Equivalent representations for Dirac algebra

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 3K views
Wledig
Messages
69
Reaction score
1
Homework Statement
Consider the following representations that satisfy the Dirac algebra:

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

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

and

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


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

Show that they are equivalent, that is write a 4x4 unitary matrix U such that:
$$\gamma^{\mu}_B = U\gamma^{\mu}_A U^\dagger$$
Relevant Equations
Dirac algebra: ##\{ \gamma^\mu , \gamma^\nu \} = 2\eta^{\mu \nu}##
Where ##\eta^{\mu \nu} ## is the metric tensor from special relativity.
One thing I was thinking about doing was to consider these representations as a basis for the gamma matrices vector space, then try to determine what the change of basis from one to the other would be. However I'm unsure if it's correct to treat the representations as a basis, or whether this is the right approach at all.
 
Physics news on Phys.org
Alright, so after searching a bit I managed to find U in an appendix in the book by Itzykson:
$$ U = \dfrac{1}{\sqrt{2}}(1+\gamma_5\gamma_0) = \dfrac{1}{\sqrt{2}}
\begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix}
$$
I've tested for ##\gamma^0##, so I'm convinced it works, but I still don't know how to reach this matrix. Like I've said it was found in an appendix, without much explanation to go along with it.
 
Forget it, I figured it out. Just needed to find the eigenvalues of the Weyl representation then normalize the eigenvector matrix to make it unitary, it's simpler than I thought.
 
Wledig said:
Alright, so after searching a bit I managed to find U in an appendix in the book by Itzykson:
$$ U = \dfrac{1}{\sqrt{2}}(1+\gamma_5\gamma_0) = \dfrac{1}{\sqrt{2}}
\begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix}
$$
I've tested for ##\gamma^0##, so I'm convinced it works, but I still don't know how to reach this matrix. Like I've said it was found in an appendix, without much explanation to go along with it.
It's hard to tell how to find if you've seen the answer, which I did when I looked up the definitions. The spatial ##\gamma^i## should remain unchanged, so it's probably an idea to focus on ##\gamma^0##.

The hard way to find ##U## is probably to solve the equations ##U(\{\,\gamma^i,\gamma^j\,\})=\{\,U(\gamma^i),U(\gamma^j)\,\}##.
 
  • Like
Likes   Reactions: Wledig
That's alright, thanks for the tip.