# Confusion in linear independence of matrices argument in Bjorken and Drell

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

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.

• Thank you! Realised this just before you posted! Jul 21, 2021 at 19:54