What is negative about negative energy states in the Dirac equation?

Question

This question is a follow up to What was missing in Dirac's argument to come up with the modern interpretation of the positron?

There still is some confusion in my mind about the so-called "negative energy" solutions to the Dirac equation. Solving the Dirac equation one finds the spectrum of allowed energies includes both positive and negative solutions. What does this negativity refer to? Given that the Dirac equation is symmetric under charge conjugation, the convention to call one positive and the other negative appears perfectly arbitrary. Would it be therefore correct to refer to the electrons as "negative energy positrons" ?

In a similar spirit, physicists used to be worried about the "negative energy" solutions decaying into infinity through emission of photons. By the same symmetry argument, this should also be a problem for photons. It is not entirely clear to me how the quantization of the field supresses this issue : is the "photon emission" for the negative state re-interpreted as photon absorption by a positron ?

My understanding is that the whole discussion about "positive" or "negative" energy solutions is misleading : what matters is the physical content through the QED interaction hamiltonian, which does not predict this infinite descent. Is this correct?

Edit: I think I understand the source of my confusion after the comments. If I get it right, the Dirac sea picture is equivalent to the freedom of choice in the formally infinite vacuum energy one observes after quantization of the QED Hamiltonian. Holes in the sea are positive-energy positrons, equivalent to the action of the positron creation operator on the vacuum. Is this correct?

Answer 1

Symmetric under charge conjugation (which gives us positrons) and symmetric under the sign of the energy are two different things, which is where I think you are getting confused.

Negative energy electrons aren't positrons, they are negative energy electrons. The absence of a negative energy electron in the "sea of charge" can be viewed as a positive energy particle via charge conjugation, but it doesn't take a negative energy particle into a positive energy particle.

There are two conventions here: positron/electron where there is no such thing as negative energy states, and electrons, where there are positive and negative energy states.

Answer 2

The four-component wave function $\Psi$ in the Dirac equation may be viewed as a counterpart of $\psi(x)$ in non-relativistic Schrödinger's equation. The Dirac equation may be written (and, in fact, was originally written by Dirac) in the Schrödinger's form $$ i\hbar \frac{\partial}{\partial t} \Psi = H \Psi $$ where $H=\vec\alpha\cdot \vec p + m\beta$ where $\beta=\gamma_0$ and $\alpha_i=\gamma_i\gamma_0$ and $\vec p = -i\hbar \nabla$. So the four components are viewed as four different "spin polarizations" of the particle and there exists a Hamiltonian operator which is totally well-defined, including the sign. This operator $H$ has eigenstates with eigenvalues that may be both positive as well as negative. That's what we mean by saying that the energy may be both positive and negative. That's different from the non-relativistic energy operator $p^2/2m$ which is never negative.

The free Dirac equation has solutions for any $E$ obeying $E^2=p_i^2 + m^2$ for both signs of $E$. In fact, the opposite-sign solutions may be obtained by a certain complex conjugation procedure from the positive ones. This charge conjugation effectively exchanges components of $\Psi$ and $\Psi^*$ but they're totally independent and separated in the Schrödinger's equation I started with; that equation only contains $\Psi$.

The negative-energy states would be a problem because the energy of a system would be unbounded from below. In reality, Nature actually finds its lowest energy state and identifies it with the "vacuum". This state has no electrons in the positive-energy solutions of the Dirac equation but the negative-energy solutions/states/boxes are occupied by the maximum occupation numbers allowed by the Fermi-Dirac statistics, namely $N=1$. A missing particle i.e. hole in this "Dirac sea" of negative-energy electrons manifests itself as a positive-energy, positive-charge particle, namely the positron.

At the end, when we want to describe electrons and positrons, we switch to quantum field theory. In that framework, $\Psi$ obeying the Dirac equation is no longer the state vector. It is a field operator. When we want to construct ket vectors in the multiparticle Hilbert space of quantum field theory, we get it by acting on the vacuum by combinations of $\Psi$ components if we want to add electrons or $\Psi^*$ if we want to add positrons. Note that the ket vector maybe obtained both from $\Psi$ and its complex conjugate even though only $\Psi$ was considered to be "the ket" at the beginning of this answer.