Invariance of Wess Zumino Action under SUSY

  • #1
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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?
 
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Answers and Replies

  • #2
haushofer
Science Advisor
Insights Author
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Without going through the explicit calculation, chapter 5 of Labelle's susy demystified goes through this in great detail.
 
  • #3
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Without going through the explicit calculation, chapter 5 of Labelle's susy demystified goes through this in great detail.
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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