2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.