I'm trying to recover the Einstein energy-momentum relation from the Dirac equation. I'm given a solution wavefunction,

$$\psi = u(E,\vec p) e^{i(\vec p\cdot\vec x - Et)}$$ with $$\vec u = N\begin{pmatrix}1\\0\\ \frac{p_z}{E+m} \\ \frac{p_x + ip_y}{E+m}\end{pmatrix}$$

The equation is given to me as $$(\gamma^\mu p_\mu - m)u=0$$

The exercise says that I should verify that when $u$ is substituted into the equation, I should get out the energy-momentum relation.

My question is not really how to do this, but more how to read the equation. I know that $\gamma^\mu$ is a $(4\times 4)$ matrix that I can look up. But what is ment by $\gamma^\mu p_\mu$? From my understanding of Einstein notation you should read this as a sum when the same index is used twice, but a sum over what? Would I not end up with $$\sum_{i=0}^{3} \gamma^i p_i = \text{Sum of }(4\times 4)\text{matrices }$$
because $p_i$ is a scalar and $\gamma^i$ is a matrix, and then need to subtract by the scalar $m$, which is not defined?

What I would really like to do is to write it all out in its full form. The notation is probably very handy when you know what it means, but I would like to just do this the most explicit way possible. After that I think it would be easier to go back to the shorthand notation.

I fear that this confusion roots in that I have never understood what is the difference between contra- and covariant vectors, $v^\mu$ vs. $v_\mu$. Is one a row vector and the other a column vector? (I have tried searching and reading all that I can find, but I seem to get no closer to understanding this).

Sorry for all the confusion, hoping someone of you can help clear some of it up!

  • $\begingroup$ The special-relativity tag has a few very good answers about index notation. I'll see if I can find them. $\endgroup$ – Javier Mar 12 '16 at 17:11
  • $\begingroup$ Here are a couple: physics.stackexchange.com/q/105347 physics.stackexchange.com/q/91593 physics.stackexchange.com/q/226531 Please read the answers and tell us if you have more questions. $\endgroup$ – Javier Mar 12 '16 at 17:26
  • $\begingroup$ And thanks @AccidentalFourierTransform, should have thought about that in the first place. Still not comfortable with the significance of a superscript index vs. subscript index, but for the purposes of this exercise I got it to work now! $\endgroup$ – Bendik Mar 12 '16 at 17:52

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