- #1
AlphaNumeric
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- 0
It's been bugging my for ages, but I cannot see to show the following supersymmetry algebra :
[tex]\delta_{\epsilon} X^{\mu} = \bar{\epsilon}\bar{\psi}[/tex]
[tex]\delta_{\epsilon} \psi^{\mu} = \rho .\partial X^{\mu}\epsilon[/tex]
Using these show that
[tex][\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]X^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}X^{\mu}[/tex]
[tex][\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]\psi^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}\psi^{\mu}[/tex]
using [tex]\rho . \partial \psi^{\mu}=0[/tex] and [tex]\epsilon[/tex] 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:
[tex]\delta_{\epsilon} X^{\mu} = \bar{\epsilon}\bar{\psi}[/tex]
[tex]\delta_{\epsilon} \psi^{\mu} = \rho .\partial X^{\mu}\epsilon[/tex]
Using these show that
[tex][\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]X^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}X^{\mu}[/tex]
[tex][\delta_{\epsilon_{1}},\delta_{\epsilon_{2}}]\psi^{\mu} = 2\bar{\epsilon}_{1}\rho^{\alpha}\epsilon_{2}\partial_{\alpha}\psi^{\mu}[/tex]
using [tex]\rho . \partial \psi^{\mu}=0[/tex] and [tex]\epsilon[/tex] 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: