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

[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?

 

[Vanadium 50]

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

 

[Anchovy]

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

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?

 

[Vanadium 50]

I'm not sure where to look to check this.

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

 

[mfb]

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

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

 

[Anchovy]

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

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

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

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

 

[mfb]

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

 

[Anchovy]

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

Gluon-mediated?

 

[mfb]

Yes.

 

[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?

 

[Vanadium 50]

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?

 

[Anchovy]

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?

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 ...?

 

[Vanadium 50]

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

 

[Anchovy]

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

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.

 

[Garlic]

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.

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?

 

[Vanadium 50]

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?

 

[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?

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.

 

[Garlic]

I don't think posting wrong things will help Anchovy.

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

 

[Vanadium 50]

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?

 

[Anchovy]

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?

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

 

[Vanadium 50]

Just the opposite. Isospin is a symmetry.

 

[Anchovy]

Just the opposite. Isospin is a symmetry.

OK, so...

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?

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?)

 

[Vanadium 50]

K0 + S -> anything and K0bar + S --> anything do NOT go at the same rate

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?

 

[Anchovy]

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?

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?

 

[Vanadium 50]

Well, let's not worry about CP violation for the moment, since it's well below a 1% effect.

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?

Right, because where do the K+ and K- come from? Among other places from charge exchange reactions like K0bar + p --> n + K+. Knowing now what you do about cross-sections and the K0L composition, can you see now why you get an asymmetry in K+ vs. K-?

 

[Anchovy]

Well, let's not worry about CP violation for the moment, since it's well below a 1% effect.
Right, because where do the K+ and K- come from? Among other places from charge exchange reactions like K0bar + p --> n + K+. Knowing now what you do about cross-sections and the K0L composition, can you see now why you get an asymmetry in K+ vs. K-?

Ah I deleted that CP violation comment earlier, didn't want to confuse things.

I'm not 100% confident yet but let me see:

(K+, K0) in one isospin doublet. (K-, K0bar) in another.
These objects are each symmetric under isospin rotations?
(ie. the strong interaction sees K+ and K0 as the same thing, and it sees K- and K0bar as the same thing) ?

But it does not see K0 and K0bar as the same thing, since they don't appear in the same isodoublet ?

--> The K0 and K0bar have different nucleon interaction cross-sections, so if the K0bar has a bigger cross-section than K0 then the charge exchange interactions that have K0bar in the initial state and K+ in the final state happen more than the those with K0 in the initial state and K- in the final state?

 

[Vanadium 50]

That's the basic idea. There are other reactions than charge exchange, but the basic idea remains the same: kaons and antikaons (both charged and neutral) have different cross-sections, and that difference manifests itself as an asymmetry in outgoing kaons from a K-long beam.

Now, to see if you understand: if one repeats this epxeriment with antirock - rock made of antimatter, would the asymmetry stay the same or would it flip signs?

 

[Anchovy]

That's the basic idea. There are other reactions than charge exchange, but the basic idea remains the same: kaons and antikaons (both charged and neutral) have different cross-sections, and that difference manifests itself as an asymmetry in outgoing kaons from a K-long beam.

Now, to see if you understand: if one repeats this epxeriment with antirock - rock made of antimatter, would the asymmetry stay the same or would it flip signs?

If we replace the nucleons (p, n) in the matter with antinucleons (pbar, nbar), there's nothing to say that they would have the same interactions, seeing as these two kaon doublets that are antiparticles of each other interact differently, so I'm going to say the asymmetry would flip signs?

 

[Vanadium 50]

Right answer, but you are not taking it far enough. There's more than just "there's nothing to say that they would have the same interactions" - it's that symmetry forces (e.g.) K+ on matter to look like K- on antimatter.

Particle physics is largely a study of symmetries and conservation laws. The best way to approach a problem is by asking "what symmetries apply, and what do they tell me about the system".