Hermitianity of gamma matrices

[ismaili]

Dear guys,

I know that gamma matrices have some relations, like

\gamma^0{\gamma^\mu}^\dagger\gamma^0 = \gamma^\mu \quad---(*)

And I am wondering if this is representation independent?
Consider,

S\gamma^0S^{-1}S{\gamma^\mu}^\dagger S^{-1}S\gamma^0 S^{-1} = S\gamma^\mu S^{-1}

\Rightarrow \gamma'^0 \big({\gamma^\mu}^\dagger\big)' \gamma'^0 = \gamma'^\mu

, hence, the condition that this relation is representation independent requires that

({\gamma^\mu}^\dagger)' = {\gamma'^\mu}^\dagger

, and this implies that

S^{-1} = S^{\dagger} 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?

 

[ismaili]

Dear guys,

I know that gamma matrices have some relations, like

\gamma^0{\gamma^\mu}^\dagger\gamma^0 = \gamma^\mu \quad---(*)

And I am wondering if this is representation independent?
Consider,

S\gamma^0S^{-1}S{\gamma^\mu}^\dagger S^{-1}S\gamma^0 S^{-1} = S\gamma^\mu S^{-1}

\Rightarrow \gamma'^0 \big({\gamma^\mu}^\dagger\big)' \gamma'^0 = \gamma'^\mu

, hence, the condition that this relation is representation independent requires that

({\gamma^\mu}^\dagger)' = {\gamma'^\mu}^\dagger

, and this implies that

S^{-1} = S^{\dagger} 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?

The solution seems that the similarity transformation matrix is really unitary.
Since, if we consider the bilinear which should be representation invariant, since it contracts all the Dirac indices:

\bar{\psi}\gamma^\mu\psi = \psi^\dagger\gamma^0\gamma^\mu\psi

In another representation,

\bar{\psi}'\gamma'^\mu \psi' = \psi'^\dagger S^\dagger S \gamma^0 S^{-1}S \gamma^\mu S^{-1} S\psi = \psi^\dagger S^\dagger S\gamma^0\gamma^\mu \psi

The above expression would equal to \bar{\psi}\gamma^\mu\psi if S^\dagger S = 1

-------------

But, I came up with another question..
If we consider the Majorana condition:

\psi^* = B\psi,

In another representation:

(\psi')^* = B' \psi' \Rightarrow (S\psi)^* = SBS^{-1} S\psi \Rightarrow S^*\psi^* = SB\psi

Since \psi^* = B\psi in the old representation, we conclude that

S = S^*

, if the Majorana condition is representation independent.

I still confused by the discovery that the representation-free relations constrain the similarity transformations between different representations. Could anyone elucidate on this?
Thank you so much!

 

[eduardooftheh]

The solution seems that the similarity transformation matrix is really unitary.
Since, if we consider the bilinear which should be representation invariant, since it contracts all the Dirac indices:

\bar{\psi}\gamma^\mu\psi = \psi^\dagger\gamma^0\gamma^\mu\psi

In another representation,

\bar{\psi}'\gamma'^\mu \psi' = \psi'^\dagger S^\dagger S \gamma^0 S^{-1}S \gamma^\mu S^{-1} S\psi = \psi^\dagger S^\dagger S\gamma^0\gamma^\mu \psi

The above expression would equal to \bar{\psi}\gamma^\mu\psi if S^\dagger S = 1

-------------

But, I came up with another question..
If we consider the Majorana condition:

\psi^* = B\psi,

In another representation:

(\psi')^* = B' \psi' \Rightarrow (S\psi)^* = SBS^{-1} S\psi \Rightarrow S^*\psi^* = SB\psi

Since \psi^* = B\psi in the old representation, we conclude that

S = S^*

, if the Majorana condition is representation independent.

I still confused by the discovery that the representation-free relations constrain the similarity transformations between different representations. Could anyone elucidate on this?
Thank you so much!

Hi, the Majorana condition is certainly not representation independent but as you found real transformation matrices preserve this condition. Ultimately the key component to the theory is the clifford algebra that only requires invertible transformations. Anything else is up to you.