- #1
karlzr
- 129
- 2
I tried to verify that the SYM lagrangian is invariant under SUSY transformation, but it turned out there is a term that doesn't vanish.
The SYM lagrangian is:
\mathscr{L}_{SYM}=-\frac{1}{4}F^{a\mu\nu}F^a_{\mu\nu}+i\lambda^{\dagger a}\bar{\sigma}^\mu D_\mu \lambda^a+\frac{1}{2}D^a D^a
the infinitesimal SUSY transformations of the gauge multiplet are:
\delta A^a_\mu =-\frac{1}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+\lambda^{\dagger a} \bar{\sigma}_\mu \epsilon)=-\frac{1}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+h.c.
\delta\lambda^a_\alpha =-\frac{i}{2\sqrt{2}}(\sigma^\mu \bar{\sigma}^\nu\epsilon)_\alpha F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon_\alpha D^a
\delta \lambda^{\dagger a}_{\dot{\alpha}} =\frac{i}{2\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\nu\sigma^\mu)_{\dot{\alpha}} F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon^\dagger_{\dot{\alpha}} D^a
\delta D^a =\frac{-i}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a-D_\mu \lambda^{\dagger a}\bar{\sigma}^\mu \epsilon)=\frac{-i}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a+h.c.
I derived that
\delta F^a_{\mu\nu}= -\frac{1}{\sqrt{2}}\epsilon^\dagger(\bar{\sigma}_\nu D_\mu-\bar{\sigma}_\mu D_\nu)\lambda^a+h.c.
Then when I calculated the variation of the total lagrangian, I obtained that the D^a and F^a_{\mu\nu} terms cancel out separately. But there is a term from the variation of A^b_\mu in the definition of covariant derivative in D_\mu \lambda^a that remains after all the other cancellation. It looks like
i g f^{abc}\lambda^{\dagger a}\bar{\sigma}^\nu \lambda^c (\epsilon^\dagger \bar{\sigma}_\nu \lambda^b+h.c.)
This is the only term with three \lambdas. So it seems there is no term to cancel this one. I have checked my calculation over and over again but couldn't find the mistake. So any idea what 's wrong?
Thanks a lot,
The SYM lagrangian is:
\mathscr{L}_{SYM}=-\frac{1}{4}F^{a\mu\nu}F^a_{\mu\nu}+i\lambda^{\dagger a}\bar{\sigma}^\mu D_\mu \lambda^a+\frac{1}{2}D^a D^a
the infinitesimal SUSY transformations of the gauge multiplet are:
\delta A^a_\mu =-\frac{1}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+\lambda^{\dagger a} \bar{\sigma}_\mu \epsilon)=-\frac{1}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+h.c.
\delta\lambda^a_\alpha =-\frac{i}{2\sqrt{2}}(\sigma^\mu \bar{\sigma}^\nu\epsilon)_\alpha F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon_\alpha D^a
\delta \lambda^{\dagger a}_{\dot{\alpha}} =\frac{i}{2\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\nu\sigma^\mu)_{\dot{\alpha}} F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon^\dagger_{\dot{\alpha}} D^a
\delta D^a =\frac{-i}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a-D_\mu \lambda^{\dagger a}\bar{\sigma}^\mu \epsilon)=\frac{-i}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a+h.c.
I derived that
\delta F^a_{\mu\nu}= -\frac{1}{\sqrt{2}}\epsilon^\dagger(\bar{\sigma}_\nu D_\mu-\bar{\sigma}_\mu D_\nu)\lambda^a+h.c.
Then when I calculated the variation of the total lagrangian, I obtained that the D^a and F^a_{\mu\nu} terms cancel out separately. But there is a term from the variation of A^b_\mu in the definition of covariant derivative in D_\mu \lambda^a that remains after all the other cancellation. It looks like
i g f^{abc}\lambda^{\dagger a}\bar{\sigma}^\nu \lambda^c (\epsilon^\dagger \bar{\sigma}_\nu \lambda^b+h.c.)
This is the only term with three \lambdas. So it seems there is no term to cancel this one. I have checked my calculation over and over again but couldn't find the mistake. So any idea what 's wrong?
Thanks a lot,