Dirac Equation Charge Density

[jb35len]

I am confused about the coupling of the Dirac equation to electromagnetism. The 4-current that is the source for Maxwell's equation that arises from the Lagrangian
\begin{equation}
\mathcal{L}=i\overline{\psi}\gamma^\mu(\partial_\mu+ieA_\mu)\psi-m\overline{\psi}\psi
\end{equation}
is $j^\mu=e\overline{\psi}\gamma^\mu\psi$.

This $j^\mu$ will give a charge density $j^0$ that has the same sign as $e$everywhere. Given that the positive and negative energy components of $\psi$ are supposed to represent particles and antiparticles respectively, if $\psi$ has only positive or only negative energy parts then the sign of $e$ can be chosen appropriately. However, if $\psi$ is a general superposition of positive and negative energy components then what are you supposed to do?

I was under the impression that the fact that particles and antiparticles have opposite sign for their charge was a consequence of the theory and not something that is put in by hand. Is this something that can be seen at the level of the classical fields or does it require quantization before you can see that the particle and antiparticle components have opposite sign of charge?

 

[stevendaryl]

Given that the positive and negative energy components of $\psi$ are supposed to represent particles and antiparticles respectively, if $\psi$ has only positive or only negative energy parts then the sign of $e$ can be chosen appropriately.

It's not exactly true that the negative energy components represent anti-particles. In Dirac's initial interpretation of his theory, an anti-particle was a "hole" in an otherwise filled sea of negative energy states. The negative energy states have the same charge as the positive energy states, but a hole in the negative energy states has the opposite charge.

In the modern way of doing things, it is handled in quantum field theory by reinterpreting once again. The annihilation operator for negative energy states is interpreted as a creation operator for positive energy anti-particles. The creation operator for negative energy states is interpreted as an annihilation operator for positive energy anti-particles. In this reinterpretation, all particles have positive energy, but some particles have opposite charge.

 

[jb35len]

Thank you. I'm still a little confused.

On page 9 in the notes http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px435/dirac.pdf
it talks about the Feynman-Stuckelberg interpretation, where negative energy solutions represent positive energy antiparticles going forward in time. Is this related to your explanation?

Does the fact that these should have the opposite sign of charge only come once you quantize things, or can it be seen at the level of the classical fields and charge density?

 

[stevendaryl]

Thank you. I'm still a little confused.
On page 9 in the notes http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px435/dirac.pdf
it talks about the Feynman-Stuckelberg interpretation, where negative energy solutions represent positive energy antiparticles going forward in time. Is this related to your explanation?

That's a different interpretation. They're probably related, although I'm not sure how.

Does the fact that these should have the opposite sign of charge only come once you quantize things, or can it be seen at the level of the classical fields and charge density?

Well, Dirac's notion of a filled "sea" of negative-energy electrons depends on the Pauli exclusion principle, which states that you can't have two electrons in the same state. But classically, you can certainly do the following with charge densities:

Write the total charge density, $\rho_{total}(\vec{r}, t) = \rho_{average} + \rho_{diff}(\vec{r}, t)$ where $\rho_{diff}$ is the difference between the local charge density and the average charge density, over all time and space. Even if $\rho_{total}$ and $\rho_{average}$ have the same sign everywhere, the difference, $\rho_{diff}$ can be both positive and negative in different regions.

Then the issue is: Rewriting the equations so that only $\rho_{diff}$ is mentioned, not $\rho_{total}$ or $\rho_{average}$. Eliminating $\rho_{total}$ is necessary to get sensible equations, since it has to be infinite. So Dirac had to assume that this infinite charge density was unobservable, which is not very plausible.

In quantum field theory, the background charge density is redefined to be zero.

 

[vanhees71]

Forget about the Dirac hole theory. It's very complicated compared to the quantum-field theoretical way. You start with the classical Lagrangian of QED and then quantize the Dirac and the Maxwell field. As it turns out, in order to have a positive definite Hamiltonian for the free particles/fields, you have to quantize the electrons as fermions and the Maxwell field as bosons.

Now you work in the interaction picture. It turns out that there is a global symmetry under change of the phase of the Dirac field. This defines a conserved Noether current. Written in the naive way it reads
$$j^{\mu}=\overline{\psi} \gamma^{\mu} \psi.$$
The total charge, when written in this naive way, $Q=\int \mathrm{d}^3 \vec{x} j^0(t,\vec{x})$, however diverges. This is the first time, where you encounter a renormalization issue! It's very simple to solve, because the flaw is due to the mathematics: You must not multiply field operators at the same space-time point, because the field operators are distributions and not true operator valued functions. The solution is to use the mode decomposition of the Dirac field and then subtract an infinite constant, the vacuum expectation value of the total charge, because we measure charge against the vacuum, which has by definition 0 charge. Formally this is done by normal ordering, i.e., in the mode decomposition of the Dirac field in the current, you shift all annihilation operators to the right of all creation operators, which leads to the subtraction of the vacuum charge. Since the Dirac field is a fermionic field, when changing the order of the creation and annihilation operators you have to introduce the correct sign changes. This implies a negative sign from the contribution of the anti-particles (i.e., the reinterpreted negative-frequency modes as a creation contribution to the field for particles with positve energy moving in the opposite direction (NOT "backwards in time" as is often claimed even in otherwise good textbooks). This makes the charge of the antiparticles opposite to the charge of the particles as it should be.

From the Lagrangian, from the coupling of the em. field it follows that (up to the coupling constant, $e$) is the electromagnetic current, and for electrons and positrons, it's negative, $e<0$ (one netative elementary charge), i.e., the particles (electrons) carry negative and the antiparticles (positrons) positive charge of exactly the same absolute value of the charge. This is a reflection of CPT invariance of all local microcausal QFTs with stable ground state, the Pauli-Lueders theorem (for QED also C, P, and T all are separately symmetries; in the Standard Model only the weak interaction violates these symmetries as well as CP).