Derivation of operator version of the classical wave equation I have the following summarised derivation for the operator version of the classical wave equation for massless and material particle.


My question is about the statement:


*

*However, a problem is that in the probabilistic interpretation of its solution as representing a single particle.

*Difficulty comes from the two possible signs of the square root operations: $E=±\sqrt{\left(pc\right)^2+\left(mc^2\right)^2}$.

*This can be shown to give rise to antiparticles which must be included for self-consistency.


Wouldn't the same problem occur for massless particles as well as it is still to the $E^2$?
 A: There is no unique choice of $\hat{E}$ in the first place.
You are correct in that there is a flaw in the argument from the very beginning as follows:
We know the solution for monochromatic electromagnetic plane waves as $\sim \exp(\pm i(\omega t - kx))$, not only $\sim \exp(- i(\omega t - kx))$! Furthermore, from the photoelectric effect we know that $E = \hbar \omega$. That means that we might try to define the energy operator as $\hat{E} = \pm i \hbar \partial_t$.
However, there is a problem with both of these signs from the very beginning, since we deal with real waves in electromagnetism. By choosing one fixed sign for $\hat{E}$ (and thus also $\hat{p}$), a real plane wave $\sim \exp(i(\omega t - kx)) + \exp(-i(\omega t - kx))$ would be interpreted as a superposition of particles with both positive and negative energies. Consequently, for any sign choice of $\hat{E}=\pm i \hbar \partial_t$, every real wave has zero energy!!! That is obviously unacceptable, since we derived the form of the operator from observations about real waves (the photoelectric effect).
But notice the following. If we apply $\hat{E}^2 = -\hbar^2 \partial_t^2$, it does give the right answer for a real wave. In other words, we seem to know what $\hat{E}^2$ should be, but we have no idea what $\hat{E}$ should be from the very beginning.
That is, the excerpt in the OP is misleading in omitting that the choice for the unsquared energy operator is simply unknown at the outset, be it for massive or massless particles.

Negative mass-energy currents - the historical misunderstanding
This being said, the real problem with massive particles as compared to massless ones was the following. For electromagnetism, it was kind of obvious that the $T^{00}$ component of the EM stress-energy tensor
\begin{align}
T^{00} = \frac{1}{\mu_0} \left(F^{0\kappa}F^0{}_\kappa + F^{\mu\nu}F_{\mu\nu} \right) \propto E^2 + B^2
\end{align}
is somehow proportional to the total energy density of the photons and it automatically stays positive definite. This also has a nice conservation law $T^{0\mu}{}_{,\mu} = 0$, i.e., $T^{00}$ is the zeroth component of a conserved current.
However, people were used to the complex Schrödinger equation for massive particles. In fact, they knew that the complex nature of this equation is what makes it possible to express the dynamics of, e.g., an electron in the Hydrogen atom. However, if you try to find a conserved current for the equation $(\hbar^2 \Box + m^2)\phi =0$ that reduces to the conserved matter-density current, you would only find a current that exhibited both positive and negative matter densities!
Now in retrospect we know that the conserved "matter-density current" of the Schrödinger equation is in fact the non-relativistic limit of a charge-density current. Since in the non-relativistic limit, the energies of motion are such that a single electron always stays a single electron, the charge per mass $m_e/e$ always stays the same. So we were able to multiply the charge current with $m_e/e$ and simply use it as an equivalent of a density current. However, in the relativistic case the electron can in principle be superposed with positrons, which have opposite charge $-e$. By multiplying the conserved charge current by $m_e/e$, we then got seemingly negative mass (or probability-of-occurence) densities. There is, however, a stress-energy tensor of the scalar field $T^{\mu\nu}$ which does have a (classically) positive-definite $T^{00}$ with the meaning of mass-energy density and that has been ignored because it would not correspond as easily to the Schrödinger-equation current.
So that is the whole deal with massive particles.
(P.S.: I kind of skip the part where electrons have spin and their Schrödinger equation is the non-relativistic limit of the Dirac equation. But the gist is the same.)
