Proving an equation built up out of Dirac-$\gamma$ matrices

Question

Given the following Feynman Amplitude:

$$\mathscr{M}=\bar{u_s} (\vec p') \Gamma u_r (\vec p) \tag 1$$

Where:

$\bar u_s, u_r$ are Dirac spinors ($1\times 4$ and $4 \times 1$ matrices respectively)

$\Gamma$ is a $4\times4$ matrix containing Dirac-$\gamma$-matrices (which are of course $4\times4$ matrices)

We're also given the (unpolarized cross-section) formula:

$$X= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2 \tag 2 $$

Where we've averaged over initial spins (i.e $\sum_r$) and summed over final spins (i.e $\sum_s$)

Defining:

$$\tilde \Gamma = \gamma^0 \Gamma^{\dagger} \gamma^0 \tag 3 $$

Where $\dagger$ stands for conjugate transpose.

Prove that $(2)$ can be rewritten as follows:

$$X= \frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big ) \Big( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p') \Big) \tag 4 $$

My proof:

I assumed that

$$X= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2=\frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}\tag 5$$

Where by $\dagger$ I mean the conjugate transpose. Thus explicitly we get:

$$X= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}$$ $$=\frac 1 2 \sum_{r=1} \sum_{s=1}\Big ( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big ) \Big( \bar{u_s} (\vec p) \Gamma u_r (\vec p')\Big)^{\dagger}$$ $$=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big )\Big( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big) \tag 6$$

Now it is about working out $\bar{u_s}^{\dagger} (\vec p)$ and $u_r^{\dagger} (\vec p')$

We know that the adjoint is, by definition:

$$\bar{u_s} (\vec p) = u_s^{\dagger} \gamma^0 $$

Taking $\dagger$ on both sides of such equation we get:

$$\bar{u_s}^{\dagger} (\vec p) = (u_s^{\dagger} \gamma^0)^{\dagger}=\gamma^{0\dagger} u_s \tag 7$$

Where :

$$\gamma^{0\dagger}=\gamma^{0}$$

And here comes the key step: I assumed that $\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$ Thus we get

$$u_r^{\dagger} (\vec p') = \bar u_r (\vec p') \gamma^{0} \tag 8$$

Plugging $(7)$ and $(8)$ into $(6)$ we get the desired $(4)$

$$X=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big ( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big ) \Big ( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big)$$ $$=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big ( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big )\bar u_r (\vec p') \gamma^{0}\Gamma^{\dagger}\gamma^{0} u_s(\vec p')$$ $$=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p)\Big )\Big ( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p') \Big)$$

Note that in my proof I assumed $\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$.

Do you agree with it? If yes, why $\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$ is OK?

Source: Second edition Mandl & Shaw, QFT page 132

Answer 1

Thanks to innisfree I got it!

We prove $\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$ is true based on the anticommutation relation (for instance, see page 452; Mandl & Shaw)

$$\{\gamma^\mu,\gamma^\nu\}=2g^{\mu \nu}$$

Using the convention $(+, -, -, -)$ for the Minkowski metric, we get for $\mu=0, \nu=0$

$$\Big(\gamma^0\Big)^2=1$$

Which implies

$$\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$$