# CP Operators Mesons

1. May 20, 2014

### jono90one

Hi,
I am currently going over this and got me thinking about a scanario where you have A -> BC
Where A is S = 0, L=0, B is S = 1 L=0, C is S=1 L =0
(I'll use S = intrinsic spin, L = angular momentum, J = Total Angular momentum, |L-S|=< J =< L+S)

Maybe such a decay doesn't exist, but I'm just trying to test the "boundaries" of what I know about this topic.

For this reaction to happen the total angular moment must be J=0 (ie 2-2 = 0 as J = 0 on LHS so L=2 on RHS for total system)
$\hat{P}|B> = (-1)(+1)(-1)^{L}$
L = 0
$\hat{P}|B> = -1$
and like wise:
$\hat{P}|C> = -1$

For charge conjugation

$\hat{C}|B> = (-1)^{L+S}$
$\hat{C}|B> = (-1)^{2} = +1$
$\hat{C}|C> = (-1)^{2} = +1$

Now put them together:
$\hat{P}|BC> = (-1)(-1)(-1)^{L}=(-1)^{2}= +1$
$\hat{C}|BC> = (-1)(-1)(-1)^{L+S} = (-1)^{4} = +1$

Hence CP:
$\hat{C}\hat{P}|BC> = (+1)(+1) = +1$

BUT
If we do the LHS using L = 0, S = 0 we get CP(A) = -1

I could be doing these calculations wrong, though this method has worked for easier examples (maybe by chance..). Or maybe it's not suppose to work ...?

[Ps this isn't homework, i'm just reading a book on it!]

Thanks for any help

2. May 20, 2014

### Staff: Mentor

How?
(-1)0=1

$H \to \gamma \gamma$ (yes I know those are all not mesons...)?

3. May 20, 2014

### ChrisVer

heavy quark mesons could do this kind of decay...
$\bar{D}^{0} \rightarrow K^{+} \pi^{-}$
$\bar{c}u \rightarrow u \bar{s} + \bar{u} d$

4. May 21, 2014

### jono90one

In terms of the LHS I did,
$\hat{P}|a> = (+1)(-1)(-1)^{L} = (-1)(-1)^{0} = -1$
$\hat{C}|a> = (-1)^{L+S} = (-1)^{0} = +1$

$\hat{C}\hat{P}|a> = (-1)(+1) = -1$

5. May 21, 2014

### Einj

I think there is a little bit of confusion here. For example, if A, B and C are particles how can they have angular momentum L? Relative to what? You can definitely have a relative angular momentum L between the final products B and C, but having L just for A doesn't make sense to me.

Particles are usually characterized by their $J^{PC}$ quantum numbers, meaning their intrinsic spin, parity and charge conjugation. The charge conjugation is actually well defined only for neutral particles. For example the neutral pion $\pi^0$ is a $J^{PC}=0^{-+}$, the well-known $J/\psi$ is a $J^{PC}=1^{--}$.

What are the these intrinsic quantum number for your particles A, B and C? Only knowing these quantities you can determine whether or not the decay can happen and what the final value of the relative angular momentum must be. Once you also know the value of the final L you can compute the action of the CP operator.

Consider for example, as mfb said, the H→γγ decay. The Standard Model Higgs should be (if I remember correctly) a $J^{PC}=0^{++}$, while the photon is a $J^{PC}=1^{--}$. Therefore our decay is basically:
$$0^{++}\to 1^{--}1^{--}$$

As you can see, if you add together the spins of the final photons you have 1+1=0,1,2 and so you can obtain the initial J=0 configuration even without any relative angular momentum between the final photons.

Now you could ask: can the two final photons be in an L=1 configuration? We have:
$$CP|H\rangle=+|H\rangle,$$
$$P|\gamma\gamma\rangle=(-1)(-1)(-1)^L|\gamma\gamma\rangle=-|\gamma\gamma\rangle,$$
and finally
$$C|\gamma\gamma\rangle=(-1)(-1)(-1)^{L+S}|\gamma\gamma\rangle=+|\gamma\gamma\rangle.$$

Therefore the CP of the final state is -1 and, if CP is a good symmetry for this decay, then it's not allowed.

6. May 21, 2014

### Meir Achuz

If B and C are two distinct particles, charge conjugation isn't relevant in the decay.
A spin zero particle can decay into two spin one particles.