Mastering Supersymmetry: Solving the Algebra and Understanding Its Applications

[AlphaNumeric]

It's been bugging my for ages, but I cannot see to show the following supersymmetry algebra :

$$\delta_{\epsilon} X^{\mu} = \bar{\epsilon}\bar{\psi}$$
$$\delta_{\epsilon} \psi^{\mu} = \rho .\partial X^{\mu}\epsilon$$

Using these show that

$$[\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]X^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}X^{\mu}$$
$$[\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]\psi^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}\psi^{\mu}$$

using $$\rho . \partial \psi^{\mu}=0$$ and $$\epsilon$$ being a Grassman spinor.

I can do the first one but I cannot do the second one. Every textbook I check just says "It can be shown that..." but I can't actually show it! :uhh:

 

[Jhero]

Thank you for bringing up this interesting topic. it is important for us to not just accept results, but to understand and be able to show them as well. I would be happy to help you with the second part of the supersymmetry algebra.

Firstly, let's recall the definition of the supersymmetry transformation:

\delta_{\epsilon} X^{\mu} = \bar{\epsilon}\bar{\psi}
\delta_{\epsilon} \psi^{\mu} = \rho .\partial X^{\mu}\epsilon

Now, we can use the definition of the commutator [\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}] to show the desired result:

[\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]\psi^{\mu} &= \delta_{\epsilon_{1}}(\delta_{\epsilon_{2}} \psi^{\mu}) - \delta_{\epsilon_{2}}(\delta_{\epsilon_{1}} \psi^{\mu}) \\
&= \delta_{\epsilon_{1}}(\rho .\partial X^{\mu}\epsilon_{2}) - \delta_{\epsilon_{2}}(\rho .\partial X^{\mu}\epsilon_{1}) \\
&= \bar{\epsilon}_{1}\bar{\psi}\partial_{\alpha}(\rho .\partial X^{\mu}\epsilon_{2}) - \bar{\epsilon}_{2}\bar{\psi}\partial_{\alpha}(\rho .\partial X^{\mu}\epsilon_{1}) \\
&= \bar{\epsilon}_{1}\bar{\psi}(\partial_{\alpha}\rho .\partial X^{\mu}\epsilon_{2}) + \bar{\epsilon}_{1}\bar{\psi}(\rho .\partial_{\alpha}\partial X^{\mu}\epsilon_{2}) - \bar{\epsilon}_{2}\bar{\psi}(\partial_{\alpha}\rho .\partial X^{\mu}\epsilon_{1}) - \bar{\epsilon}_{2}\bar{\psi}(\rho .\partial_{\alpha}\partial X^{\mu}\epsilon_{1}) \\
&= \bar{\epsilon}_{1}\bar{\psi}\rho^{\beta}\partial_{\beta}\partial_{\alpha}X^{\mu}\epsilon