# Expanding gamma matrices

$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{q}} = i \hbar \gamma^0$$

How do I expand

$$i\hbar \gamma^0$$

the matrix in this term, I am a bit lost. All the help would be appreciated!

zje2009
$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{q}} = i \hbar \gamma^0$$

How do I expand

$$i\hbar \gamma^0$$

the matrix in this term, I am a bit lost. All the help would be appreciated!

It's itself.Using Mathematica.

Expanding that, gives back that?

What is mathematica??

The gamma matrices are 4 x 4 matrices whose values depend on the basis ("representation") you decide to use in spinor space. For a list of possibilities, see "gamma matrix" in Wikipedia.

so you don't know how to expand the terms I asked of?

so you don't know how to expand the terms I asked of?
I thought the Wikipedia article explained it pretty clearly. It gives the explicit form of γ0 in the Dirac, Weyl and Majorana representations. Isn't that what you want?

But if that's not what you mean by "expanding" it, the only other thing to do is this...

iħγ0

I thought the Wikipedia article explained it pretty clearly. It gives the explicit form of γ0 in the Dirac, Weyl and Majorana representations. Isn't that what you want?

But if that's not what you mean by "expanding" it, the only other thing to do is this...

iħγ0

Why have that when I started off with that... my equation I really thought was simple. HOW do you expand $$i \hbar \gamma^0$$

The answer I was looking for was not a go to wiki one! And no wiki does not explain it well for me, I am new at this stuff.

Can you show me, in plane mathematical language, in an equation, how to expand it please.

kloptok
You will have to define what you mean by "expand". So far we have only been able to guess, and apparently this was not what you intended. So define "expand" please.

PhilDSP
Do you mean this? $\qquad \gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} \qquad \qquad$ (which is the Dirac representation)

so that $\ \ \qquad \qquad i \hbar \gamma^0 = \begin{pmatrix} i \hbar & 0 & 0 & 0\\ 0 & i \hbar & 0 & 0\\ 0 & 0 & -i \hbar & 0\\ 0 & 0 & 0 & -i \hbar \end{pmatrix} \qquad \qquad$

Last edited:
Do you mean this? $\qquad \gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} \qquad \qquad$ (which is the Dirac representation)

so that $\ \ \qquad \qquad i \hbar \gamma^0 = \begin{pmatrix} i \hbar & 0 & 0 & 0\\ 0 & i \hbar & 0 & 0\\ 0 & 0 & -i \hbar & 0\\ 0 & 0 & 0 & -i \hbar \end{pmatrix} \qquad \qquad$

Why is then when $$\gamma^{0}^2$$ is equal to 1?

PhilDSP
$$\qquad (\gamma^0)^2 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$