# Supersymmetry notation question

1. Oct 26, 2008

### StatusX

In super yang mills theories in 4,6,10 dimensions, the supersymmetry transformation is often written as (ignoring color indices):

$$\delta A_\mu = i \bar \alpha \gamma_\mu \lambda - i \bar \lambda \gamma_\mu \alpha$$

$$\delta \lambda = c F^{\mu \nu} [ \gamma_\mu, \gamma_\nu] \alpha$$

where c is some constant depending on dimension, and $\alpha$ is the parameter of the transformation, a fermionic c-number spinor.

I have a few questions about this. First of all, are we supposed to assume $\alpha$ has the same properties as $\lambda$, ie, wrt majorana and weyl -ness? It seems like we should to get the right number of supercharges, and maybe to preserve the corresponding property of $\lambda$ after a transformation, but this is never mentioned.

Second, what exactly do these transformations mean in terms of the susy generators Q? Do these generators fit into a spinor with the same properties as $\alpha$ and $\lambda$? If so, and if we call this spinor Q, can we write:

$$\delta \mathcal{O} = [ \bar \alpha Q, \mathcal{O} ]$$

This doesn't seem right, because if $\alpha$ is Weyl, the RHS doesn't depend on the conjugates of Q. Maybe it's something like:

$$\delta \mathcal{O} = [ \bar \alpha Q + \bar Q \alpha, \mathcal{O} ]$$

Is this right? And in any case, what would be the easiest way to determine the anticommutators of the Q's given the transformations in the form of the first equations above?

2. Oct 26, 2008

### nrqed

What I have seen is this, instead:
$$\delta \mathcal{O} = [ \alpha Q + \bar Q \bar \alpha, \mathcal{O} ]$$

There are two ways to determine the anticommutation rules: finding the supercurrent generating the transformation, getting the explicit charges (as the integral of the zeroth component of j^mu ) and then calculating the explicit anticommutators.

The other way is to calculate the commutator of two variations (each with a different parameter) on the fields $$(\delta_\alpha \delta_\beta - \delta_\beta \delta_\alpha) A_\mu$$ and then you do the same thing using the supercharges, and then set the two results equal to one another.

3. Oct 30, 2008

### BenTheMan

The $$\alpha$$ is there to get the indices right. As to your question about "majorana-ness" or "weyl-ness", the infinitessimal $$\alpha$$ transforms as a spinor of the Lorentz group as your $$\lambda$$, so in that sense, yes. Again, you can see this by just putting in the spinor indices.