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Infinitesimal SUSY transformation of SYM lagrangian

  1. Sep 27, 2013 #1
    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:
    [tex]\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[/tex]
    the infinitesimal SUSY transformations of the gauge multiplet are:
    [tex]\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.[/tex]
    [tex]\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[/tex]
    [tex]\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[/tex]
    [tex]\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.[/tex]
    I derived that
    [tex]\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.[/tex]
    Then when I calculated the variation of the total lagrangian, I obtained that the [itex]D^a[/itex] and [itex]F^a_{\mu\nu}[/itex] terms cancel out separately. But there is a term from the variation of [itex]A^b_\mu[/itex] in the definition of covariant derivative in [itex]D_\mu \lambda^a[/itex] that remains after all the other cancellation. It looks like
    [tex]i g f^{abc}\lambda^{\dagger a}\bar{\sigma}^\nu \lambda^c (\epsilon^\dagger \bar{\sigma}_\nu \lambda^b+h.c.)[/tex]
    This is the only term with three [itex]\lambda[/itex]s. 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,
  2. jcsd
  3. Sep 27, 2013 #2
    Does the extra term vanish on the equations of motion?
  4. Sep 27, 2013 #3
    That term is proportional to the current of the gaugino. But since all [itex]D_\mu F^{a\mu\nu}[/itex] terms have already been canceled, this current term doesn't vanish by itself.
  5. Sep 27, 2013 #4
    Actually I do have this question. Is it ok if the infinitesimal transformation of lagrangian vanishes only after we apply the equations of motion to it? since I don't think so. I expect it vanishes off-shell.
  6. Sep 27, 2013 #5
    Lagrangian should be invariant off-shell, you are right.

    I am not an expert in this. But probably you need to introduce new field with purely algebraic equations of motion, like F in WZ model, just to cancel this term.
  7. Sep 27, 2013 #6
    I think the [itex]D^a[/itex] is introduced to do the same thing as [itex]F[/itex] in WZ model. But this [itex]D^a[/itex] doesn't solve this problem. Many textbooks I know discussed the infinitesimal SUSY transformation of WZ lagrangian but not this SYM lagrangian.
    Thanks though.
    I guess I will check my calculation again. I couldn't think of a way to cancel the term containing three [itex]\lambda[/itex]s.
  8. Sep 27, 2013 #7
    I think I know what's going on there. The remaining term vanishes by itself by using Fierz identity.
    Last edited: Sep 28, 2013
  9. Sep 27, 2013 #8
    I think the answer to your question is related to the normed division algebras [tex] tri(\lambda)[/tex] law. It is also the reason SYM only takes place in dimensions 3,4,6,10 same with super strings and why there only superbranes in 4 different dimensions due to the fact there are only 4 normed division algebras. Check out this paper.
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