# Supersymmetry transformation of auxiliary scalar in Wess-Zumino model

This question is related to my earlier question "Error bringing in the auxiliary scalar field in the Wess Zumino model".

In equation (3.1.13) of "A Supersymmetry Primer", arXiv:hep-ph/9709356, the supersymmetry transformation of the auxiliary scalar field is given as, $$\delta F=-i\epsilon^{\dagger}\bar{\sigma}^{\mu}\partial_{\mu}\psi,\\ \delta F^{*}=i\partial_{\mu}\psi^{\dagger}\bar{\sigma}^{\mu}\epsilon. \tag{3.1.13}$$ My question is "Why isn't there an additional sign change in $\delta F^{*}$ due to the interchange of order of the anti-commuting spinors $\epsilon$ and $\psi$ ?" In other words, starting from $\delta F$, working in components in the same notation as hep-ph/9709356, evaluate $\delta F^{*}$. $$\delta F=-i\epsilon^{\dagger}_{\dot{\alpha}}(\bar{\sigma}^{\mu})^{\dot{\alpha}\beta}\partial_{\mu}\psi_{\beta}\\ \delta F^{*}=i\epsilon^{T}_{\alpha}(\bar{\sigma}^{*\mu})^{\alpha\dot{\beta}}\partial_{\mu}\psi^{*}_{\dot{\beta}}=i\epsilon_{\alpha}(\bar{\sigma}^{*T\mu})^{\dot{\beta}\alpha}\partial_{\mu}\psi^{*}_{\dot{\beta}}=i\epsilon_{\alpha}(\bar{\sigma}^{\dagger\mu})^{\dot{\beta}\alpha}\partial_{\mu}\psi^{*}_{\dot{\beta}}=i\epsilon_{\alpha}(\bar{\sigma}^{\mu})^{\dot{\beta}\alpha}\partial_{\mu}\psi^{*}_{\dot{\beta}}$$ Now swop the order of the anti-commuting spinors, and write the result in matrix notation. $$\delta F^{*}=-i\partial_{\mu}\psi^{*}_{\dot{\beta}}(\bar{\sigma}^{\mu})^{\dot{\beta}\alpha}\epsilon_{\alpha}=-i\partial_{\mu}\psi^{\dagger}\bar{\sigma}^{\mu}\epsilon.$$ Notice that this result has the wrong sign compared to equation (3.1.13) in hep-ph/9709356. Why does the paper ignore the anti-commuting nature of the spinors in this case? On page 15 of the paper, immediately below equation (2.18), the paper states,

"Note that when taking the complex conjugate of a spinor bilinear, one reverses the order."

Why is this reversal not accompanied by a sign change due to the anti-commuting property of the spinors?

In genaral, for any non-commuting objects, $A$ and $B$, the conjugate of the product, $AB$, is NOT the products of each one's conjugate in the same order, but instead in the reverse order, i.e., $$(AB)^* = B^* A^*$$

So, for the variation of the auxillary field, $\delta F^*$, your calculation has the reverse order of what it should be. Therefore, you don't swap anything or have an extra sign.

Also check the second and fourth expressions in Eq.(2.18) of the paper in question:

$$... = \;- \chi \sigma^\mu \xi^\dagger \; = ...= \; -(\xi \sigma^\mu \chi^\dagger )^*$$

where they have the same sign.

• Ramond's book "Field Theory", second edition eqn (1.4.31) writes the involution as $(\psi\chi)^{*}=\psi^{*}\chi^{*}$. However, other references (e.g. Berezin) define the involution in your way. Is it just a matter of a free choice and the theories can be made to work either way? I found that the Wess-Zumino worked out with your definition of the involution. – Stephen Blake Jul 15 '18 at 10:33
• I don't think Ramond's book is correct (or the definitions or the notation of spinors are different and $\psi$ and $\chi$ are their components?). Please see this post in math.SE: math.stackexchange.com/questions/49506/… – Oktay Doğangün Jul 15 '18 at 11:21
• In Ramond's book, I think he means $\psi$ and $\chi$ are components of spinors. – Stephen Blake Jul 15 '18 at 12:22