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A Invariance of Wess Zumino Action under SUSY

  1. May 7, 2016 #1
    Hi guys,
    I have a very basic question about the WZ model. I want to show that it is invariant under SUSY transformations.

    The action is [tex]\int{d^4 x} \partial^\mu \phi* \partial_\mu \phi +i\psi^† \bar{\sigma}^\mu \partial_\mu \psi [/tex]

    The SUSY transformations are [tex] \delta\phi = \epsilon \psi [/tex], [tex] \delta\phi* = \epsilon^† \psi^† [/tex], [tex] \delta\psi_\alpha = -i(\sigma^\nu \epsilon^†)_\alpha \partial_\nu \phi [/tex], [tex] \delta\psi^†_\alpha =i(\epsilon \sigma^\nu)_\alpha\partial_\nu \phi* [/tex]

    I know that I should get a total derivative that vanishes but there is one minus sign that is preventing me from completing the proof. It comes from the term [tex]i\psi^† \bar{\sigma}^\mu \partial_\mu \delta\psi [/tex] in the variation. For this term I get the following:

    [tex]\psi^† \bar{\sigma}^\mu \sigma^\nu \epsilon^† \partial_\mu \partial_\nu \phi [/tex]

    Then by using [tex]\bar{\sigma}^\mu \sigma^\nu + \bar{\sigma}^\nu \sigma^\mu = 2\eta^{\mu\nu}[/tex] I have

    [tex]\psi^† \epsilon^† \partial^\mu \partial_\mu \phi [/tex]

    I rewrite this as a total derivative minus another term

    [tex]\partial^\mu(\psi^† \epsilon^† \partial_\mu \phi)-\partial^\mu\psi^† \epsilon^† \partial_\mu \phi [/tex]

    This is where the problem is. Now, in order for the second term to cancel a term from the variation of φ*, it should have exactly this form but with ψ† and ε† in reverse order. However, doing this introduces a minus sign as the spinors are grassmann odd. What have I done wrong in the manipulation, and how do I get rid of this extra minus sign?
    Last edited: May 7, 2016
  2. jcsd
  3. May 7, 2016 #2


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    Without going through the explicit calculation, chapter 5 of Labelle's susy demystified goes through this in great detail.
  4. May 7, 2016 #3
    Hi, thanks a lot for your answer. I'll have a look at that when I get to the library today. From the above though, can you see an error in my working or is it ok so far?
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