# Trivial SUSY question

1. May 3, 2015

### the_pulp

Im reading Adel Bidal Introduction to SUSY. It is all pretty much clearly deduced from the asumptions, but I dont understand the asumptions... In particular, Im talking about the Qs commutation:
1 with Ps: why are they assumed to be 0.
2 with Ms: why are the proportional of Qs but not of Ms?
3 the sigmas: I sort of see why are the sigma mattrixes in the commutation relations but I cannot put it in words, so I will ask it anyway... Why are the sigmas there and not another mattrices?

Thanks in advance and sorry for my english!

2. May 3, 2015

### fzero

SUSY is introduced explicitly as a symmetry that takes bosons to fermions and vice versa and works within the frame work of quantum field theory. Therefore our SUSY generator $Q$ is supposed to take a boson (say represented by a scalar field $\phi$) into a fermion (say spin 1/2 $\psi$). So morally $\psi_\alpha = Q_\alpha \phi$. Now we know that fermions and bosons are different. In particular we have the spin-statistics theorem that says fermions must have wavefunctions that are odd under interchange of fermions at the same points. In QFT this is reflected in the convention that fermion field operators anticommute: $\psi_1\psi_2 = - \psi_2 \psi_1$. We say that fermion fields are "odd", while boson fields are "even". This means that the SUSY generators $Q$ must also be odd, while the Poincare generators are even. These properties lead to restrictions on what terms can appear in the SUSY algebra.

Consider $[P_\mu,Q_\alpha]$. The whole point of an algebra is to rewrite this as a linear combination of operators. Since $P$s are even and $Q$s are odd, ionly odd operators can appear on the RHS if it is nonzero. However, let's think about how this combination would act on a field

$$[P_\mu,Q_\alpha] \phi = P_\mu Q_\alpha \phi - Q_\alpha P_\mu \phi .$$

In the first term, we do the SUSY transformation and then an infinitesimal translation. The 2nd term just has the order reversed. We take it as a definition of the SUSY transformation that this order doesn't matter. Therefore this commutator should vanish.

Here we have $[M_{\mu\nu},Q_\alpha]$. Again, the RHS must be odd, so it has to be a linear combination of $Q$s only.

Again $[M_{\mu\nu},Q_\alpha]$. Here we're doing a SUSY transformation and then a Lorentz transformation or vice-versa. If we act on a scalar field, it's clear that we only get a contribution from the term where the LT acts on the $Q_\alpha$, which transforms as a spin 1/2 fermion. The $\sigma$ matrices are precisely the matrices that appear in the spin 1/2 representation of the generators of the Lorentz group, so they must appear in this formula.

One can verify that there are no other terms by demanding that the algebra closes. That is, we verify that this commutator is consistent with the other relations in the algebra.

3. May 3, 2015

### the_pulp

Thanks fzero and thanks everyone else around here that make this forum such an incredible source of knowledge!!!!!! I think I get ehat you are saying.

4. May 4, 2015

### haushofer

The fact that [Q,P] = 0 means that SUSY is a global symmetry. In Supergravity, where SUSY becomes local, one trades the P's for gct',s and then this commutator changes. This also holds for {Q,Q} (the algebra becomes "soft"). However, [M,Q] doesn't change because the Q's are still fermionic; [M,Q] ~ Q means that Q transforms as a spinor ( [M,Q] is the adjoint action of M on Q ). Hope this helps :)