Nested Matrix Elements: Define \Gamma^{\dagger}?

[topsquark]

I posted this elsewhere and was sort of able to figure out a result myself, but 1) I didn't do it right, and 2) No one answered it anyway. I thought I'd give it a shot over here.

The problem deals with nested matrices. The gamma matrices can be found here.

My question deals with a "vector" of these gamma matrices:
Define
\(\displaystyle \Gamma = \left ( \begin{matrix} \gamma ^0 \\ \gamma ^1 \\ \gamma ^2 \\ \gamma ^3 \end{matrix} \right )\)

What is \(\displaystyle \Gamma ^{\dagger}\) ? (The dagger is the conjugate transpose.)

So far I've been able to treat \(\displaystyle \Gamma \) as a 4-vector and so we should have
\(\displaystyle \Gamma = \left ( \begin{matrix} A \\ B \\ C \\ D \end{matrix} \right )\)

\(\displaystyle \Gamma ^{\dagger} = \left ( \begin{matrix} A^* & - B^* & - C^* & - D^* \end{matrix} \right )\)

(Treating \(\displaystyle \Gamma\) as a SR 4-vector gives the negative signs.)

But: Should \(\displaystyle \Gamma ^{\dagger} = \left ( \begin{matrix} A^{\dagger} & - B^{\dagger} & - C^{\dagger} & - D^{\dagger} \end{matrix} \right )\) instead?

For the work I'm doing I don't care about the \(\displaystyle \gamma^2\) part so either method yields the same result for me. But I was wondering if there is a general rule for this sort of thing.

Thanks!

-Dan

 

[I like Serena]

Hey Dan,

The inner product of a finite dimensional space $\mathbb C^n$ has the property that:
$$\langle x, y \rangle = y^\dagger G x$$
where $G$ is the Gramian matrix, a hermitian positive-definite matrix.

In our case $G$ is the matrix of the Minkowski metric.
In particular that means that the minus signs that come from the Minkowski metric are not part of the conjugate transpose.Furthermore, we have the property that $LL^\dagger$ and $L^\dagger L$ are hermitian positive-definite.
So $\Gamma^\dagger$ must be such that we have this property.
Suppose we pick $\Gamma ^{\dagger} = \begin{bmatrix} \bar A & \bar B & \bar C& \bar D \end{bmatrix}$.
Then:
$$\Gamma^\dagger \Gamma = \begin{bmatrix} \bar A & \bar B & \bar C& \bar D \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = \bar A A + \bar B B + \bar C C + \bar D D$$
(I'm using $\bar A$ for the component wise conjugate instead of $A^*$ to avoid the ambiguity with the conjugate transpose $A^\dagger$.)
We cannot guarantee that this will be hermitian unless for instance the matrices are symmetric.
To ensure the hermitian positive-definite property, I believe we need to pick $\Gamma^\dagger$ such that:
$$\Gamma^\dagger \Gamma = \begin{bmatrix} A^\dagger & B^\dagger & C^\dagger & D^\dagger \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = A^\dagger A + B^\dagger B + C^\dagger C + D^\dagger D$$
At the very least we can be sure that this will be hermitian positive-definite.