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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?

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The choice of polarization is arbitrary (in the same sense as a choice of basis is arbitrary - the most convenient (under the circumstances) basis is chosen). It is called polarization because it describes spin projection on some axis. Remember that light polarization is related to photon spin projection.

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  • $\begingroup$ Thank you. I actually don't know that light polarization is related to the spin of the photon, I only studied it from a classical point of view. How advanced is the topic? Do I need QFT or I can find references where this is explained in the good old QM framework? $\endgroup$
    – RenatoRenatoRenato
    Commented Dec 14, 2016 at 17:08
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    $\begingroup$ @Runlikehell: Depending on how much detail you need. You don't need QFT for that. You may look at en.wikipedia.org/wiki/Photon_polarization or google: photon polarization spin $\endgroup$
    – akhmeteli
    Commented Dec 14, 2016 at 17:19
  • $\begingroup$ The choice of polarization is not arbitrary. The polarization vector is a spacelike vector orthogonal to the four momentum. See Greiner's relativistic quantum mechanics book page 175. Also I think the OP is asking about the polarization vector in Dirac's equation. $\endgroup$
    – Simplyorange
    Commented Aug 11, 2022 at 20:54
  • $\begingroup$ @Simplyorange : So you don't think one can choose spinors for electrons polarized in the $x$ direction, rather than the $z$ direction? $\endgroup$
    – akhmeteli
    Commented Aug 12, 2022 at 1:36

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