# A Definition of a decay process

1. Dec 5, 2016

### spaghetti3451

Consider the following momentum-space Feynman diagram

This Feynman diagram is the leading-order contribution to the any of the following processes:

1. $e^{-} \rightarrow e^{+} + \gamma$
2. $e^{+} \rightarrow e^{-} + \gamma$
3. $\gamma \rightarrow e^{+} + e^{-}$

The process in 3 is clearly the decay of a photon to an electron-positron pair.

Why can't the processes in 1 and 2 be considered to be decay processes of the electron and positron respectively?

2. Dec 5, 2016

3. Dec 5, 2016

### spaghetti3451

What if I replace the photon by a massive scalar boson?

4. Dec 5, 2016

5. Dec 5, 2016

### spaghetti3451

Ok, so why is the process $h \rightarrow h + h$ allowed?

In the rest frame, the total energy is the rest energy of $h$, but the outgoing particles have twice the rest energy, so energy is not conserved.

6. Dec 5, 2016

### stoomart

Sorry, this doesn't make any sense to me, I better let the pros take this one.

7. Dec 5, 2016

### Orodruin

Staff Emeritus
It is not allowed for free Higgses.

Neither is any of your reactions listed in the OP.

8. Dec 5, 2016

### spaghetti3451

Ah! I see!

But consider the following Lagrangian

$$\mathcal{L} = \bar{\psi}_{e}(i\gamma^{\mu}{\partial_{\mu}}-y_{e}\nu)\psi_{e}-y_{e}\bar{\psi}_{e}h\psi_{e}+\frac{1}{2}(\partial_{\mu}h)(\partial^{\mu}h)-\frac{1}{2}\left(2|\kappa^{2}|\right)h^{2}-\frac{\lambda}{6}\nu h^{3}-\frac{\lambda}{24}h^{4}$$

which has the intreaction term $-y_{e}\bar{\psi}_{e}h\psi_{e}$. This term leads to an interaction vertex of the form given in the diagram:

https://i.imgsafe.org/64394a297d.jpg

This diagram describes the leading-order Feynman diagram of the process $h \rightarrow e^{-} + e^{+}$.

Can we have a process like $h \rightarrow h$, where the electron and anti-electron are in a loop? Can we call this process the decay of $h$?

9. Dec 5, 2016

### Orodruin

Staff Emeritus
No, the $h$ can decay to the electron positron pair if its mass is large enough. The $h \to h$ diagram is not a decay, it is a contribution to the self energy of the $h$.

10. Dec 5, 2016

### spaghetti3451

Ok, so (assuming that the mass of $h$ is greater than the mass of $e^{-}$) apart from $h \rightarrow e^{-} + e^{+}$, what other decays of $h$ are possible?

11. Dec 5, 2016

### Orodruin

Staff Emeritus
The mass needs to be greater than two electron masses for the decay to occur.

What other decays that will occur depends on what else you stuff into your theory.

As given, none (apart from possible higher order decays to several electron-positron pairs, but those are suppressed).

12. Dec 5, 2016

### spaghetti3451

Why are higher-order decays suppressed?

13. Dec 5, 2016

### Orodruin

Staff Emeritus
Because they contain more vertices and are therefore only available at higher order in perturbation theory. If the total mass of the decay products is close to that of the decaying particle, you will also get additional phase space suppression.

14. Dec 5, 2016

### spaghetti3451

But you can't deny that these higher-order decay processes still exist!

It's just that they are suppressed since they contain more vertices, right?

15. Dec 5, 2016

### Orodruin

Staff Emeritus
Where did I do that?

Also, they will not exist unless the sum of the masses of the outgoing particles is smaller than the $h$ mass.

This is what I said.

16. Dec 5, 2016

### spaghetti3451

Is the image in the following link the leading order Feynman diagram for the (higher-order) decay process $h \rightarrow e^{-} + e^{+} + e^{-} + e^{+}$

https://i.imgsafe.org/650622580e.jpg?

17. Dec 5, 2016

### Orodruin

Staff Emeritus
No. It has three electrons and three positrons in the out state. Also, there are several different diagrams contributing.

18. Dec 5, 2016

### spaghetti3451

How do I understand that there are several different diagrams contributing?

By several different diagrams, do you mean several interaction vertices?

19. Dec 6, 2016

### Orodruin

Staff Emeritus
Because you can draw them while respecting the Feynman rules.

No.

20. Dec 6, 2016

### spaghetti3451

Okay, so if instead, we had the Lagrangian

$$\mathcal{L} = \bar{\psi}_{e}(i\gamma^{\mu}{\partial_{\mu}}-y_{e}\nu)\psi_{e}-y_{e}\bar{\psi}_{e}h\psi_{e}+\bar{\psi}_{\mu}(i\gamma^{\nu}{\partial_{\nu}}-y_{\mu}\nu)\psi_{\mu}-y_{\mu}\bar{\psi}_{\mu}h\psi_{\mu}+\frac{1}{2}(\partial_{\mu}h)(\partial^{\mu}h)-\frac{1}{2}\left(2|\kappa^{2}|\right)h^{2}-\frac{\lambda}{6}\nu h^{3}-\frac{\lambda}{24}h^{4}$$

we can write down processes like

$$h \rightarrow e^{-} + e^{+}, \qquad\qquad h \rightarrow e^{-} + e^{+} + e^{-} + e^{+}, \dots$$

$$h \rightarrow \mu^{-} + \mu^{+}, \qquad\qquad h \rightarrow \mu^{-} + \mu^{+} + \mu^{-} + \mu^{+}, \dots$$

$$h \rightarrow e^{-} + e^{+} + \mu^{-} + \mu^{+}, \dots$$

I was wondering if a particle (e.g. muon) must always be produced with its corresponding antiparticle (e.g. antimuon).

Also, I was wondering if there are only a finite number of processes for this reaction.