Transition from positive energies to negative energies in Dirac equation

Question

In textbooks, the criticism of the Dirac equation is that it may allow transitions from the positive energy state $m c^2$ to its corresponding negative energy state by releasing one or more photons. It is mentioned that, to satisfy energy-momentum conservation, this transition might happen close to a nucleus. Once this problem is presented, it is usually stated that second quantization is needed to solve this problem, introducing electrons and positrons.

My question is, if we consider the Dirac equation with the electromagnetic field, given an initial state representing a single electron, are transitions from the positive energy states to negative energy states actually possible given the time evolution dictated by the equation ? I learned from quantum mechanics that not all transitions between energy states are possible because there are selection rules that make some matrix elements zero.

My intuition would say that such transitions should not be possible and an electron will stay in the energy level $m c^2$ because no further transitions toward negative energy states can happen, mediated by the electromagnetic field. However, I am not sure this is actually true and I am not currently able to provide a reasoned answer. On the other hand, I have never seen any demonstration that such transitions can actually occur.

Answer 1

The Dirac equation is not flawed.

A single real photon emission in an "free electron -- electron running backwards in time -- transition" (I will come to this in the next paragraph) is not possible since for energy and momentum conservation to hold (and we want to maintain that at all cost) the photon cannot be real (i.e. lightlike), i.e. its 4-momentum square would not be zero, i.e. $c\vec{p}^2 \neq E^2$ ($\vec{p}$ is the momentum and $E$ is the energy of the photon). I assume here an initial free electron (and free positron). Bound electrons are considered below.

However, the photon can be virtual and must immediately engage in another process where it is absorbed -- again under strict conservation of energy and momentum.

Now I will explain what I mean with "free electron -- electron running backwards in time -- transition". Actually, in order to make that happen, one already needs a positron running forward in time in place. Because electrons running backwards in time are observed in nature as positron running forward in time. Then the in-going electron hits the positron and the collision generates a virtual photon which engages in a second process, for instance its absorption by a nucleus.

Actually a bound electron could turn into electron running backwards in time (which means that it hits a forward in time running positron which is already around) and thereby emits a real photon, i.e. a photon fulfilling $c\vec{p}^2 = E^2$. The energy momentum balance would be different because the initial electron is bound (of course some rules apply to make that happen). Therefore in that case the photon could have 4-momentum square zero, i.e. it is a real photon.

Actually electrons hitting positrons already in place thereby generating (real or virtual) photons sounds kind of banal, however in non-relativistic classical Quantum mechanics (QM) this process does not exist, in classical QM an electron could not hit a positron already in place because positrons do not exist in classical QM. They do not exist because they correspond to negative energy states which do not exist in QM.

(notice the fine difference here between "correspond" and "be", actually positrons are not properly negative energy states)

The formalism of second quantization used in QED allows to describe this process in a rather simple manner, that's true. However, the formalism already has included all the considerations above which are called the Feynman-Stueckelberg interpretation. In particular, in classical QM the negative frequency states do not appear in the field operator development, whereas in QFT they do.

Answer 2

Imagine a free electron, in a state with compact support in space. It is accompanied by its electric Coulomb field at infinity, with invariant surface integral $-e$.

Its simply impossible to convert it to a state with surface integral of electric field of $+e$ and the difference is a travelling photon.

This is a super-selection rule by causality. Its impossible to change the electric field simultaneusly on all spheres of radius tending to infinity in a moment of time.

This fact renders the usual quantization non-relativistic procedure of the free electromagnetic field useless in the presence of charges.

The usual dipole approximation of the radiation probability does not work.

All three fields have to be quantized in a relativistic invariant manner in order to get rid of the simple single particle approximation, that is practically a cut-off in energy to momentum modes below $m c.$

Generally, by Haags theorem, the single particle approximations break down, if the representations of the complete, self interacing fields involving potentially infinite intermediating modes, is not unitarily equaivalent to the product representation of statistcally indenpendent, free particles.