I'm trying to understand the plane wave solutions of the Dirac Equation. But I'm still a newbie on indices notation and contravariant and covariant objects. What I don't understand is how to get:

$$(\gamma^{u}p_u - m) u(p)=0$$

Putting $\psi = u(p)e^{-ipx}$ in the Dirac equation:

$$(i\gamma^{u}\partial_u - m) \psi=0$$

I'm assuming $\hbar=c=1$ and

$$\begin{aligned} \partial_u&= (\partial/ \partial t ,\nabla )\\ p^u &=(E,p)\\ i\frac{\partial \psi}{\partial t} &=E \end{aligned} $$

When I do the derivative and multiply by $i$ I get:

$$(\gamma^{0}E + \gamma^{k}p - m) u(p)=0 \qquad \text{with}\quad k=1,2,3$$

Which is equal to $(\gamma^{u}p^u - m) u(p)=0$ but I know this doesn't make sense. What did I do wrong?

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    $\begingroup$ the rule of thumb is to never split time and space indices. Writing $\partial_\mu=(\partial_t,\nabla)$ is in general a bad idea. In this case, all you need is $\partial_\mu \mathrm e^{-ipx}=-ip_\mu \mathrm e^{-ipx}$. Easy peasy! $\endgroup$ Apr 15, 2016 at 17:59
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    $\begingroup$ Ok, interpreting px as $p_{u}x^{u}$ I got what you said and the right result. Also, converting it to classical terms (lol) $-i(p_{u}x^{u})= -i(Et-\vec p \cdot \vec x)$ what gives the right expanded result. I guess what I was doing wrong was doing the derivatives carelessly of the type of objects in the exponential argument. $\endgroup$ Apr 15, 2016 at 18:54
  • $\begingroup$ Yep. You could answer to your own question below so that this post becomes useful to future users that may have the same problem as you :-) $\endgroup$ Apr 15, 2016 at 19:13

1 Answer 1


So, the problem was I was thinking they were considering a case with only momentum in x, without loss of generality, such that $\vec p\cdot \vec r = p\cdot x$.

That is not the case. It's only a notation used in some papers and books, where they define $ e^{-ip_u x^u} \equiv e^{-ipx}$. Expanding it you get $−i(p_ux^u)=−i(Et−p⃗ ⋅\vec r )$ what gives the right and expected result. The other way you get a wrong minus sign because $p_x\cdot x= - p_1x^1$ (!!!)

So, doing $\partial_u ( u(p)e^{-i(p_ux^u)} ) = -ip_u u(p)e^{-i(p_ux^u)}$ and applying it in the Dirac Equation:

$$(\gamma^{u}p_u - m) u(p) =0$$

You get the Dirac Equation in the momentum representation.

The lesson is : don't mix covariant notation with old classical notation!


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