# A Invariance of Wess Zumino Action under SUSY

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1. May 7, 2016

### fa2209

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 $$\int{d^4 x} \partial^\mu \phi* \partial_\mu \phi +i\psi^† \bar{\sigma}^\mu \partial_\mu \psi$$

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

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 $$i\psi^† \bar{\sigma}^\mu \partial_\mu \delta\psi$$ in the variation. For this term I get the following:

$$\psi^† \bar{\sigma}^\mu \sigma^\nu \epsilon^† \partial_\mu \partial_\nu \phi$$

Then by using $$\bar{\sigma}^\mu \sigma^\nu + \bar{\sigma}^\nu \sigma^\mu = 2\eta^{\mu\nu}$$ I have

$$\psi^† \epsilon^† \partial^\mu \partial_\mu \phi$$

I rewrite this as a total derivative minus another term

$$\partial^\mu(\psi^† \epsilon^† \partial_\mu \phi)-\partial^\mu\psi^† \epsilon^† \partial_\mu \phi$$

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. May 7, 2016

### haushofer

Without going through the explicit calculation, chapter 5 of Labelle's susy demystified goes through this in great detail.

3. May 7, 2016

### fa2209

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?