# Why don't a charm-anticharm meson decay to pair of leptons?

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1. May 7, 2015

### unscientific

If you consider the $\chi_0$ with a mass of $3.4 GeV/c^2$ meson, why doesn't it decay to a pair of charged leptons? Technically it is possible though the weak interaction (Z boson) or EM interaction, right?

Is it because it is so heavily suppressed because the strong interactions are favoured? For example production of $D+D-$ mesons.

2. May 7, 2015

### fzero

$\chi_{c,0}$ has angular momentum $J=0$ like the neutral pion, so at tree-level we can only have EM decays $\chi_{c,0}\rightarrow 2\gamma$ or $\chi_{c,0}\rightarrow \gamma \ell^+\ell^-$. So EM decays are suppressed by a factor of around 100 compared to the $J=1$ states.

3. May 7, 2015

### unscientific

Why can't weak decay via Z boson be possible?

Also even if $\chi$ decays electromagnetically, can't the photon couple to a pair of charged leptons?

What other $J=1$ states?

4. May 7, 2015

### fzero

The photon and Z boson are spin 1, and it is impossible to put a single spin 1 particle into a $J=0$ state. We need to conserve angular momentum at every step of the process, so we can't have a single photon or Z intermediate state anywhere in the decay process. The $J=1$ charmonium states can indeed decay via single photon or Z.

5. May 7, 2015

### unscientific

$\chi_{c0} = 0^{+}, \chi_{c1} = 1^{+}, \chi_{c2} = 2^{+}$. Ok, so the first can only decay electromagnetically and the rest can decay both via weak and EM interactions.

Even then, why can't the photon/Z boson decay to pairs of charged leptons?

6. May 7, 2015

### fzero

With a pair of spin 1/2 particles, it is possible to construct a $J=1$ angular momentum state. Like $\chi_{c1}$ and $\chi_{c2}$ we can also have some orbital angular momentum if necessary with a pair of particles. When there is a single particle in an intermediate state is when you are more likely to run into problems.

7. May 7, 2015

### unscientific

That makes sense. But surely leptons are fermions as well, so they won't have the issue with spin. Why don't they decay to a pair of charged leptons then? Is it because strong interaction is preferred?

8. May 7, 2015

### fzero

The final $\ell^+\ell^-$ state can indeed have $J=0$, but the problem is that however we draw the diagram in your OP, there is always some reference frame in which we have a intermediate state consisting of a single spin 1 particle, which cannot be in a $J=0$ state. If we cooked some other diagram up to avoid this (like by adding a loop), then the additional interaction vertices we'd have to add would effectively suppress the process.

9. May 7, 2015

### unscientific

Sorry I didn't understand that at all. Based on the lowest order feynman diagram, I don't see why this following weak interaction is suppressed:

10. May 7, 2015

### RGevo

11. May 7, 2015

### unscientific

I know they do, but this isn't J/psi. This is $\chi_c$ meson. My question is why doesn't it decay to a pair of charged leptons, e.g. $\mu+\mu-$.

12. May 7, 2015

### RGevo

Oops , sorry. I didn't read the text in any detail.

13. May 7, 2015

### Staff: Mentor

The graphical version of fzero's post.

14. May 7, 2015

### fzero

It is necessary to conserve angular momentum at each vertex. The $\chi_{c0}$ is a 1P orbital state and has $J=0$. This implies that the $c$ and $\bar{c}$ spins are aligned. If the orbital angular momentum was zero, then $1/2 +1/2=1$ would work. However, $L=1$. If there was a second particle in the intermediate state then it would be possible for the vector + other particle system to have orbital angular momentum to make up the difference. However, we only have the vector and so cannot do so.

15. May 7, 2015

### Staff: Mentor

The decay is possible at tree-level with an intermediate Higgs boson (it has spin 0), but the contribution from this process is completely negligible.

16. May 7, 2015

### fzero

Yes, it's worth noting that, by comparing coupling constants, EM (radiative) decays tend to be suppressed by a factor of $10^{-4}$ with respect to strong decays. Weak decays are suppressed by more like 12 orders of magnitude compared to strong. There are a lot of other interesting effects in charm systems that can enhance some channels vs others, but the rarer the channel, the more likely it will be lost in the noise.

17. May 7, 2015

Staff Emeritus
I'm sorry, but most of what has been written in this thread is simply incorrect.

The picture in #13 is not right. If it were true, the charged pion could never decay, since the pion is spin-0 but thge W is spin-1. Internal lines do not need to have the same spin as real particles.

The whole premise is also not right. The chi_0 probably does decay to lepton pairs. However, the partial width of this electromagnetic process is probably of order a few eV, and it competes with 10 MeV of other decays. So only one in a million - or a few million - chi_0's decay this way. The Higgs contribution to this is completely negligible.

Finally, the chi_0 is too light to decay into open charm.

18. May 7, 2015

### unscientific

So the strong interaction is preferred over the weak/EM interaction in this case?

19. May 7, 2015

The EM process is actually forbidden by C-parity conservation, which is in fact related to the angular momentum state. This is the same reason that $\pi^0\rightarrow \gamma^*\rightarrow e^+e^-$ doesn't occur. $\Gamma(\pi^0\rightarrow e^+e^-)/\Gamma_\mathrm{tot}$ is around $10^{-7}$ where you'd expect it to be via an intermediate Z.