# Gamma matrices and how they operate

## Homework Statement

Just a matter of convention (question)

## Homework Equations

$$\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}$$

## The Attempt at a Solution

If then,

$$\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}$$

and $$\gamma^0$$ is just $$\beta$$ and $$\beta \alpha^k = \gamma^k$$ is it true then that

$$\gamma^k = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\alpha^k$$

Can no one confirm I have done this right?

dextercioby
Homework Helper
Isn't the particular form of a Dirac matrix representation dependent ? So then only the general relations will hold, i.e.

βrepαkrepkrep

where i/o <rep> one has the Dirac, Majorana or Weyl/chiral representations.

Isn't the particular form of a Dirac matrix representation dependent ? So then only the general relations will hold, i.e.

βrepαkrepkrep

where i/o <rep> one has the Dirac, Majorana or Weyl/chiral representations.

I think so. I think you have to work with $$D(\psi(x,t))$$ on the three matrices $$\gamma^1,\gamma^2,\gamma^3$$ to get back the matrix $$i\gamma^0 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$

which when squared gives you the chirality.

Now I am really confused: consider the matrix form of $$a^k$$ and calculate it all out we have

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{pmatrix}$$

A nullified matrix?

Have I got my $$a^k$$ matrix right... ?

$$a^k$$ is just a submatrix, right? of

0_2 sigma^k

\sigma^k 0_2

k=1,2,3

in my case, 1 and 3

I just don't understand why the relationship

$$\beta \alpha^k = \gamma^k$$

would be important if it spat out a zero matrix, which makes me wonder strongly whether I even have the right conditions down.

TSny
Homework Helper
Gold Member
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{pmatrix}$$

A nullified matrix?

You will not get the null matrix. For example, check the element in the first row, third column of the resultant matrix.

I'm sorry, I did it all wrong didn't I? I now get

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}$$

I am an idiot some times lol