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

diracsgrandgrandson
Messages
6
Reaction score
1
Homework Statement
Peskin and Schroeder Exercise 3.4 Majorana Fermions part b wants me to show that the variation of the action ##S## with respect to ##\chi^\dagger## gives the Majorana equation.
Relevant Equations
The action is given by: $$S = \int d^4 x \left[ \chi^\dagger i \sigma \cdot \partial \chi + \frac{im}{2} \left( \chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^* \right) \right]$$

The Majorana equation is $$i \bar{\sigma} \cdot \partial \chi - im \sigma^2 \chi^* = 0$$
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.
 
Physics news on Phys.org
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^*##.
 
Last edited:
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^*##
 
  • Like
Likes PhDeezNutz and diracsgrandgrandson
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. :oldsmile:
 
  • Like
Likes diracsgrandgrandson, PhDeezNutz and haushofer
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top