- #1
ismaili
- 160
- 0
Dear guys,
I know that gamma matrices have some relations, like
[tex] \gamma^0{\gamma^\mu}^\dagger\gamma^0 = \gamma^\mu \quad---(*)[/tex]
And I am wondering if this is representation independent?
Consider,
[tex] S\gamma^0S^{-1}S{\gamma^\mu}^\dagger S^{-1}S\gamma^0 S^{-1} = S\gamma^\mu S^{-1} [/tex]
[tex] \Rightarrow \gamma'^0 \big({\gamma^\mu}^\dagger\big)' \gamma'^0 = \gamma'^\mu [/tex]
, hence, the condition that this relation is representation independent requires that
[tex] ({\gamma^\mu}^\dagger)' = {\gamma'^\mu}^\dagger [/tex]
, and this implies that
[tex] S^{-1} = S^{\dagger} [/tex] if the relation eq(*) is representation independent.
However, I've never seen any books or literature stress that the similarity transformation between different representations of gamma matrices must be unitary. But, we often used such equations, say, eq(*) as a representation independent formula.
I'm confused.
Could anybody clarify this?
I know that gamma matrices have some relations, like
[tex] \gamma^0{\gamma^\mu}^\dagger\gamma^0 = \gamma^\mu \quad---(*)[/tex]
And I am wondering if this is representation independent?
Consider,
[tex] S\gamma^0S^{-1}S{\gamma^\mu}^\dagger S^{-1}S\gamma^0 S^{-1} = S\gamma^\mu S^{-1} [/tex]
[tex] \Rightarrow \gamma'^0 \big({\gamma^\mu}^\dagger\big)' \gamma'^0 = \gamma'^\mu [/tex]
, hence, the condition that this relation is representation independent requires that
[tex] ({\gamma^\mu}^\dagger)' = {\gamma'^\mu}^\dagger [/tex]
, and this implies that
[tex] S^{-1} = S^{\dagger} [/tex] if the relation eq(*) is representation independent.
However, I've never seen any books or literature stress that the similarity transformation between different representations of gamma matrices must be unitary. But, we often used such equations, say, eq(*) as a representation independent formula.
I'm confused.
Could anybody clarify this?