In Bjorken and Drell, Relativistic Quantum Mechanics, the following argument is constructed to show that a set of matrices derived from the $\gamma$ matrices are linearly independent. The matrices of concern (though I do not think that is is particularly important in the argument):

$$\Gamma^S = \mathrm{I},\space \space \space \space \space \space \space \ \Gamma^V_\mu = \gamma_\mu \space \space \space \space \space \space \space \Gamma^T_{\mu\nu} = \sigma_{\mu\nu}$$ $$\Gamma^P = \gamma_5 :=\gamma^5 \space \space \space \space \space \space \space \Gamma^A_\mu=\gamma_5\gamma_\mu,$$

where the $\gamma_\mu$ and $\sigma_{\mu\nu}$ have the usual meaning.

Then the following argument is made to show that the above matrices form 16 linearly independent matrices:

  1. $\forall\space \Gamma^n,\space (\Gamma^n)^2 = \pm \text{I}$
  2. $\forall \space \Gamma^n \neq\Gamma^S, \exists \space\Gamma^m \space\text{such that} \space \{\Gamma^n,\Gamma^m\} = 0$ where $\{A,B\}$ is the anticommutator, which implies Tr$(\Gamma^n) = 0$
  3. For $a \neq b$, $\exists \space \Gamma^n \neq\Gamma^S$ such that $\Gamma^a \Gamma^b = \Gamma^n$.

To those point I am okay with what has been said, but it is the fourth point that confuses me:

  1. Suppose there exists $a_n$ such that $$\sum_na_n\Gamma^n = 0$$ Then by multiplying by $\Gamma^m \neq\Gamma^S$, and taking the trace, using (3) we find that $a_m = 0$.

If someone could explain these steps in some more detail, that would be greatly appreciated.


1 Answer 1


Suppose there exist $\{a_n\}$ such that $\sum_n a_n \Gamma^n = 0$.

Choose some $\Gamma^m$ other than the identity, and left-multiply both sides by $\Gamma^m$. The result is

$$ \sum_n a_n \Gamma^m \Gamma^n = 0 $$

We can rewrite this by considering separately $n = m$ and $n \ne m$:

$$ a_m \Gamma^m \Gamma^m + \sum_{n \ne m} a_n \Gamma^m \Gamma^n = 0$$

Use (1) and (3):

$$ a_m (\pm I) + \sum_{n \ne m} a_n \Gamma^{p_n} = 0$$

Here, $\Gamma^{p_n}$ is some matrix other than $\Gamma^S$, and is equal to the product $\Gamma^m \Gamma^n$. We know this exists, thanks to (3).

Finally, take the trace of both sides, and use (2), according to which each $\operatorname{tr} \Gamma^{p_n}$ vanishes:

$$ \pm 4 a_m = 0$$

This shows that 15 of the 16 $a_m$ coefficients vanish. To show that the coefficient of the identity $a_S$ also vanishes, use the fact that $\Gamma^S$ is the only one of the 16 matrices with trace.

  • $\begingroup$ Thank you! Realised this just before you posted! $\endgroup$
    – FizzKicks
    Jul 21, 2021 at 19:54

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