P&S Exercise 3.4 Majorana Fermions Derivative of ##\chi##

I am stuck at the final part where one is supposed to show that the derivative of the second term of the action gives the mass term in the Majorana equation. For $$\chi^T\sigma^2\chi = -(\chi^\dagger\sigma^2\chi^*)^*$$ we get $$\frac{\delta}{\delta\chi^\dagger}(\chi^\dagger\sigma^2\chi^*)^*$$ which is supposed to give $$\sigma^2\chi^*.$$ I don't see how. Suppose $$f(\chi) = \chi^*,$$ and now $$\frac{d}{d\chi}f(\chi) = \frac{d\chi^*}{d\chi}$$ which would be zero due to the field and its complex conjugate being zero.

diracsgrandgrandson said:

I am stuck at the final part where one is supposed to show that the derivative of the second term of the action gives the mass term in the Majorana equation.

You can get the Majorana equation by varying ##S## with respect to ##\chi_1^*## and ##\chi_2^*##.

Note that ##\chi^T \sigma^2 \chi## does not contain either ##\chi_1^*## or ##\chi_2^*##.
So, this expression in the Lagrangian will not contribute when doing the variation with respect to ##\chi_1^*## and ##\chi_2^*##.

Write out the expression ##\chi^\dagger \sigma^2 \chi^*## explicitly in terms of ##\chi_1^*## and ##\chi_2^*##. Then you can look at its variation with respect to ##\chi_1^*## and ##\chi_2^*##.

TSny said:

Write out the expression χ†σ2χ∗ explicitly in terms of χ1∗ and χ2∗. Then you can look at its variation with respect to χ1∗ and χ2∗.

Doesn't that give zero?

$$
\begin{align*}
\chi^\dagger \sigma^2 \chi^* &=
\begin{pmatrix}
\chi_1^* & \chi_2^*
\end{pmatrix}
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\begin{pmatrix}
\chi_1^* \\
\chi_2^*
\end{pmatrix} \\
&= \begin{pmatrix}
\chi_1^* & \chi_2^*
\end{pmatrix}
\begin{pmatrix}
- i \chi_2^* \\
i \chi_1^*
\end{pmatrix} \\
&= i (-\chi^*_1\chi^*_2+\chi^*_2\chi_1^*)
\end{align*}
$$

which gives zero for each derivative.

According to the problem statement in the textbook, ##\chi_1## and ##\chi_2## are to be treated as anticommuting quantities (Grassmann numbers) with the following properties

##\chi_1 \chi_2## = -##\chi_2 \chi_1 \,\,## and ##\,\, (\chi_1 \chi_2)^* \equiv \chi_2^* \chi_1^* = - \chi_1^* \chi_2^*##

Ahh I see, that clarifies it, thank you. I find the problem is formulated in a very confusing way, at least for a beginner like me :)

Yes, it's a difficult subject. I'm also a beginner. I've been a beginner for years.