# Why does electron-positron annihilation conserve parity?

I think I'm missing something quite basic here but consider the process:

$$e^- + e^+ \rightarrow 2\gamma$$

Fermions have opposite parity to antifermions so the parity quantum number before the process is $$P=-1 \times (-1)^L$$ where $$L$$ is the relative orbital angular momentum, which should vanish in the zero momentum frame where the collision is head on. So we start with $$P=-1$$. The photons on the other hand have the same parity as each other so have parity $$P=+1$$ (after again noting that they must be heading in opposite directions to each other in the ZMF so have $$L=0$$).

I think my reasoning about the angular momentum must be wrong, but I don't see why. (Additionally, the same argument says that if this is $$L:0\rightarrow 0$$ then charge congugation $$C$$ is not a conserved quantity either.)

• Working in the ZMF does not imply L=0. The two momenta (incoming or outgoing) can be equal and opposite but with a transverse offset between them. – RogerJBarlow Jun 1 at 15:17
• @RogerJBarlow so am I right that this can't happen if the collision is perfectly head on, but that just doesn't happen in practice (no phase space volume)? – jacob1729 Jun 1 at 15:24
• @RogerJBarlow I think you're right, but I can't visualise this for the QM case where I have infinite extended plane waves impinging on each other from opposite directions. I feel like I shouldn't be needing to rely on specific wave packet shapes just to know if something is possible at all. – jacob1729 Jun 1 at 15:26
• Thinking of such a collision like two trucks colliding head on is not helpful because of Heisenberg: $\Delta y$ is zero so $\Delta p_y$ is infinite. Two extended plane waves moving in opposing directions and meeting is a much more valid way to think about it. Then you remember that a plane wave can be expressed as a sum of partial waves en.wikipedia.org/wiki/Plane_wave_expansion. and so the plane wave collapses into a particular $L$ component. – RogerJBarlow Jun 1 at 16:04
• you are forgetting the polarization of the photons. similar argument to physics.stackexchange.com/questions/483588/… . see link imsc.res.in/~taushif/pdfs/… – anna v Jun 1 at 17:34

The electron-positron collision state in question has the same quantum numbers as positronium, which may exist in two states. The first of these is the short-lived singlet state $$^1S_0$$, so $$L=S=J=0$$, hence $$P=-$$ and $$C=(-)^{L+S}=+$$, and this is your parapositronium two-photon mode (must have even number of photons, from C), with symmetric parity-odd wavefunction, $$\propto (|\hat \epsilon _1\rangle |\hat \epsilon _2\rangle - |\hat \epsilon _2\rangle |\hat \epsilon _1\rangle)(e^{ik(\hat {\mathbf k} \cdot ({\mathbf r}_1- {\mathbf r}_2)-2t)}- e^{ik(\hat {\mathbf k} \cdot ({\mathbf r}_2- {\mathbf r}_1)-2t)} ) ,$$ for respective photon polarizations $$\hat \epsilon_1, \hat\epsilon_2$$.

Should remind you of the neutral pion (a pseudoscalar), and decays just like it into an odd-parity couple of photons.

The other, not sought here, long-lived bound state is orthopositronium, triplet and longer lived, $$^3S_1$$, so $$P=-, ~ C=-$$. Hence, it must decay to 3γ, unless it were a virtual γ in production through collision of electron-positron.

(So, does this remind you of the ρ or the J/ψ ?)

• In the $L=0$ state is the parity not $-1$? Because the electron has $P=1$, the positron $P=-1$ (which way around is irrelevant obviously) and so the product has $P=-1$. This differs from the 2-photon decay products. – jacob1729 Jun 1 at 21:11
• My question is not directly concerning $e^-e^+$ bound states but free particles that collide and annihilate. Perhaps if I understood the bound states that would answer my question though, although I think the issue might be my understanding of what $L$ means for incoming beams. – jacob1729 Jun 1 at 21:13
• The odd-parity 2γ wavefunction is explicitly and simply given here. – Cosmas Zachos Jun 1 at 21:58
• Would you mind copying the odd-parity two photon wavefunction into the answer so I can accept it? My confusion was (at least in part) that I was assuming the two photon wavefunction needed to be even and so the process was parity violating. (The other issue was that I was forgetting partial wave expansions as noted in the comments to my OP.) – jacob1729 Jun 16 at 20:33
• Done. I have taught neutral pion decay too many times for any sticking points to still stick... – Cosmas Zachos Jun 16 at 21:13