What are the theoretical / mathematical problems in discarding negative solutions of Dirac equation?

I read some Q&A about it, but my question is why Dirac was so sure that he could not discard negative energy solutions.

It seems so natural that energy must be positive, that I suppose that if we use only positive solutions we get some theoretical problems. The plane wave $$\psi = e^{-ip_{\mu}x^{\mu}}$$ is a solution of the Dirac equation if $$p_0^2 = E^2 = |\mathbf p|^2 + m^2$$. What comes from the relativistic invariance of the mass: $$E^2 - P^2 = m^2$$. And nobody thinks of negative energies when looking at that equation in special relativity. Moreover, he had to deal with the strange notion of an infinite sea of electrons.

Of course, positrons were discovered soon after his work, and gave experimental support to not discard them.

But besides the experimental confirmation, are there any theoretical problems if we discard them?

• In Quantum Mechanics, everything that is not forbiden is compulsory. en.wikipedia.org/wiki/Totalitarian_principle – Cham Mar 16 '20 at 20:06
• – Cham Mar 16 '20 at 20:08
• Everything not forbidden is compulsory. One example of forbidden thing is infinity, Solutions that blow up are discarded without regret. Are not negative energies so strange as infinities? – Claudio Saspinski Mar 16 '20 at 21:56
• They are not negative energy but negative frequency solutions. – my2cts Mar 16 '20 at 22:13
• @my2cts, negative frequency and negative energy are the same. Both four-vectors ($\boldsymbol{p}$ and $\boldsymbol{k}$) are pointing to the observer's past (or backward traveling in spacetime). – Cham Mar 16 '20 at 22:19

The problem is that Dirac equation can't be written as two equations, where one would only refer to positive-energy components, and the other to the negative ones. E.g. $$\partial_t\psi_1$$ component depends on $$\psi_3$$ and $$\psi_4$$ in the equation. The result is that, if you find the general solution of the equation, you'll see that for nonzero momenta the components are intermixed, and you only get pure positive/negative solutions for particle at rest (see this post for explicit solutions).

All this makes rejection of negative energy solutions not only "physically unwanted", but mathematically impossible.

Besides the experimental confirmation, are there any theoretical problems if we discard the negative energy states?

One major problem is that quantum Hamiltonians should be bounded from below to guarantee stability of the system. This principle is what ensure stability of matter in condensed matter.

The reason is that, according to statistical mechanics, at equilibrium (and at zero temperature) the system goes in the state of minimum energy. So clearly we must require such minimum to exist.

Dirac's theory of antiparticles solves this problem too.

Ruslan is correct, and there are also further problems when interactions are introduced. The time ordered product in the Dyson expansion means that consistency under Lorentz transformation imposes that the locality condition on the field operators, that (anti-)commutators are zero outside the light cone. This condition is also needed to ensure that no observable effect can propagate faster than the speed of light. Satisfying it requires field operators which annihilate particles and create particles, as part of a single physical process.

This is best understood. imv, using the Stückelberg-Feynman interpretation. Energy is the time component of the energy momentum vector. Negative energy refers to a particle going back in time. The creation of a negative energy particle going back in time is logically the same as the annihilation of a positive energy antiparticle, which is of course what we actually observe.