The plane wave solution for the Dirac equation for a free particle is of the form $ \psi = U e^ {i (xP_x + yP_y + zP_z -Et)/\hbar}$, where $\psi$ is the Dirac spinor with four components and $U$ is a constant four-dimensional vector in space so that $U^T U$ is the probability density. However, if we integrate the probability density over all space, the result is greater than 1; in fact the usual lecture notes give $2E$ particles per unit volume which is certainly greater than 1. If we wish to consider one particle in the whole space so that the integrated probability density over all space is 1, we will need to let $U$ to vary over space so that $U$ at infinity is 0. However, I cannot find such a solution to the Dirac equation; such a solution probably does not exist. But this raises the question of the capacity of the Dirac equation to accommodate the physical scenario of one single free particle in space.

  • $\begingroup$ Do you normalise the plane wave solution? Its total energy should be $\hbar \omega $. $\endgroup$
    – my2cts
    Dec 27, 2019 at 18:39
  • $\begingroup$ My concern is not about the energy. My concern is that with $U$ constant in space the integrated probability density over infinite space will be infinite. Hence, it does not represent the scenario of one particle where the integrated probability density over infinite space is one. $\endgroup$
    – Damon
    Dec 27, 2019 at 19:09
  • $\begingroup$ You are overlooking the fact that the energy and other expectation values depend on the normalisation. $\endgroup$
    – my2cts
    Dec 27, 2019 at 22:05


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