I have question that I've been thinking about for some time now, and that I can't get my head around. I an experimentalist without education in quantum field theory, and my quantum mechanics introduction courses were a long time ago, so bear with me please.(adsbygoogle = window.adsbygoogle || []).push({});

As far as I understand, neutral kaons oscillate between [itex]{K}^0[/itex] and [itex]\bar{K}^0[/itex], and this process comes about because of box-type Feynman diagrams. They decay either very quickly as [itex]{K}_\text{S}[/itex] or after a long lifetime as [itex]{K}_\text{L}[/itex]. This is sometimes explained as: The short-lived component of the wave-function decays quickly, after which only the long-lived component remains.

Does that mean that at the time the particle is created, it is not created as a [itex]{K}_\text{S}[/itex] or [itex]{K}_\text{L}[/itex], and its lifetime is not fixed? As far as I know, the particles are produced, e.g., as [itex]{K}^0[/itex] (a flavor eigenstate?), which can be written as a mixture of [itex]{K}_\text{S}[/itex] and [itex]{K}_\text{L}[/itex]. These two have different masses.

My problem is: why does this not contradict energy conservation? I guess it's a quantum effect, having to do with not knowing the particle's mass until it is measured. But then, I don't understand how it is produced in the first place. For producing a particle, I need energy equal to its mass. If, say, I look at the decay [itex]K^*(892) \to K^0 \pi^0[/itex], knowing the [itex]K^*(892)[/itex] mass and the pion's energy and momentum should let me calculate the kaon's mass. So how can I have a chance of measuring different masses later (even though the difference is on the scale of μeV)? Does the particles' width/mass uncertainty give the necessary wiggle room? Or would an exact measurement of the pion "collapse" the state and determine whether I have a [itex]{K}_\text{S}[/itex] or [itex]{K}_\text{L}[/itex]?

The same question bugs me with neutrinos, which oscillate between flavors but are assumed to have (as far as I know) slightly different masses.

I'm sorry for the long post. Here are a few additional points that confuse me, mainly concerning nomenclature:

- Is it true that the [itex]{K}^0[/itex] is a flavor eigenstate, but not a mass eigenstate?
- Does that mean that I cannot assign a mass to it at all? Then again, the [itex]{K}^0[/itex] and [itex]\bar{K}^0[/itex] are the only particle-antiparticle pair here, so does the principle of the two having equal masses not apply here?

- Does the strong interaction always produce flavor eigenstates?
- How do the terms in these groups relate to each other? (I'm not expecting an answer here, I just want to demonstrate my confusion):

- Kaons

- [itex]{K}^0[/itex] and [itex]\bar{K}^0[/itex]
- [itex]{K}_1[/itex] and [itex]{K}_2[/itex] (often used in explanations of CP violation as the not-quite CP eigenstates)
- [itex]{K}_\text{S}[/itex] and [itex]{K}_\text{L}[/itex]
- Eigenstates:

- Mass eigenstate
- CP eigenstate
- Strong eigenstate
- Weak eigenstate
- Flavor eigenstate
- Strangeness eigenstate
- Isospin eigenstate

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Neutral particle oscillation and energy conservation

**Physics Forums | Science Articles, Homework Help, Discussion**