# Neutral pion decay: JPC conservation

1. May 27, 2013

### bayners123

$\pi^0$s decay to two photons via the EM interaction. The $J^{PC}$ of the pion is $0^{-+}$ and of a $\gamma$ is $1^{--}$.
$\gamma\gamma$ therefore has $J^{PC} = 0^{++}, 1^{-+}, 2^{++}$.
This does not match the pion, so how can this decay occur?

2. May 27, 2013

Staff Emeritus
The polarization state is E1 x E2, and as you can see, this is odd under parity.

3. May 27, 2013

### bayners123

Sorry, I don't quite understand your answer. Are you referring to the polarization of a single photon? If so, then yes I agree: that's why I put the Parity eigenvalue of the photon as -1, making the parity of a two photon system $(-1) \times (-1)^L = (-1)^{L+1}$

4. May 27, 2013

### Bill_K

The parity of a two-photon state depends on their relative polarizations. They are both transverse, of course, and in the case of a pion decay the photon polarization vectors are also perpendicular to each other. This is what V50 means by E1 x E2, and such a state has odd space parity.

5. May 27, 2013

### bayners123

Hmm ok. Is it possible to think of this in terms of combining quantum numbers? So adding two $1^{--}$ systems and obtaining a $0^{-+}$? If not, what's special about photon which makes this possible?

Also, does this imply that a two photon system can have either parity depending on what decayed?

6. May 27, 2013

Staff Emeritus
The states you wrote down have the photons polarization vectors all pointing in the same direction. In this case, they are perpendicular to each other.

7. May 28, 2013

### Bill_K

Consider the case where the pion decays at rest, and the two photons are emitted along the ± z-axis. A photon can either have helicity + (spin parallel to momentum) or helicity - (spin antiparallel to momentum) Since the total Jz of the two photons must be zero, they are either both helicity + or both helicity -. Call these states |++> and |-->.

Now the parity operation reverses helicity, so neither of these states is an eigenstate of parity. Rather the eigenstates are (|++> + |-->)/√2 (even parity) and (|++> - |-->)/√2 (odd parity). The odd parity state is the one we want.

So yes.

If you write the helicity states in terms of the states of transverse polarization, i.e. |±> = (|x> ± i |y>)/√2, you'll see that in the odd parity state the polarizations of the two photons come out perpendicular.

Last edited: May 28, 2013
8. May 28, 2013

### bayners123

Brilliant, thanks for both your help in understanding this!