# Help with a supersymmetry problem 3.5b in Peskin and Schroeder

I am self studying Quantum Field Theory and I am using the book Introduction to Quantum Field Theory by Peskin and Schroeder along with the solution manual by Zhong Zhi Xianyu. I am currently working on problem 3.5b, which is on supersymmetry. In the solutions for (b), the first line (which I was able to successfully understand) says

$$\delta (\Delta L) = -mi\epsilon^T \sigma^2 \chi F - im\phi \epsilon^T \bar{\sigma}^{\mu}\partial_\mu\chi + \frac{1}{2}im(\epsilon^T F + \epsilon^+(\sigma^2)^T (\sigma^\mu)^T\partial_\mu \phi)\sigma^2 \chi + \frac{1}{2}im\chi^T \sigma^2(\epsilon F + \sigma^\mu \partial_\mu\phi \sigma^2\epsilon^*).$$

The next line says,

$$-\frac{1}{2}imF\epsilon^T \sigma^2 \chi + \frac{1}{2}imF\chi^T \sigma^2 \epsilon - im\phi \epsilon^T \bar{\sigma}^{\mu}\partial_\mu\chi - \frac{1}{2}im(\partial_\mu \phi)\epsilon^+ \bar{\sigma}^\mu \chi + \frac{1}{2}im(\partial_\mu \phi) \chi^T (\bar{\sigma^\mu})^T\epsilon^*$$

I am completely lost on how the second line follows from the first. I see that the $$- im\phi \epsilon^T \bar{\sigma}^{\mu}\partial_\mu\chi$$ cancel. I also tried using the identities written beneath the derivation in the solutions manual, but I still cannot get the second line's results. I found that I often got the same letters but in a different order, and order matters here, so I am lost. Can anyone explain how the manual author got from the first line to the second?

I just did that problem two weeks ago to review my qualification exam. Let me first explain his purpose. He wants to prove the following Lagrangian

$$\Delta\mathcal{L}=[m\phi F+\frac{1}{2}im\chi^T\sigma^2\chi]+c.c.$$

is invariant under the supersymmetry transformation

\begin{align} \delta\phi&=-i\varepsilon^T\sigma^2\chi\\ \delta\chi&=\varepsilon F+\sigma\cdot\partial\phi\sigma^2\varepsilon^*\\ \delta F&=-i\varepsilon^\dagger\bar{\sigma}\cdot\partial\chi. \end{align}

I think the complicated part is $$\frac{1}{2}im\delta\chi^T\sigma^2\chi$$ and $$\frac{1}{2}im\chi^T\sigma^2\delta\chi$$. Let's do it step by step. Firstly,

$$\frac{1}{2}im\delta\chi^T\sigma^2\chi=\frac{1}{2}im(\varepsilon F+\sigma\cdot\partial\phi\sigma^2\varepsilon^*)^T\sigma^2\chi.$$

As he have seen, the first term of the above equation $$\frac{1}{2}im(\varepsilon F)^T\sigma^2\chi$$ cancels $$m(\delta\phi)F$$ and give you $$-\frac{1}{2}im\varepsilon^T F\sigma^2\chi$$. While the second part of it

$$\frac{1}{2}im(\sigma\cdot\partial\phi\sigma^2\varepsilon^*)^T\sigma^2\chi= \frac{1}{2}im(\varepsilon^\dagger(-\sigma^2)\sigma^T\cdot\partial\phi)\sigma^2\chi= -\frac{1}{2}im\varepsilon^\dagger\bar{\sigma}\cdot\partial\phi\chi,$$

where at the second equal sign $$\sigma^2\sigma^T\sigma^2=\bar{\sigma}$$ is used. And this is exactly the fourth term in "the next line". Take transpose on both sides of $$\sigma^2\sigma^T\sigma^2=\bar{\sigma}$$ can you get $$\sigma^2\sigma\sigma^2=\bar{\sigma}^T$$. Then do the same thing to $$\frac{1}{2}im\chi^T\sigma^2\delta\chi$$ and you will get the last term in "the next line".

There is a typo in "the next line". The third term should be $$-im\phi\varepsilon^\dagger\bar{\sigma}\cdot\partial\chi.$$ In "the next line", the first two terms cancel and the last three terms give a total differential.