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Neutral particle oscillation and energy conservation

  1. Apr 28, 2015 #1
    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.

    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):
    1. 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]
    2. Eigenstates:
      • Mass eigenstate
      • CP eigenstate
      • Strong eigenstate
      • Weak eigenstate
      • Flavor eigenstate
      • Strangeness eigenstate
      • Isospin eigenstate
     
    Last edited: Apr 28, 2015
  2. jcsd
  3. Apr 28, 2015 #2

    mfb

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    Total energy has to equal its mass, but you have rest energy and kinetic energy.
    The differences between the two particles are way too tiny to see them with precise measurements of other particles - just have a look at the K*(892) width, it is tens of MeV.

    Right.
    You can consider the CP eigenstates K1 and K2, both are their own antiparticles. They are close to the KL and KS (but not exactly -> CP violation).
    Yes.
     
    Last edited: Apr 29, 2015
  4. Apr 28, 2015 #3

    Orodruin

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    Correct. K0 and K0bar are the interaction states and KL and KS the mass eigenstates.

    The point, as you suspect, is that flavour eigenstates are not propagation eigenstates. However, the created mass eigenstates are not created with definite energy or momentum, but rather as wave packets containing a range of momenta around the expectation value. It takes time for the wave packets of different mass eigenstates to separate and thus decohere. As long as coherence between mass eigenstates is maintained, the fact that different mass eigenstates pick up different phases will lead to oscillations. This is the same for kaons as well as neutrinos.

    It would also be the case for quarks if not for the larger mass separation between quark mass eigenstates. The larger mass gap leads to faster decoherence, so fast that it makes more sense to consider the mass eigenstates only and have CC interactions violate flavour in the quark sector.
     
  5. Apr 29, 2015 #4
    OK, thanks for your answers. This clears a lot of things up for me.
    Rest and kinetic energy should be precisely constrained in the K* rest frame by four-momentum conservation, right? So it really comes down to the width of the K*, and the energy not being exactly defined?
    OK, so the K1 and K2, as well as the KL and KS, are mass eigenstates?

    I'm still not clear about the terms that are used, especially in superficial explanations of the CP-violation mechanism in some text books. I guess one has to do all the math do grasp the gritty details. I hope to get around to doing that at some point. For now, please help me to understand this:
    • When talk is of "strong eigenstates", are these identical to "flavor eigenstates"? Could these also be considered "strangeness eigenstates", "isospin eigenstates", and "color eigenstates"?
    • Are "weak eigenstates" in that sense identical to "CP eigenstates", even though the weak interaction violates CP-invariance?
    • Are "mass eigenstates" an additional concept? How does the term "propagation eigenstate", which you mentioned, fit into this?
    Ah, that's very interesting. I've never heard the concept stated in this way, especially the last part about flavor change.

    I have two more scenarios I'd like to discuss:
    1. Look at the decay of KL and KS into two or three muons. Suppose we could measure all decay products with arbitrary precision. The electrons, photons, and neutrinos should all have width zero, so suppose we could measure the invariant masses down to the kaon mass differences. For the Cronin-Fitch-scenario with a "pure KL beam", would I always measure the KL mass, even for its CP-violating decay into two pions?
    2. If I managed to collide two pion beams at the kaon resonance, would the weak interaction produce mainly KS that decay very quickly, with a few long-living KL from CP violation? Would that be considered the time-inverted process to the decay, and does T-symmetry violation come into play here?
     
  6. Apr 29, 2015 #5

    mfb

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    Constrained to what? To something that is not too far away from the K* mass: yes.

    K0 and ##\bar{K^0}## are flavor eigenstates (they have a well-defined quark content), I guess "strong eigenstate" means the same. They are also eigenstates for all quark counting numbers (strangeness, isospin).
    K1 and K2 are the CP eigenstates (they have well-defined CP state and decay mode)
    KS and KL are the weak eigenstates and mass eigenstates (they have well-defined mass and lifetime in the weak interaction) <- sorry, got the mass part wrong in the previous post (now fixed)

    All those particles are color neutral as we have hadrons.

    Pions, not muons. You would get different masses for KL and KS, independent of the number of pions in the decay.

    Should work like that I think.
     
  7. Apr 30, 2015 #6
    OK, that helped me a lot. Thank you!

    Of course, my bad.
     
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