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So we have the commonly quoted momentum space version of the Dirac equation and the adjoint Dirac equation:

$$ (\gamma^{\mu}p_{\mu}-m)u=0 $$ Often, we are asked to show that the adjoint momentum Dirac equation can be written as: $$ \bar{u}(\gamma^{\mu}p_{\mu}-m)=0 $$ I'm not too sure on the method. However, I have attempted something.

I multiply the top equation by $\bar{u}\gamma^{\nu}\times$ and the bottom equation by $\times\gamma^{\nu}u$ giving:

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

$$ \bar{u}(\gamma^{\mu}p_{\mu}-m)\gamma^{\nu}u=0 $$ Taking the sum of these two gave me: $$ \bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u-\bar{u}\gamma^{\nu}mu+\bar{u}\gamma^{\mu}p_{\mu}\gamma^{\nu}u-\bar{u}m\gamma^{\nu}u=0 $$ I then use $\gamma^{\mu}\gamma^{\nu}=-\gamma^{\nu}\gamma^{\mu}$, giving me: $$ \bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u-\bar{u}\gamma^{\nu}mu-\bar{u}\gamma^{\nu}\gamma^{\mu}p_{\mu}u+\bar{u}\gamma^{\nu}mu=0 $$ And which point I can say that it is true. I'm just wondering if this is sufficient or if there is another more correct way of getting the adjoint Dirac equation...

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

Using the property $\overline{AB}=\bar{B}\bar {A}$, we have the following: $$ 0=\overline{(\gamma^\mu p_\mu-m)u}=\bar u \overline{(\gamma^\mu p_\mu-m)} $$

Now, use $\overline{A+B}=\bar A+\bar B$: $$ 0=\bar u (\overline{\gamma^\mu p_\mu}-\bar m) $$

Next, as both $m$ and $p_\mu$ are real numbers, we have $\bar m=m$ and $\bar p_\mu=p_\mu$: $$ 0=\bar u (\overline{\gamma^\mu} p_\mu- m) $$

Finally, use the fact that the gamma matrices are self-adjoint, that is, $\bar \gamma^\mu=\gamma^\mu$: $$ 0=\bar u (\gamma^\mu p_\mu- m) $$ and we are done.

Your attempt cannot work because you are trying to prove the adjoint equation by using the adjoint equation, which means that your argument is cyclical.

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