# Feynman diagram of K0 transmutation

1. Aug 8, 2011

### Gulli

1. The problem statement, all variables and given/known data

I need to find the Feynman diagram of the transmutation of a K0 particle (one anti-strange quark and one down quark) into its anti-particle (one strange quark and one anti-down quark).

2. Relevant equations

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3. The attempt at a solution

s- has a charge of +1/3, d has a charge of -1/3, so together they are neutral and they both have spin 1/2, so maybe they could annihilate into a gluon (0 charge, spin 1) which could then decay into an s and d- , so maybe it's just:

Although maybe an interaction involving the quarks changing flavour via Z0 particles (s- into d- and d into s) is also possible...

Last edited: Aug 8, 2011
2. Aug 8, 2011

### sgd37

strangeness is conserved in the strong interaction but broken by the weak

3. Aug 9, 2011

### Gulli

So the interaction involving Z0 particles is impossible, while an interaction involving gluons is possible. Can you tell me if the diagram I drew is correct?

4. Aug 9, 2011

### sgd37

the other way round since the s quark has strangeness -1 and the sbar has strangeness 1 so in the interaction you have strangeness is not conserved so the gluon interaction is not possible

5. Aug 9, 2011

### Gulli

Oh yeah, strangeness is inverted for the anti-particle, so then I have to use a Z0 instead of a gluon in my diagram?

6. Aug 9, 2011

### vela

Staff Emeritus
The Z doesn't change the flavor of quarks. The only vertex that does involves the W.

Hint: The diagram won't be a simple one where the two incoming quarks simply exchange a W. It's a bit more complicated.

7. Aug 9, 2011

### Gulli

Do they have to change flavour to change into their own anti-particles?

8. Aug 9, 2011

### vela

Staff Emeritus
Of course. The $K_0$ and $\bar{K}_0$ aren't made up of the same flavors of quarks, so the flavors have to change when the neutral kaon turns into its antiparticle.

9. Aug 9, 2011

### Gulli

Well, all I know is that the strangeness and bottomness of the meson become reversed, while total charge, mass and spin remain the same. The thing is this excercise comes from an undergraduate course that didn't delve into feynman diagrams too deep, so I probably don't need a lot of vertices.

10. Aug 9, 2011

### sgd37

11. Aug 9, 2011

### Gulli

Is that the simplest diagram possible? If so then I give up. Feynman diagrams are definitely not covered that thoroughly in the course.