# Antiparticles identical in qft

1. Apr 22, 2013

### copernicus1

Has anyone ever heard of treating a particle and antiparticle as identical based on the formalism of quantum field theory? The argument given is that the creation operator for a particle is the annihilation operator for its antiparticle, but I can't find this idea of treating them as identical anywhere else, and of course it seems strange to call particles which may have opposite charge and other quantum numbers "identical." Can anyone shed some light on this?

Thanks!

2. Apr 22, 2013

### vanhees71

Of course only particles of 0 charge can be their own antiparticles. They are called strictly neutral particles (there are of course electrically neutral particles what are not their own antiparticles, but then they differ by the sign of some other charge-like quantity, e.g., a neutral Kaon differs from the neutral anti-Kaon by the strangeness (-1 and +1 respectively)).

An example in the standard model is the photon, which is its own antiparticle. It's not clear yet whether neutrinos are their own antiparticles or not. In the standard model they are treated as Dirac particles and thus are different from the antiparticles, but they could as well be Majorana fermions. The question, what's right is unsettled. There are experiments looking for the socalled neutrinoless double-beta decay of nuclei, but there's no conclusive evidence for this yet.

3. Apr 22, 2013

### copernicus1

So you've never heard of a situation where you would treat, for example, the electron and positron as identical? A set of course notes I'm currently reading uses this sort of concept throughout, with the explanation referring to creation and annihilation operators I mentioned in my first post.

4. Apr 22, 2013

### mpv_plate

The electron and positron have different creation operators, so they don't seem to be identical in this picture. Really identical particles should have identical creation operators (above other things).

Anyway, I have seen something similar, but I'm not sure if it's what you are looking for. Check out Robert Klauber's online book, chapter 3, page 63, botom of the page. He says:

5. Apr 22, 2013

### copernicus1

Thanks. It sounds as though treating particles and antiparticles as identical in general is probably unjustified.

6. Apr 22, 2013

### Chopin

Two things that might be what your course notes are saying:

• While the creation/annihilation operators don't work the way you described, the field operators do. The operator $\psi$ contains both $a$ and $b^\dagger$, so it will both create an electron and annihilate a positron, and vice versa.
• Particles and antiparticles do behave very similarly in a lot of ways--they have the same mass, and the same coupling constant to other fields, etc. So in many cases it makes sense to talk about them as being identical, if you are able to ignore the areas where they do differ. These things are consequences of the U(1) symmetry that connects them (and which is responsible for the charge quantum number being a conserved quantity). In a similar way, one often talks about protons and neutrons as being identical particles, because of their SU(2) isospin symmetry.

7. Apr 22, 2013

### Bill_K

Whoa! There is no U(1) symmetry connecting particles with antiparticles. Certainly not the U(1) symmetry of electromagnetic gauge invariance.

8. Apr 22, 2013

### tiny-tim

an anti-particle is identical to its particle, except that it "lives backwards"

this symmetry is fundamental to quantum field theory!

9. Apr 22, 2013

### Bill_K

Sorry, tiny-tim, this is popsci horseradish!

10. Apr 22, 2013

### Chopin

Sorry, I think I mixed up some concepts here. I guess the symmetry that shows that particles and anti-particles have the same mass and coupling constants is symmetry under charge conjugation, not the U(1) gauge symmetry. Is that right, or am I just spouting nonsense now?

11. Apr 23, 2013

### vanhees71

In general particles and antiparticles are distinguishable by at least one intrinsic quantum number. E.g., electrons and positrons are distinguishable through their electric charges ($-e$ for the electron $+e$ for the anti-electron=positron).

The free-field operator of a Dirac field can be written as a superposition of creation and annihilation operators wrt. to the single-particle momentum-spin eigenstates:
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \sum_{\sigma=\pm 1/2} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^{3/2} \sqrt{2 E(\vec{p})}} \left [\hat{a}(\vec{p},\sigma) u(\vec{p},\sigma)\exp(-\mathrm{i} p \cdot x) + \hat{b}^{\dagger}(\vec{p},\sigma) v(\vec{p},\sigma) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=+E(\vec{p})}.$$
It is crucial that the particles come with an annihilation an the antiparticles with a creation operator. This reinterprets the modes with negative frequency as antiparticles with positive frequency.

Particle or anti-particle number is by itself not a "good quantum number", because it's not conserved. This is the case for net-particle number (or charge), i.e., the particle number - the antiparticle number. The field operator changes lowers this net-particle number by 1.

12. Apr 23, 2013

### andrien

It is convenient to decompose ψ and ψ-(free field) into two parts as
ψ=ψ+-,similarly for ψ_,where for example ψ+ is
ψ+=1/√VƩp,s√(mc2/E)bpsus(p)e(ip.x-iEt/h-),this ψ+ is linear in electron annihilation operators,it will not do anything except annihilating electrons while ψ- the other part of ψ will create positrons.ψ_+ will annihilate positrons while ψ_- will create electrons.