This is regarding exercise 3.4b in Peskin and Schroeder's Introduction to Quantum Field Theory. This problem asks us to find the equations of motion for the Majorana field (the equations of motion were given in part a to be $i\bar{\sigma}^\mu \partial_\mu \chi - im \sigma^2 \chi^* = 0$). The action is given by $S* = \int d^4x L$, where $L = $ Lagrangian density $= \chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi + \frac{im}{2}\bigg(\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^* \bigg)$.

To find the equations of motion I decided to use $\frac{\partial L}{\partial \chi^\dagger} = \partial_\mu \bigg(\frac{\partial L}{\partial(\partial_\mu \chi^\dagger)} \bigg)$ and got $i\bar{\sigma}^\mu \partial_\mu \chi - \frac{im}{2}\sigma^2 \chi^* = 0$.

The solution's manual said to vary $S$ and I think that I did that (the euler lagrange equations are derived from varying S and L)

What did I do incorrectly? What would be the correct way to find the equations of motion?

  • 1
    $\begingroup$ Is the problem in the fact that you used the Euler Lagrange equations instead of varying $S$? Or did you get the wrong answer? Because the Euler Lagrange equations are basically a fast way to vary the action. $\endgroup$ Commented Jul 13, 2021 at 12:30
  • 1
    $\begingroup$ @AccidentalTaylorExpansion I got the wrong answer but I think I figured out what is wrong (I will post answer below) $\endgroup$
    – user261609
    Commented Jul 13, 2021 at 12:31

2 Answers 2


There is not really a standard notion of an operation that takes $\xi\to \xi^*$ for a Grassmann variable and in a Euclidean signature Grassmann path integral $\psi$ and $\psi^*$ are independent objects. I don't know how widespread P&S's $(ab)^*=-b^*a^*$ is but I find it confusing. Also appearences of $i\sigma_2$ as in P&S are dependent on the choice of representation for the Gamma matrices.

I think is much simpler just to take the Grassmann-variable Majorana action to be $$ S_{\rm Majorana} ={\textstyle \frac 12} \int \psi^T C (\gamma^\mu \partial_\mu +m)\psi\, d^4x $$ where $C$ is the charge conjugation matrix defined by $C\gamma^\mu C^{-1}= -(\gamma^\mu)^T$. (Peter van Nieuwenhuizen calls $\psi^T C$ the Majorana adjoint, and his discussion seems simpler than P&S.)

In 4 Euclidean dimensions the expression $C (\gamma^\mu \partial_\mu +m)$ is skew symmetric in the combined spin and $x$ space, so just as the "variation" of the quadratic Grassmann expression $$ S={\textstyle \frac 12} \xi^TA\xi $$ with $A$ a skew-symmetric matrix gives $A\xi=0$, the variation of gives the equation of motion $$ (\gamma^\mu \partial_\mu +m)\psi=0. $$

In Minkowski signature the four-component spinor is no longer a Grassman variable but is operator-valued with $\bar\psi= \psi^\dagger \gamma^0$ with $\dagger$ implying the adjoint operation on the many-body Hilbert space as usual. The field $\psi$ however, satisfies the Majorana condition $$ \psi=\psi^c\equiv C^{-1} \bar\psi^T $$ and the anticommutation commutation relations $$ \{\psi_\alpha(x), \psi_\beta(x')\}_{t=t'} = [\gamma^0 C^{-1}]_{\alpha\beta} \delta^3(x-x'). $$ In the space-time dimensions ($d=2,3,4$ mod 8) where Majorana's exist the matrix $\gamma^0 C^{-1}$ is symmetric --- as it needs to be for the anticommutator to be consistent with the RHS.

I think that in the $\gamma^5$ diagonal representation of the Gamma matrices we have $$ C= \left[\matrix{0& i \sigma_2\\ -i\sigma_2&0}\right], $$ so it should be possible to match my basis-independent notation to that of P&S.


$\chi^T \sigma^2 \chi = (-\chi^* \sigma^2 \chi^\dagger)^*$ and then differentiate to get another $-\frac{im}{2}\sigma^2 \chi^*$

Note that the values $\chi$ takes are Grassmann numbers and so, as noted in the exercise description, $(\alpha \beta)^* = - \alpha^* \beta^*$.

Thus: $\chi^T \sigma^2 \chi = (-\chi^* \sigma^2 \chi^\dagger)^* = (\chi^*)^*(\sigma^2 \chi^\dagger)^* = -(\chi^*)^*(\chi^\dagger)^*\sigma^2 = (\sigma^2\chi^\dagger \chi^*)^*$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.