A problem with QED

Question

I have a small problem with the understanding of QED.

The equations of motion in QED are

$\square A^\mu=e\bar{\psi}\gamma^\mu\psi$

$\left(i\gamma^\mu\partial_\mu-m\right)\psi=e\gamma^\mu A_\mu\psi$

If we analyze the first equation we get that both particles and antiparticles create the same electric potential because

$\bar{\psi}\gamma^0\psi=|\psi_1|^2+|\psi_2|^2+|\psi_3|^2+|\psi_4|^2$

Also from the second equation we get that particles and antiparticles interact by the same way with electric potential.

Is this just some misconception or it's just some weird effect we observed?

Answer 1

You should take into account that $\bar{\psi}\gamma^0\psi=u^2-v^2$, which is positive for particles and negative for antiparticles.

Answer 2

One has to make a case distinction: 1) Bound states & 2) Scattering of free particles.

1. Bound states:

As already observed, the positive and negative energy solutions couple to the same coupling $e$ with the same sign, i.e. both describe electrons. So if one wants to describe positrons in a bound state one has indeed to use the charged conjugated solution:

$$\psi^c = C\bar{\psi}^T$$

where $C$ is the charge conjugation matrix which can vary depending on the representation of the Dirac spinor. In one of the simplest representations it is

$$\psi^c = \gamma^2 \psi^\ast$$

The charge conjugation makes negative energy electron solutions to positive energy positron solutions.

2. Scattering

For the description of scattering the formalism of $2^{nd}$ quantization can be used. In this case the solution of the Dirac equation $\psi$ is considered as an (field) operator $\hat{\psi}$ acting on the Fock space:

$$\hat{\psi}(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}} \sum_s \left( a^s_\mathbf{p} u^s(p) e^{-ipx} + b^{s\dagger}_\mathbf{p} v^s(p) e^{ipx}\right)$$

where $a^s_\mathbf{p}$ and $b^{s\dagger}_\mathbf{p}$ are annihilation operators for electrons and a creation operators for positrons. Actually choosing as "coefficient" of the negative energy solution $v(p)$ a creation operator (note an annihilation operator would be the other choice) makes the corresponding solution $v^s(p)$ to an outgoing particle (of different kind, because it is another operator ($b$) and not an $a$-operator) which can then be interpreted as in time backwards running "in-going" --- i.e. from the future into the presence --- electron which can by virtue of interpretation be seen as an out-going positron. The formalism of $2^{nd}$ quantization in combination with Feyman-diagrams makes this possible. The rules of Feynman diagrams have to be applied strictly. In particular positron scattering is described in this formalism as ($p$ is the 4-momentum of the in-going particle, whereas $p'$ is the 4-momentum of the scattered out-going particle).

$$\bar{v}(p)\gamma^\mu v(p')$$

where $\bar{v}(p)$ describes the in-going "positron" and $v(p')$ the out-going "positron" whereas in case of electron scattering the scattering current looks like:

$$\bar{u}(p')\gamma^\mu u(p)$$

i.e., the in-going electron is described by $u(p)$ and the scattered out-going electron described by $\bar{u}(p')$, i.e. the positions of $u$ and $v$ in the current are swapped. Therefore solutions $v(p)$ are not genuine -- let's say --- positive charged particles. Genuine opposite charged particles would have the same scattering current as the electron current $\bar{u}(p')\gamma^\mu u(p)$ (Think of up- and down-quarks which are both described by the Dirac equation, but are of opposite sign (of charge) and not a particle-anti-particle pair).

Actually, it is also possible to use a description involving the charge conjugation operator $C$ to describe scattering currents, but in the above described formalism of $2^{nd}$ quantization the same can be achieved much easier.

The Feynman interpretation of the negative energy solutions as positrons has prevailed so strongly that most people consider $v(p)$ as positron solutions, but there are not. They are electron solutions of negative energy (or frequency) solutions, because they couple with same coupling constant to the EM-field as the positive energy solutions.

Finally I would say, the Dirac equation is not problematic, one only has to deal with the information it provides appropiately.

See also my post Does the Dirac equation put charge and spin on the same footing?

Answer 3

we get that both particles and antiparticles create the same electric potential

No, we don't. One-particle Dirac equation is problematic, so one needs some tweaking. Either you assume that components of the Dirac spinor are Grassmannian (or satisfy canonical anti-commutation relations), or you say that antiparticles are not negative-energy states but holes in the sea of negative-energy states.