Why does the photon have no antiparticle?

Question

Considering the QED Lagrangian we get a solution of the form $A^{\mu}$. This solution has four degrees of freedom (dof). With the weak Lorenz gauge we reduce it to two dof. With this condition we also solve the problem of negative energies in our Hamiltonian.

But I think that if Dirac had thought the same way after the discovery of his Dirac equation, he would never have found the theory of antiparticles. But he was brave enough and interpreted things differently.

I'm taking a course in QFT right now, and every other result has to be rearranged or something like that - it's confusing for me - I mean, yes, in the end we see that the experiment is in line with the theory, but maybe you know what I mean, why I think that's bad style.I wouldn't say I can do better, I'm much too stupid for that and not in a position to complain, I just want to know if anyone has a nice view on it.

So to get back to the question in the title: One of these four dof will lead to negative energies, why can't we say we have four kinds of photons - maybe an intrinsic property that hasn't been discovered yet?

Answer 1

It is related to the use of anticommutators for fermions and commutators for bosons (the possibility of unobservable timelike and longitudinal polarisation states is not an issue here). A necessary condition for a valid QFT is that the locality (or microcausality) condition is satisfied by field operators. This is required to ensure the consistency of the perturbation expansion under Lorentz transformation. For space-like $x-y$ for fermions we require

$$ \left\{ \psi(x), \bar\psi(y) \right\} = 0$$

Satisfying this relation requires that the field operator annihilates a particle or creates an antiparticle. For bosons, locality applies to commutators. The commutator of the $A$ field always vanishes (only derivatives are observable). Instead we have for space-like $x-y$

$$ \left [ \partial_c A^a(x), A^b(y) \right ] = 0$$

This is satisfied when $A$ creates or annihilates a photon, meaning that the photon is its own antiparticle.

Answer 2

I mean, yes, in the end we see that the experiment is in line with the theory,

This is the basic problem. The cart before the horse. It is not the experiment that has to be in line with the theory, but the theory should model the experiment.

The standard model of particle physics ( a quantum field theory model) models very well the great majority of experimental data and observations up to now. Part of these observations are the masses and quantum numbers of the observed particles, and the experimental fact, that to every elementary particle there exists an antiparticle, having the equal and opposite quantum numbers to the particle. When they scatter , or make a vertex in a calculation with Feynman diagrams, particle and antiparticle annihilate and only the four vector of their summed energy-momentum four vectors remains.

Of the elementary particles ,

The exact same table exists with the antiparticles

The photon, the Z0 and the standard model Higgs (for the gluon see here ) are antiparticles of themselves. The model describes nature and is predictive of new data.

So the answer to

Why does the photon have no antiparticle?

is because that is what data tells us and it has been axiomatically assumed in the standard model.

Had we observed antiparticles of the photon, a different model would have been developed.

Answer 3

The choice of gauge while quantising Electromagnetic field is not just to make the Hamiltonian positive definite but it is necessary to describe the Hamiltonian consistently. The problem arises due to the constrained nature of Electromagnetic field which also is responsible of the gauge freedom in the theory.

Consider the EM Lagrangian density: $$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu}, $$

now if one tries to go to Hamiltonian picture by defining conjugate momenta as

$$\Pi^{\mu}(x) = \frac{\partial\mathcal{L}}{\partial\dot{A}_{\mu}(x)} = - F^{0\mu} $$

$$ \Pi^{0}(x) = - F^{00} = 0 $$

So now one cannot eliminate $ \dot{A}^{0}$ from the Hamiltonian without choosing some gauge fixing condition e.g. $ {A}^{0} = 0 $ and $\nabla \boldsymbol{A} = 0 $. One is forced to gauge conditions to remove the non-physical degrees of freedom as the dynamics of the system is constrained. All constraints do not require to choose the gauge conditions, Dirac gave a systematic procedure to distinguish such constraints and to effectively quantize such systems.

In Dirac field there are no such constraints present so there is no need to fix gauge conditions. Moreover even if your Hamiltonian is positive it does not mean that there are no anti-particles. Since quanta of EM field is changeless you cannot distinguish the particles with anti-particles because except charge rest of properties are same for both of them.

Answer 4

Photon is a boson for bosons they are the antiparticles it self but collisions of them do not cause any annihilation