What was missing in Dirac's argument to come up with the modern interpretation of the positron?

Question

When Dirac found his equation for the electron $(-i\gamma^\mu\partial_\mu+m)\psi=0$ he famously discovered that it had negative energy solutions. In order to solve the problem of the stability of the ground state of the electron he invoked Pauli's exclusion principle and postulated that negative energy states well already filled by a "sea" of electrons. This allowed him to predict the positron, viewed as a hole in the sea.

This interpretation was ultimately discarded owing to its inapplicability to bosons and difficulties with explaining the invisibility of the infinite charge of the sea.

According to my understanding, the modern argument goes something like this. There is a discrete symmetry of the lagrangian called charge conjugation $\psi \rightarrow \psi^c$ which allows the negative energy solutions to be interpreted as positive energy solutions for a second mode of excitation of the electron field with opposite charge, called positrons. The decay of electrons to positrons is then suppressed by the $U(1)$ gauge symmetry of the lagrangian forcing conservation of electrical charge.

According to this interpretation What Dirac would have missed was the lagrangian formalism. Is this historically and physically correct?

Answer 1

Dirac's derivation of the existence of positrons that you described was a totally legitimate and solid argument and Dirac rightfully received a Nobel prize for this derivation.

As you correctly say, the same "sea" argument depending on Pauli's exclusion principle isn't really working for bosons. Modern QFT textbooks want to present fermions and bosons in a unified language which is why they mostly avoid the "Dirac sea" argument. But this fact doesn't make it invalid.

The infinite potential charge of the Dirac sea is unphysical. In reality, one should admit that he doesn't know what the charge of the "true vacuum" is. So there's an unknown additive shift in the quantity $Q$ and of course that the right additive choice is the choice that implies that the physical vacuum $|0\rangle$ (with the Dirac sea, i.e. with the negative-energy electron states fully occupied) carries $Q=0$. The right choice of the additive shift is a part of renormalization and the choice $Q=0$ is also one that respects the ${\mathbb Z}_2$ symmetry between electrons and positrons.

It is bizarre to say that Dirac missed the Lagrangian formalism. Dirac was the main founding father of quantum mechanics who emphasized the role of the Lagrangian in quantum mechanics. That's also why Dirac was the author of the first steps that ultimately led to Feynman's path integrals, the approach to quantum mechanics that makes the importance of the Lagrangian in quantum mechanics manifest.

It would be more accurate to say that Dirac didn't understand (and opposed) renormalization so he couldn't possibly formulate the right proof of the existence of the positrons etc. that would also correctly deal with the counterterms and similar things. Still, he had everything he needed to define a consistent theory at the level of precision that was available to him (ignoring renormalization of loop corrections): he just subtracted the right (infinite) additive constant from $Q$ by hand.

Your sentence

The decay of electrons to positrons is then supressed by the U(1) gauge symmetry of the lagrangian forcing conservation of electrical charge.

is strange. Since the beginning – in fact, since the 19th century – the U(1) gauge symmetry was a part of all formulations of electromagnetic theories. It has been a working part of Dirac's theory from the very beginning, too.

The additive shift in $Q$, $Q=Q_0+\dots$, doesn't change anything about the U(1) transformation rules for any fields because they're given by commutators of the fields with $Q$ and the commutator of a $c$-number such as $Q_0$ with anything vanishes: $Q_0$ is completely inconsequential for the U(1) transformation rules. All these facts were known to Dirac, too. The fact that the U(1) gauge symmetry was respected was the reason that there has never been anything such as a "decay of electrons to positrons" in Dirac's theory, not even in its earliest versions.

