# Adjoint Dirac equation (in momentum form) from Dirac eq in momentum form method

I just wanted to check the method I have formulated for the derivation for the adjoint Dirac equation using Gamma matrice notation. This is a problem from the very excellent "Modern Particle Physics" by Mark Thomson book. Incidentally, I fully recommend purchasing this book if you are studying at either Cambridge, Oxford or UCL. Very good book, pedagogically speaking.

So I start from the familiar Dirac equation in momentum form:

$$(\gamma^{\mu}p_{\mu}-m)u=0$$

I take the Hermitian of this, $^{\dagger}$: $$(\gamma^{\mu}p_{\mu}-m)^{\dagger}u^{\dagger}=0$$

I multiply $\times\gamma^{0}$ of this: $$(\gamma^{\mu}p_{\mu}-m)^{\dagger}u^{\dagger}\gamma^{0}=0$$ Noting that $m^{\dagger}=m$ and $(\gamma^{\mu})^{\dagger}=-(\gamma^{\mu})$ I get the following: $$(-\gamma^{\mu}p_{\mu}-m)u^{\dagger}\gamma^{0}=0$$ This is where it gets a little sketchy, apologies for the colloquialism. I go to here: $$u^{\dagger}\gamma^{0}(\gamma^{\mu}p_{\mu}+m)=0$$ I feel this is a little bit of an illegal step. But this comes out as, using the definition of the adjoint spinor $\bar{u}=u^{\dagger}\gamma^0$: $$\bar{u}(\gamma^{\mu}p_{\mu}+m)=0$$ If anyone could suggest a better method to swop the ordering of the $u^{\dagger}\gamma^{0}$ rather than just fudging it that would be very helpful!

Many thanks!

In the second step, when you take the hermitian conjugate, you should have $$u^\dagger(\gamma^{\mu}p_{\mu}-m)^{\dagger}=0\tag{1}$$ instead of $$(\gamma^{\mu}p_{\mu}-m)^{\dagger}u^{\dagger}=0\tag{2}$$ because $(AB)^\dagger=B^\dagger A^\dagger$.
Also, you wrote $$(\gamma^\mu p_\mu-m)^\dagger=(-\gamma^{\mu}p_{\mu}-m)\tag{3}$$ which is wrong. The correct result is $$(\gamma^\mu p_\mu-m)^\dagger=(\gamma^0\gamma^{\mu}\gamma^0p_{\mu}-m) \tag{4}$$ because $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu \gamma^0$.
• Perfect, yes. We were told to derive $(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}$ in a previous question. This makes a lot more sense now, I will review my derivations in light of this and check this off as the correct answer once the time has elapsed. Thank you! – DarthPlagueis Apr 23 '16 at 13:18
• @DarthPlagueis you probably got confused because $\gamma^{i\dagger}=-\gamma^i$, but this only works for $i=1,2,3$. On the other hand, $\gamma^{0\dagger}=+\gamma^0$. – AccidentalFourierTransform Apr 23 '16 at 13:19
• Absolutely. I think had I also known about $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$ too, then this would have helped! Really pleased, very much appreciated! – DarthPlagueis Apr 23 '16 at 13:20
• Just to follow up, what do I then do with $-\gamma^{0}\gamma^{\mu}$ after taking on of the $\gamma^{0}$'s out...? And then isn't the $\gamma^{0}$ acting on $m$ as well? is this what flips the sign? – DarthPlagueis Apr 23 '16 at 13:35