Susy transformation for gauge multiplet

Question

How can the supersymmetrie transformation $\delta A_\mu = \frac{1}{2} \overline{\epsilon}\gamma_\mu \psi $ be derived from the susy algebra ( or group ). Where $ (A_\mu , \psi)$ are in a gauge multiplet and $\overline{\epsilon}$ is the susy parameter.

Under which representation transforms $A_\mu$ under the susy algebra ? Under Lorentz it would be in the vector repr. so $ A_\mu^\prime \rightarrow g A_\mu$ and g can be expand to find $\delta A_\mu$. Is something like this possible for the supersymmetrie transformation ? And what is the group element g?

Answer 1

There is a long and formal way, and also an easy and dirty way. I will tell you the easy option. The algebra tells you that

$[\delta_Q (\epsilon_1), \delta_Q (\epsilon_2)] = \delta_{P}(\xi^\mu_3)$

where $\epsilon$ is your SUSY parameter and $\xi^\mu_3 = \bar\epsilon_1 \gamma^\mu \epsilon_2$ is your translation parameter. Now, the only Lorentz vector that you can form with $\epsilon$ and $\psi$ is $\bar\epsilon \gamma_\mu \psi$ thus you must have

$\delta A_\mu = \bar\epsilon \gamma_\mu \psi$

Note that the overall factor is not important as you can always rescale your fields to get the above transformation. Next, the commutator of Q's, as shown above, tells you that if you act on $\delta A_\mu$ with another $\delta$ it must give rise to the field strength of $A_\mu$, which implies that

$\delta \psi = a_1 \gamma^{\mu\nu} F_{\mu\nu} \epsilon$

where $a_1$ is an undetermined coefficient, which can be fixed by exact implementation of the algebra, and $F_{\mu\nu} = 2 \partial_{[\mu} A_{\nu]}$.

In the transformation rule above, you have only given the SUSY transformation rule. $A_\mu$ transforms under the full super-Poincare group, which includes translations (P), Lorentz $(M_{ab})$ as well as $Q$.