# Solution of Dirac equation: arbitrary polarizations

In my lecture notes (signature $-+++$) we find the free Dirac equation solutions. We proceed in this way:

Dirac equation:

$$(i\,\displaystyle{\not} p +m)\psi(x) = 0$$

We make the following ansatz:

$$\psi_p(x) \sim \omega(p) e^{ip\mu x^\mu}$$

with the polarization

$$\omega(p) = \left( \begin{array}{c} w_1(p)\\ w_2(p)\\ w_3(p)\\ w_4(p)\\ \end{array} \right)$$

For a particle at rest: $p^\mu=(E,0,0,0)$, inserting the ansatz we obtain

$E\, \omega(p) = m\beta\, \omega(p)$ $\, \, \,$ with $\beta = i \gamma^0 = \,\left(\begin{matrix} \mathbb{1}_2 & 0 \\ 0 & -\mathbb{1}_2 \end{matrix}\right)$

therefore $$E \,\omega(p) \, = \, \left(\begin{matrix}m & 0 & 0 & 0 \\ 0 & m & 0 & 0 \\ 0 & 0 & -m & 0 \\ 0 & 0 & 0 & -m\end{matrix}\right) \omega(p)$$

Then it says that we got two solutions $\psi_{1,2}$ for positive energy $E=m$ and two solutions $\psi_{3,4}$ for negative energy, which are

$\psi_1(x) \sim \left( \begin{array}{c} 1\\ 0\\ 0\\ 0\\ \end{array} \right) e^{-imt} \, ;\quad \psi_2(x) \sim \left( \begin{array}{c} 0\\ 1\\ 0\\ 0\\ \end{array} \right) e^{-imt} \, ;\quad \psi_3(x) \sim \left( \begin{array}{c} 0\\ 0\\ 1\\ 0\\ \end{array} \right) e^{imt} \, ;\quad \psi_4(x) \sim \left( \begin{array}{c} 0\\ 0\\ 0\\ 1\\ \end{array} \right) e^{imt}$

Now what I dont' get is the choice of the polarizations, is it arbitrary or does it follow from something in the discussion above? Why they have that form?

Thank in advance for any answer, I know this may be a stupid question but i just don't get it.

And another question, why $\omega$ is called polarization? Is it related with the polarization of the an electromagnetic wave?