Given the following Feynman Amplitude
$$\mathscr{M}=\bar{u_s} (\vec p') \Gamma u_r (\vec p) \ \ \ \ (1)$$
Where
$\bar u_s, u_r$ are Dirac spinors ($1\times 4$ and $4 \times 1$ matrices respectively) and $\Gamma$ 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
$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2 \ \ \ \ (2)$$
Where we've averaged over initial spins (i.e $\sum_r$) and summed over final spins (i.e $\sum_s$)
Defining:
$$\tilde \Gamma := \gamma^0 \Gamma^{\dagger} \gamma^0 \ \ \ \ (3)$$
Where $\dagger$ stands for conjugate transpose.
Prove that $(2)$ can be rewritten as follows
$$X:= \frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p') \Big) \ \ \ \ (4)$$
My proof:
I assumed that
$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2=\frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}$$
Where by $\dagger$ I mean the conjugate transpose. Thus explicitly we get
$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( \bar{u_s} (\vec p) \Gamma u_r (\vec p')\Big)^{\dagger}=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big) \ \ \ \ (6)$$
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
$$\bar{u_s}^{\dagger} (\vec p) := (u_s^{\dagger} \gamma^0)^{\dagger}=\gamma^{0\dagger} u_s \ \ \ \ (7)$$
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
$$u_r^{\dagger} (\vec p') = \bar u_r (\vec p') \gamma^{0} \ \ \ \ (8)$$
Plugging $(7)$ and $(8)$ into $(6)$ we get the desired $(4)$
$$X=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big)=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\bar u_r (\vec p') \gamma^{0}\Gamma^{\dagger}\gamma^{0} u_s(\vec p')=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p') \Big)$$
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
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
$$\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$$