Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I would be glad if someone can explain to me the argument as to why supersymmetry breaking is necessarily accompanied by appearance of a massless fermion, namely the goldstino. (and also why this is a non-perturbative effect)

Let me quote here the two lines from the third volume of the QFT books by Weinberg where he tries to explain this phenomenon,

  • "(when supersymmetry is broken)..any n-particle state is accompanied with 2 states of the same energy and momentum and opposite statistics, containing an additional 0-momentum goldstino of spin up or down and with another state of the same energy and momentum and the same statistics, containing two additional 0-momentum goldstinos of opposite spin."

  • "In particular when supersymmetry is spontaneously broken the vacuum state has non-zero energy, so it must be paired with a fermionic state of the same energy and zero-momentum; more precisely, the vacuum and the state containing two zero-momentum goldstinos are paired with the two states of a single zero-momentum goldstino."

I am unable to understand the above two arguments and will be glad if someone can help.

share|improve this question

1 Answer 1

The propositions by Weinberg above are just tautologies. When SUSY is broken it means that $$ Q_\alpha|0\rangle \neq 0$$ the vacuum is not annihilated by the supercharges. It means that $Q_\alpha |0\rangle$ has to be equal to something. Because $Q_\alpha$ is Grassmann-odd, it must be a state with an odd number of fermions. But the momentum is still zero because the momentum of $|0\rangle$ vanishes and $Q_\alpha$ commutes with the energy-momentum vector.

So for broken $N=1$ SUSY, one has four states, $$|0\rangle, Q_1 |0\rangle, Q_2 |0\rangle, Q_1 Q_2 |0\rangle $$ which contain 0,1,1,2 fermions, respectively. This proof is somewhat heuristic because we haven't really proved that there also exist states with a nonzero momentum goldstino. To prove that, one has to define the supergenerators dressed with a momentum. The proof is then totally analogous to the Goldstone theorem for bosonic symmetries.

http://en.wikipedia.org/wiki/Goldstone_theorem#Goldstone.27s_theorem

share|improve this answer
    
Thanks for your reply. Can you kindly tell me in your notation which is the "n-particle" state that Weinberg refers to? You are showing the pairings for the vacuum state but Weinberg seems to arguing for such a pairing for any n-particle state. Kindly explain what I am confusing! –  user6818 Feb 22 '11 at 11:28
    
@Anirbit: the n-particle state Weinberg is using replaces the vacuum in Lubos's answer. The argument is exactly the same, and shows that you can always scatter a soft goldstino from the matter, and derive its scattering from the supersymmetry transformations. Weinberg pioneered these methods for soft-pion scattering in the 1960s. –  Ron Maimon Sep 18 '11 at 9:15
    
The reason Weinberg considers the vacuum paired with two goldstinos is he wants to show you an object closed to supersymmetry transformations, with equal numbers of bosonic and fermionic states. The vacuum and the 1 goldstino states by themselves are not enough to close the algebra. You need a two goldstino states too. As far as being "non-perturbative", every spontaneous symmetry breaking involves a field with a VEV, which is infinitely many particles in a Feynman diagram description, so it is beyond perturbation theory. This is not a serious non-perturbativeness. –  Ron Maimon Sep 18 '11 at 9:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.