I Why would interacting K0L produce K+ more often than K-?

1. Aug 20, 2016

Anchovy

I am simulating lots of $K^0_{L}$ going into a volume of rock, and they sometimes produce a $K^{+}$ or a $K^{-}$. However, this does not happen in equal amounts - I am seeing roughly 3 times as many $K^{+}$ as I am $K^{-}$. What is the explanation for this?

2. Aug 20, 2016

Staff Emeritus
Are the K0-nucleon and K0bar-nucleon cross-sections identical?

3. Aug 21, 2016

Anchovy

I'm not sure where to look to check this. If they are different then that would be a simple explanation I think, but what if they are identical?

Also, can you tell me if the following diagrams are the correct interpretation of what's happening?

4. Aug 21, 2016

Staff Emeritus
OK, take one step back. Are the K+ and K- cross-sections identical? Based on that, what can you say about K0 and K0bar?

5. Aug 21, 2016

Staff: Mentor

The weak interaction is only relevant if there is no strong process doing the same.

6. Aug 21, 2016

Anchovy

The K+ and K- must not be identical if I'm seeing a disparity between them, so the K0, K0bar must differ too?

Is that as simple as replacing the W bosons in those diagrams with charged pions?

7. Aug 21, 2016

Staff: Mentor

There is an easier process, but the relative contribution might depend on the kaon energy.

8. Aug 21, 2016

Anchovy

Gluon-mediated?

9. Aug 21, 2016

Staff: Mentor

Yes.

10. Aug 21, 2016

Garlic

Is this how the reaction diagrams would look like?

Note: I would argue that the neutral Kaon is instead the one with the strange quark, and the anti neutral Kaon is the one with the strange antiquark. Is that right?

11. Aug 21, 2016

Staff Emeritus
Feynman diagrams are the wrong tool to solve this problem, especially if they use the wrong interaction or in Garlic's case don't conserve charge.

Are the K+ and K- cross-sections identical? Based on that, what can you say about K0 and K0bar? Is there some symmetry that relates them?

12. Aug 21, 2016

Anchovy

Not sure if you missed my reply earlier but I said the K+, K- cross-sections mustn't be identical seeing as different amounts are produced?

And regarding symmetries... do you perhaps mean that K0 and K0bar are both odd-parity states? $P | K^0 \rangle = - | K^0 \rangle \hspace{3 mm}$, $P | \overline{K}^0 \rangle = - | \overline{K}^0 \rangle$ ? Or that K0 and K0bar are related by charge conjugation $C | K^0 \rangle = | \overline{K}^0 \rangle$ ...?

13. Aug 21, 2016

Staff Emeritus
Hint 1: I am not talking about production cross-sections for K+ and K-; I am talking about the K+-nucleon and K- -nucleon cross-sections.
Hint 2: Isospin

14. Aug 21, 2016

Anchovy

Hmm, ok, so there are two $\pm\tfrac{1}{2}$ isospin doublets related to kaons: $(K^-, \overline{K}^0)$ and $(K^+, K^0)$, and the $\pm\tfrac{1}{2}$ isospin doublet for the nucleons $(p, n)$... and if isospin is conserved in strong interactions... maybe there are fewer ways for an isospin-conserving K+ - nucleon interaction to occur than there are for K- - nucleon... so more survive? I feel very lost here.

15. Aug 22, 2016

Garlic

Where did I make the mistake?
In the left diagram, a neutral particle (kaon) emits a negative charged particle (pion) and becomes positive. A positive charged particle (proton) captures a negative charged particle and becomes neutral.
Is the mistake in the right diagram?

16. Aug 22, 2016

Staff Emeritus
Garlic, the right diagram doesn't conserve charge, and everywhere you have the K0 and K0bar switched. I don't think posting wrong things will help Anchovy.

Anchovy, what does isospin tell you about the relation between K+ n reactions and K0 p reactions? And what does it tell you about the relation between K- p reactions and K0bar n reactions?

17. Aug 22, 2016

Anchovy

The K+ has isospin +1/2, neutron has -1/2, so a total of zero in the initial state so the final state should be zero overall too?
The K0 has -1/2, proton has +1/2, so final state should be zero?

The K- has +1/2, proton has +1/2, so final state should be +1 overall.
The K0bar has -1/2, neutron has -1/2, so final state should be -1 overall.

The K+ n and K0 p reactions involve two objects with opposite isospin, and the K- p and K0bar n reactions involve two objects with the same isospin... I'm still confused.

18. Aug 22, 2016

Garlic

I understand, you are right. But I was only asking if they were right, I wasn't claiming anything.

19. Aug 22, 2016

Staff Emeritus
Let's go back a bit. Calculating isospin is not the important part. What's important is that there is an isospin symmetry. Let's consider two reactions,
p+ + S -> anything and n + S --> anything. Here S is an isospin scalar, like 4He. What does isospin tell you about their rates?

Now consider:
K+ + S -> anything and K0 + S --> anything. What does isospin tell you about their rates?

Now consider:
K0bar + S -> anything and K- + S --> anything. What does isospin tell you about their rates?

Finally, does isospin tell you anything about
K0+ S -> anything and K-bar + S --> anything?

20. Aug 22, 2016

Anchovy

Is it that their rates have to differ because p and n have different isospin 3rd components?

21. Aug 22, 2016

Staff Emeritus
Just the opposite. Isospin is a symmetry.

22. Aug 22, 2016

Anchovy

OK, so...

p + S --> anything and n + S --> anything go at the same rate then. (ie. Nucleon is invariant under rotations in isospin space?)

K+ + S --> anything and K0 + S --> anything go at the same rate (since (K+, K0) are in the same isospin doublet?)

K0bar + S -> anything and K- + S --> anything go at the same rate (since (K-, K0bar) are in the same isospin doublet?)

and...

K0 + S -> anything and K0bar + S --> anything do NOT go at the same rate (since they're not members of the same isospin doublet?)

23. Aug 22, 2016

Staff Emeritus
Exactly.

Now, a K0L beam is half K0 and one half K0bar, right? If the K0 - nucleon cross-section is smaller than the K0bar - nucleon cross-sectioon, what will happen when a K0L beam interacts with the rock?

24. Aug 22, 2016

Anchovy

Is it that the charged kaon partners that are in each kaon isospin doublet are produced at different rates? Are we saying the doublet members, say (K+, K0), are interchangeable when a strong interaction occurs (and the same being true for (K-, K0bar) )? But if the K0 and K0bar don't interact at the same rate then the formation of K+ and K- occur at different rates also?

Last edited: Aug 22, 2016
25. Aug 22, 2016