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

+ 4 like - 0 dislike
1395 views

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?

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Paul
The Higgs has a non-zero vev, but there's no reason to suppose this applies to other scalars. It's a special property of the Higgs.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user John Rennie

I wished someone could explain this in a language more comfortable for laymen. Isn't that possible? Or are physicists just conditioned to think this way?

+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user innisfree
answered Apr 12, 2014 by (295 points)
Hi @innisfree, thanks for your answer. But, it will be very helpful, if you could explain it a bit more mathematically?

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Paul
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.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Robin Ekman
@Paul, is it clearer now?

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user innisfree
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.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user innisfree
@innisfree: I don't think this has to do with Lorentz invariance, but that's it is a general fact about Grassmanian fields, though I don't know the proof...

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Adam
@innisfree: In any "derivatively-coupled" theory, one cannot have a vev, right? Since there is no special point in field space.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Siva
@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.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Siva
+ 2 like - 0 dislike

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).

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Adam
answered Apr 12, 2014 by (115 points)
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

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user innisfree
Any field linear in creation and annihilation operators will give a zero vev

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user innisfree
@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 $\langle\psi\rangle\to0$.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Adam

_Any_ expectation value of a fermion field vanishes, not only the VEV.

+ 1 like - 0 dislike

In quantum field theory, the deviations of the field operators from the vacuum expectation value must be linear combinations of creation and annihilation operators satisfying, like the fields themselves, canonical commutation or anticommutation relations. The fields therefore have the form VEV + such a linear combination.

Working out commutation relations gives no restriction on the VEV, but working out anticommutation relations leads to a contradicition unlesc the VEV vanishes. Therefore bosonic fields may have nonvanishing VEV, but for fermionic fields the VEV must vanish.

The same holds in any other state for the expectation values; only the creation and annihilation operators change, being related by a Bogoliubov transformation to those of the vacuum.

answered Oct 29, 2020 by (15,488 points)
edited Oct 30, 2020
+ 0 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user TwoBs
answered Apr 12, 2014 by (315 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.