can be written in a Lorentz-invariant way and no problem would arise. My question is more general. What is the VEV of a spin-$s$ (integer or half-integer) field in a general quantum field theory? To which extent one can answer this question? $\endgroup$
    – QGravity
    Sep 21, 2017 at 23:22

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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    $\begingroup$ 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 $\endgroup$
    – innisfree
    Apr 12, 2014 at 20:42
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    $\begingroup$ Any field linear in creation and annihilation operators will give a zero vev $\endgroup$
    – innisfree
    Apr 12, 2014 at 20:44
  • $\begingroup$ @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 $\langle\psi\rangle\to0$. $\endgroup$
    – Adam
    Apr 12, 2014 at 22:46
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    $\begingroup$ @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 $\endgroup$
    – TwoBs
    Apr 24, 2016 at 7:13
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    $\begingroup$ @Adam let me add a comment on your reply to innisfree. I think that your argument there, which I think is correct, doesn't imply though a vanishing Vev, but rather that fermions and their grassmann vevs aren't physical observable. I would agree with that, but not on the vanishing of the fermionic Vevs. That doesn't mean they have no effect (think again to the Higgs boson which is charged under a gauge symmetry, which isn't physical, and yet its 'breaking' by the Higgs vev has physical consquencies) $\endgroup$
    – TwoBs
    Apr 24, 2016 at 21:29

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.


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