In relativistic quantum mechanics, the solution of the free Dirac equation is assumed to be $$\Psi(\textbf{r},t)=u(\textbf{p})e^{i(\textbf{p}\cdot \textbf{r}-Et)}$$ How do I know that this is the most general separable solution?

I was trying to derive this explicitly by the method of separation of variables as $$\Psi(\textbf{r},t)=u(\textbf{p})\Phi(\textbf{r})T(t)$$ Substituting, this into the free Dirac equation I trivially obtained $$T(t)\sim e^{-iEt}$$ but I cannot solve the space-part of the equation (in a representation, I used the Dirac Pauli representation). The equation I'm stuck with is: $$-i\frac{\alpha\cdot\nabla\Phi}{\Phi}u+\beta mu=Eu$$ where $E$ is the separation constant (which is dimensionally the energy). How to solve this part to show that $\Phi\sim e^{i\textbf{p}\cdot \textbf{r}}$. Any suggestion in this regard will be helpful.

$\bullet$ I have written the equation in natural units i.e., $c=\hbar=1$.


1 Answer 1


The Dirac operator $H=-\alpha\cdot i\nabla+\beta m$ is self-adjoint on $L^2(\mathbb{R}^3,\mathbb{C}^4)$. Therefore you may write a general solution as $\psi(t,x)=e^{-itH}\psi_0$, $\psi_0\in L^2(\mathbb{R}^3,\mathbb{C}^4)$.

The problem with explicit solutions of the "eigenvector type" (for an energy $E$) is that they cannot belong to the $L^2$ space, for the spectrum of the operator is continuous. While the eigenvalue equation (in a suitable space that is not $L^2$) is easy to solve explicitly for the Laplacian and gives the usual plane waves, it is more involved for the Dirac operator, so the physicists make the assumption you wrote.

Nevertheless, the most general solution in the relevant physical space $L^2$, as I said above, is perfectly well defined mathematically to be $e^{-itH}\psi_0$, given $\psi_0\in L^2$. (it is not so explicit, obviously, but surely well-defined)


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