Showing that the variation in given action leads to Majorana equation

Question

This is a question on mathematics rather than the physics. It is based on QFT Q3.4, part b on Peskin and Schroeder.

My confusion stems from the fact that we are consider $\chi(x)$ to be a classical anticommuting field,that takes Grassmann numbers as their values and satisfy the following relation:

$$\alpha \beta = -\alpha \beta \quad \text{ for any }\quad \alpha, \beta.$$

So I am given the action which is: $$S = \int d^4x [\chi^\dagger i\bar{\sigma} \cdot\partial\chi + \frac{im}{2}(\chi^T\sigma^2\chi^*- \chi^\dagger \sigma^2 \chi^*)]$$ And then I am trying to show that varying this w.r.t $\chi^*$ yields Majorana equation.

My attempt:

I rewrote $S$ as:

$\large S = \int d^4x [\chi^*_a i\bar{\sigma}^\mu_{ab}\partial_\mu\chi_b + \frac{im}{2}(\chi_a\sigma^2_{ab}\chi_b- \chi^*_a \sigma^2_{ab}\chi^*_b]$

So then I found $\delta S$ and I am at the following step:

$\large \delta S = i \int d^4x [\bar{\sigma}^\mu_{ab} \delta\chi_a^*\partial_\mu\chi_b+\frac{im}{2}\sigma^2_{ab}(-\chi_a^* \delta\chi_b^*-\chi_b^* \delta \chi_a^*)] -----(i)$

Okay, so then this is how I proceeded:

$\large \delta(\chi_a^*\chi_b^*) = \chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^*$

But, $\large \chi_a^* \chi_b^* = -\chi_b^* \chi_a^*$ as given to us in the question.

$\large \rightarrow \delta(-\chi_b^*\chi_a^*) = \chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^* $

$\large \rightarrow -\chi_a^*\delta\chi_b^*-\chi_b^*\delta\chi_a^* =\chi_b^* \delta \chi_a^* + \chi_a^* \delta \chi_b^* $

$\large \rightarrow \chi_a^*\delta\chi_b^* = -\chi_b^*\delta \chi_a^*$

But if I did the last bit correctly, then it means the terms inside the parenthesis in $\delta S$ in eqn(i) will cancel out, and it is not going to give me the Majorana equation.

Could someone help me see where I made the mistake? For this to work out, I feel like I need to get,

$\large \chi_a^*\delta\chi_b^* = \chi_b^*\delta \chi_a^*$

and I am off by a minus sign.

Edit 1: Upon thinking more, I don’t think what I am doing makes sense to me cause I feel like I am not correctly understanding what it means for $\large \chi_b^* \chi_a^* = -\chi_a^* \chi_b^*$. But i also do not know of a way out of this. So any pointer towards right direction would be very helpful.

Answer 1

We would like to derive the majorana equation: \begin{equation} i\bar{\sigma}\cdot \partial \chi - i m \sigma^2 \chi^* = 0 \end{equation} from its action \begin{equation} S = \int d^4x \left[ \chi^\dagger i \bar{\sigma} \cdot \partial \chi+ \frac{i m}{2} \left( \chi^\mathrm{T} \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^* \right) \right] \end{equation} we use the variational approach to find the field equation such that $\delta S = 0$. Variation wrt $\chi^*$ gives \begin{equation} \begin{split} \delta S &= S \left( \chi^* + \delta \chi^* \right) - S \left( \chi^* \right)\\ &= \int d^4x \mathcal{L} \left( \chi^* + \delta \chi, \partial_\mu \chi^* + \partial_\mu \delta\chi^* \right) - \int d^4x \mathcal{L} \left( \chi^*, \partial_\mu \chi \right)\\ \end{split} \end{equation} expand the 1st term using suffix notation: \begin{equation} \begin{split} \int d^4x \mathcal{L} &\left( \chi^* + \delta \chi, \partial_\mu \chi^* + \partial_\mu \delta\chi^* \right) \\ &=\int d^4 x \left\{ i\left( \chi_a^* + \delta \chi_a^*\right) \bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2 \bigl[ \chi_a\chi_b - (\chi_a^* + \delta \chi_a^*)(\chi_b^* + \delta\chi_b^*) \bigr] \right\} \end{split} \end{equation} The 2nd term in $\delta S$ can be expanded by the same token: \begin{equation} \int d^4x \mathcal{L} \left( \chi^* , \partial_\mu \chi^* \right) =\int d^4 x \left\{ i \chi_a^* \bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2 (\chi_{a}\chi_b - \chi_a^* \chi_b^*) \right\} \end{equation} From these we simplify $\delta S$ into: \begin{equation} \delta S = \int d^4x \left[ \delta\chi_a^* i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2(-\chi_a^*\:\delta \chi_b^* - \delta \chi_a^*\: \chi_b^*) \right] \end{equation} since $\sigma_{ab}^2$ is symmetric, we can rearange indices, so that \begin{equation} \begin{split} \delta S &= \int d^4x \left[ \delta\chi_a^* i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b + \frac{i m}{2} \sigma_{ab}^2(-\delta\chi_a^*\chi_b^* - \delta \chi_a^*\: \chi_b^*) \right]\\ &= \int d^4x \;\delta\chi_a^* \Bigl(i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b - i m \sigma_{ab}^2\chi_b^*\Bigr) \end{split} \end{equation} we require $\delta S$ to vanish, hence we must have \begin{equation} i\bar{\sigma}_{ab}^\mu \partial_\mu \chi_b - i m \sigma_{ab}^2\chi_b^* = 0 \end{equation}