An electron can't decay to a positron because that would violate charge conservation while the charge has always been conserved. For historical reasons, one could mention that unlike Dirac, some other physicists were confused about these elementary facts such as the separation of 1-electron state and 1-positron states in different superselection sectors. In particular, Schrödinger proposed a completely wrong theory of "Zitterbewegung" (trembling motion) which was supposed to be a very fast vibration caused by the interference between the positive-energy and negative-energy solutions. However, there's never such interference in the reality because the actual states corresponding to these solutions carry different values of the electric charge. Their belonging to different superselection sectors is the reason why the interference between them can't ever be physically observed. The "Zitterbewegung" is completely unphysical.

Answer 2

Positrons aren't negative energy solutions in the first place. This is quite evidently shown by the ~1 MeV of energy of the photons emitted upon annihilation. If positrons had negative energy then the remaining energy would be zero.

So there is no need at all to explain the "suppression of the decay of electrons into positrons" with whatever argument.

But OK, Without his Wheelerian ideas Feynman would have had much more trouble to set up the right Feynman diagrams. After all, In the diagrams electrons do not interfere with other electrons but they do interfere with any positron. This seemingly inconsequential behavior was only accepted based on the notion that the positron was the electron itself being reversed back in time.

Hans

Answer 3

Dirac argument was make before modern proofs of the CPT theorem in quantum field theory.

The trick of removing the sea of negative energy states, and a replacing it with a positive energy state with its charges and spin reversed, is the Dirac argument above, and argument that essentially evolved in the CPT theorem.

Feynman then considered the possibility of time reversed states, in the Feynman Wheeler absorber theory. That allowed modern physicists to remove the negative energy states, and replace them with Time-symmetry reversed states. Finally T = CP brought the positron back.

A sea of negative energy states wouldn't work with Bosons, but removing negative energy states with CPT still does.

Answer 4

$\let \a=\alpha \let\b=\beta \let\eps=\varepsilon \def\vp{\vec p} \def\va{\vec\a}$ IMHO the right approach to original Dirac's theory is to interpret it as a one particle theory. In this approach it makes no sense to speak of antiparticles etc. Surely this was not entirely clear to Dirac, but after all we are under obligation to understand something more, after 90 years and after so much work of so many researchers.

A modern approach to Dirac's theory as a one particle theory should be to say: OK, Dirac's equation also allows for negative energy solutions (after all, it isn't the only one: Klein-Gordon equation suffers the same flaw). This simply means that we have to consider only the Hilbert subspace belonging to eigenvalue +1 of the observable $\eps$ (sign of energy). Note that $\eps$ commutes with Dirac's hamiltonian. In fact it can be defined as $$\eps = H E^{-1}$$ where $H=\va\cdot\vp+\b\,m$ is the Hamiltonian, and $E=\sqrt{p^2+m^2}$.

This works for a free electron, and also for an electron in a static potential, like in hydrogen atom. In fact, if you follow Dirac's calculation of relativistic corrections to the H-atom, you see that he implicitly makes the choice of positive energy eigenvalues.

Difficulties arise when the electron is subject to a varying field (e.g. an e.m. wave) but this is no surprise: QED treatment is required to deal with that problem.

A final word about Zitterbewegung (of position and of spin). It is known since Foldy's paper (1956, I believe) that the solution is simply that $x$ and $\sigma$ observables in Dirac theory are not physical observables for the one-particle theory, as they do not commute with $\eps$. Foldy gave a unitary transformation bringing them to acceptable observables.

We again get into trouble however when e.m. field enters the game, since the right Hamiltonian must be written with fields as functions of the original Dirac $x$'s, which have matrix elements between states with positive and negative energy. Of course we are confronted anew with the need of second-quantizing the Dirac field.

Answer 5

I think there is already a good discussion here so I won't add to it, but I do want to underline that the phrase "decay of electrons to positrons" is in the original question is misleading. According to the Dirac sea picture, "positron" is the name for the physical scenario of an unfilled state or "hole" in the negative-energy sea. There is no question of "decay" in either direction between electrons and such "holes" (though pair creation and annihilation are of course possible). In the QFT picture the notion of decay is also questionable when the two entities have the same mass (whether or not they have the same charge).