In the action term for Majorana fermion as given in Peskin and Schroeder's question 3.4, the first term is :

$\int d^4x \space\chi^\dagger i \bar{\mathbb{\sigma}} \cdot \partial\chi$

which can be written as

$\int d^4x \space \chi^\dagger i \bar{\mathbb{\sigma}}^\mu \partial_\mu \chi$

where $\bar{\mathbb{\sigma}}^\mu = (1,-\vec{\sigma})$

Now, $\chi$ is a two components column and $\chi ^\dagger$ is a 2 component row. But $\partial_{\mu}$ is a four derivative, so it has four component.

a) So mathematically speaking, what kind of object is $\partial_{\mu} \chi$?

Then, going one step further, there is $\bar{\mathbb{\sigma}}^\mu$ which is a 4 components row, where each component is a 2 by 2 matrix.

b) Can someone show me what then is the overall term $\chi^\dagger i \bar{\mathbb{\sigma}}^\mu \partial_\mu \chi$?

This sort of mathematical manipulation is still not my strong suit, and I would really appreciate some help.

P.S. $\chi(x)$ is defined as a 2-component field.

  • 1
    $\begingroup$ Ah. Now I see you define $\chi$ in two single components? You should ignore my answer. $\endgroup$
    – joseph h
    Oct 30, 2020 at 5:21
  • $\begingroup$ Yes, my bad. Very sorry I realized I should have clarified that earlier. $\endgroup$
    – time12
    Oct 30, 2020 at 12:47

1 Answer 1


Here is the action written with explicit sums over the indices

\begin{equation} S = \int {\rm d}^4 x \sum_{\dot{\alpha}=1}^2 \sum_{\beta=1}^2 \sum_{\mu=0}^3 \chi^\dagger_{\dot{\alpha}} i \sigma_{\dot{\alpha} \beta}^\mu \partial_\mu \chi_\beta \end{equation}

$\chi$ is a two component spinor and $\partial_\mu \chi$ is the gradient of a spinor.

At a deeper level, in this expression, $\alpha,\beta$ indices label components of objects that transform in the spinor representation of the Lorentz group, while the $\mu$ index labels an object that transforms in the fundamental representation.


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