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

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

• What’s $\gamma_0^2$? Jun 1, 2020 at 19:34
• @innisfree hi, I guess you mean we can use a property of the Dirac $\gamma$-matrices to get such a result. All I know for sure about $\gamma^0$ is the following: $\gamma^{0\dagger}=\gamma^{0}$ and $\gamma^{\mu \dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}$. May you please be more explicit? Jun 1, 2020 at 19:58
• Do you know what $\{\gamma^\mu,\gamma^\nu\}$ equals? Look at the 00 component. Jun 1, 2020 at 21:29
• @innisfree ahhh thank you, I think I got it! Based on the anticommutation relation $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu \nu}$ and with the convention $(+, -, -, -)$ for the Minkowski metric we get $\Big(\gamma^0\Big)^2=1$, which indeed proves $\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}$ :)) Do you agree? Jun 1, 2020 at 21:42
• Very good, now I encourage you to answer your own question here. You could just copy & paste your above comment, or write something longer. Jun 2, 2020 at 0:19

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

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