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.
innisfree
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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).
Adam
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1Can'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
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1Any field linear in creation and annihilation operators will give a zero vev – innisfree Apr 12 '14 at 20:44
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@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
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2@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 '16 at 07:13
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@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 '16 at 08:15
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@Adam I have run your very same argument for the bosonic $Z_2$ example, where it fails. Do you agree on this? If so, why should I believe in your argument for the fermionic example? Besides, I have already provided a weakly coupled example where $\langle\psi\rangle$ arise with no source. Not to mention the continuum limit of a lattice with the spin locked to each side, that provide a model for spontaneous symmetry breaking of rotations where a finite magnetization is induced even without sources. – TwoBs Apr 24 '16 at 08:57
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@TwoBs: of course your argument works for bosons, never said otherwise. Your "example" is not one, since you haven't done any calculation showing it works. Because it won't. And your example with magnetization doesn't work either, since spins are quadratic in fermionic operators... which of course can have a finite vev. – Adam Apr 24 '16 at 09:28
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@adam I'm quite interested in this. If you have time, maybe you could settle the matter by extending your answer with the technical details about the differences between scalar and fermion VEVs, and why twobs potential won't break Lorentz – innisfree Apr 24 '16 at 09:53
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@innisfree: put it that way: a fermionic annihilation operator $\hat c$ is a standard QM operator. Therefore, the average of $\hat c$ with respect to any physical state should give a complex number. Thus having $\langle \hat c\rangle =\psi$ with $\psi$ a grassmann number is impossible. It is in fact trivial to check that $\langle \hat c\rangle=0$ for any physical state (that is, $a|0\rangle+b|1\rangle$ with $a,b$ complex number. Stated otherwise, fermionic coherent state are not physical states, just very useful to write down a path integral. – Adam Apr 24 '16 at 16:08
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@Adam I agree that I haven't done the calculation for the example, I will try to be more explicit as soon as I get some spare time in the next few days. I also agree that the spin operators are actually bosonic, sorry for the confusion. But I insist that $\langle\psi\rangle$ can be non-zero (allowing breaking of Lorentz). For example, should somebody hand you the following non-analytic piece of functional generator $\delta Z(\eta,\bar{\eta})=\sqrt{\bar{\eta}\eta}$ that is meant to produce SSB, wouldn't you agree on a $\langle\psi\rangle\neq0$? (the $\eta$ and $\bar{\eta}$ are the sources) – TwoBs Apr 24 '16 at 21:02
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1@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) – TwoBs Apr 24 '16 at 21:29
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@TwoBs: in fact, I'm not that happy with my argument, since one can find a density matrix that allows a non vanishing expectation value. But of course, it won't be a grassmann number, whereas a mean-field approximation of the action would generate that. But I think that your $\delta Z$ is not possible, because when you expand the exponential of the sources, there is only one term, $\bar\eta \psi$, so there is no way to generate a square root (which would do the job, indeed). – Adam Apr 25 '16 at 06:17
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@Adam ok, I think I am convinced that you are right (but maybe with tiny loophole at the end). I just realized that one can actually never build non-analytic functions of grassmann constant variables fourth power would vanish. They are bound to be finite polynomial of order 4, hence analytic, which rules out the spontaneous breaking. The (maybe) loophole, I think, would be departing from constant sources and vevs, that is breaks also translations. In this way I think expect one should be able to build non- analytic function(al)s of the sources and thus construct a fermionic vev. Thanks Adam! – TwoBs Apr 25 '16 at 07:21
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@TwoBs: no problem, it was useful to me too, and forced to organize my thoughts ! Though I don't think having non constant sources would change anything. The partition function will be analytical in all the $\eta(x)$ by the same argument I think. – Adam Apr 25 '16 at 07:34
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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.
Tom Kim
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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?
– QGravity Sep 21 '17 at 23:22