Neutral pion decay: JPC conservation

[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?

 

[Vanadium 50]

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

 

[bayners123]

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

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 $(-1) \times (-1)^L = (-1)^{L+1}$

 

[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.

 

[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?

 

[Vanadium 50]

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.

 

[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 J_z 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.

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

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.

 

[bayners123]

Brilliant, thanks for both your help in understanding this!