I want to expand the kinetic term of the Majorana Lagrangian (considering only a left handed Weyl spinor) from

$$\mathcal{L}= \frac{1}{2} \bar{\psi}_M i \gamma^{\mu}\partial_{\mu} \psi$$

So I wrote :$$\bar{\psi}_M \equiv \psi^{\dagger}_M \gamma^0 = \begin{pmatrix} \psi^{\dagger}_L & -i\psi^{T}_L \sigma^2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -i \psi^{T}_L \sigma^2 & \psi^{\dagger}_L \end{pmatrix}$$

And the representation with the representation of the Dirac matrices: $$\gamma^0 = \begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma^{\mu}} & 0 \end{pmatrix}$$

So the expression for the lagrangian becomes: $$\mathcal{L}=\frac{i}{2}\begin{pmatrix} -i \psi^{T}_L \sigma^2 & \psi^{\dagger}_L \end{pmatrix}\begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma^{\mu}} & 0 \end{pmatrix} \begin{pmatrix} \partial_{\mu}\psi_L \\ \partial_{\mu} i \sigma^2 \psi^{*}_L \end{pmatrix} $$

And just by multiplying these matrices we finally obtain: $$\mathcal{L}= \frac{i}{2} \left[ \psi^{\dagger}_L \bar{\sigma^{\mu}} \partial_{\mu} \psi_L + \psi^T_L \sigma^2 \sigma^{\mu} \partial_{\mu} \sigma_2 \psi^{*}_L \right]$$

But I know that this term should be equal to

$$\mathcal{L}=i\psi^{\dagger}_L \bar{\sigma^{\mu}} \partial_{\mu} \psi_L$$

Which leads me to think that somehow the second term in the previous equation is equal to the first one, but I have not been able to derive that.Could anyone help?


1 Answer 1


For the Pauli matrices $$ \sigma_2 \sigma_i \sigma_2 = -\sigma_i^* $$ where the $*$ denotes the matrix with complex-conjuated entries.

  • $\begingroup$ Also important to note that the spinors must anticommute, otherwise you will get zero $\endgroup$
    – FrodCube
    Commented Apr 20, 2020 at 13:34

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