# What exactly makes Mesons unstable?

1. Oct 12, 2013

### 4nn4

As the title says- what is it that makes all mesons unstable? I know some quarks are less stable than others obviously, but why is there no stable meson? What is it about any combination of quark and anti-quark that makes a particle decay nearly immediately?

In a related topic- what exactly is it about uud that makes a proton stable?

Thank you in advance to any respondents.

2. Oct 12, 2013

### fzero

In general an initial state $|i\rangle$ will be unstable whenever there exists a final state $|f\rangle$, such that:

1. $|i\rangle$ and $|f\rangle$ are connected by some interaction term $\Gamma$. Namely, the matrix element $\langle f |\Gamma| i \rangle$ is nonzero.

2. The potential energy of the final state is less than the energy of the initial state. Here we really mean that the mass of the particles in the final state are less than the initial state particle. Any excess energy will contribute to the kinetic energy of the final state particles.

For the mesons, the electroweak interactions always exist between quarks and antiquarks of the same or different type. For the same type of quark, we have electromagnetic interactions $\bar{q} A_\mu \gamma^\mu q$, while for different quarks, there are weak interactions $\bar{q} W_\mu \gamma^\mu q'$. These interactions can switch a quark $q$ to the same quark $q$ plus a photon or to a different type quark $q'$ plus a W boson. They can also destroy a quark-antiquark pair $\bar{q}q$ and create a photon, or destroy a $\bar{q}q'$ to create a W. The final state photon can create an electron-positron pair in the final state, while the W bosons will themselves decay to electrons and neutrinos in the final state, both of which are much, much lighter than the observed masses of mesons.

What is not explained by the above is why the mesons are heavier than electrons and neutrinos. It turns out that this is a consequence of the strength of the strong interactions, but there is no fundamental reason why. It is just an experimental fact.

The proton is stable because it is the lowest energy (lightest) configuration of 3 quarks. Because of charge conservation and color confinement, any electroweak interaction term like the ones above applied to the proton must lead to another final state composed of 3 quarks. Remember, the electroweak interactions either switch a quark for a quark + a gauge boson or replace a quark-antiquark pair for a gauge boson. So acting on a baryon $q_1q_2q_3$, since we don't have an antiquark to destroy, we obtain another state

$$q_1q_2q_3' + W,Z, \gamma.$$

There is no state $q_1q_2q_3'$ with mass less than the proton, so a decay of this type is forbidden by energy conservation.

The strong interactions are a slightly different story. These have a form similiar to the EM interaction, between same type quarks: $\bar{q} G_\mu \gamma^\mu q$. The difference is the color indices on the quarks, which I haven't drawn in. The difference between this operator and the electroweak ones is that it operates on gluons. Since the proton is bound by the strong interactions, there are lots of virtual gluons present at any given time. So we can have strong interactions where we convert an internal gluon to a quark-antiquark pair. We can represent this as an interaction

$$q_1q_2q_3g \rightarrow q_1q_2q_3 + \bar{q}_4 q_4.$$

So the strong interactions can generate an additional meson in the final state. However, since we still have the $q_1q_2q_3$ particle in the final state, energy conservation still prevents the photon from "decaying" in this manner.

3. Oct 12, 2013

### 4nn4

Thank you for taking the time to write such a detailed and clearly explained response.

Thanks again

4. Oct 13, 2013

### kloptok

I find this statement interesting as I've always been told that it is baryon number conservation which leads to the proton being stable. Something I've never been entirely satisfied with as the baryon number conservation is in a sense not as fundamental as e.g. charge conservation.

Could you explain further why charge conservation and confinement leads to the proton being stable (and perhaps, if it is too much asked, comment on why this is different in GUT theories, where the proton is unstable? I would be grateful for any answer!

5. Oct 13, 2013

### dauto

fzero did not mention it, but Baryon number conservation is essential to explain proton stability. Baryon number is not conserved in GUT theories, hence the prediction that protons do decay if GUT turns out to be correct.

6. Oct 13, 2013

### Staff: Mentor

It is expected that baryon number conservation breaks down at high energies - there has to be some reason why we are surrounded by baryons without significant amounts of antibaryons. If there would be no such process, it would be as fundamental as charge conservation.

7. Oct 13, 2013

### fzero

Baryon number conservation is the explanation for why there are no interaction terms in the standard model that turn a quark into something else, such as $\bar{e} V_\mu \gamma^\mu q$ or the like. In the SM, baryon number is assigned to quarks directly and so is part of "charge conservation," though I didn't elaborate since my post was already very long and liable to be confusing. It's clear that confinement is necessary, else one could imagine a baryon splitting up into free quarks.

Before the quark model, baryon number conservation was much more directly connected with the fact that, in the decay chain of any heavy baryon, the end state always contains a proton. In the quark model, the end result is the same.

8. Oct 13, 2013

### arivero

Incidentally, if the mass of charged pion were equal to the mass of muon and the electron were massless, it would be stable at first order, because of "helicity suppression". Actually I am not sure if it is unstable beyond first order.

9. Oct 14, 2013

### Staff: Mentor

How does a massless, charged particle look like?

If electron and muon both had at least the mass of a pion, the charged pion would be the lightest charged particle, and electrons and muons would decay to pions (+neutrino).

10. Oct 17, 2013

### Doofy

Tonight I've been unsuccessfully googling around for a document that walks you through those nucleon decay predictions without requiring too much background knowledge, do you know of any good ones?

11. Oct 18, 2013

### dauto

I'm not aware of any good ones. The basic concept is quite simple though. In grande unified theories there are multiplets that include leptons and quarks in them. That means that quarks can turn into leptons and vice-verse and would emit a virtual boson in the process. That's analog to the way that an electron can turn in a neutrino and vice-verse while emitting a W boson in the process. The W boson is very massive suppressing decays that depend on the weak force (That is why it is weak). The Gut bosons are expected to be about a million billion times more massive than the W boson suppressing decays that depend on them (such as the proton decay) by a much larger factor.