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Neutral Kaon decay and CP violation

  1. Apr 30, 2005 #1
    First of all, many books assume first that K_S and K_L are CP eigenstates. They then tell that by observation that cannot be true. Is that relevant to the CP violation? Why don't the authors just tell that K_L is a mixed state with only a small amount of CP=+1 eigenstate?

    Second, observing CP=+1 decays at long distances is regarded as CP-Violation. Why is that? Is it because the K_0 was a perfectly balanced mixture of both CP eigenstates directly after the strong decay (eg. pi^- and Proton to Lambda and K_0) and now the decay products do not show this perfect balance any more (we have more CP=+1 decays)?

    Third and last, how exactly is CP violation defined? The whole process of the neutral Kaon decay requires us to take different decay rates for the two CP eigenstates into account. Without CP=+1 states decaying before the CP=-1 states, this sort of CP violation could not be observed because the CP-changing weak interaction would have no net effect on the CP amplitudes of the Kaon. Is that correct? If yes, how comes that the definition of CP violation takes into account the CP=+1 decay processes? Isn't that a too far-reached definition/interpretation of the CP symmetry?

    Is there any book/article where that problem is being treated with rigour? I have perused the particle data guide book but the reviews there are only very superficial (IMHO).
     
    Last edited: Apr 30, 2005
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  3. Apr 30, 2005 #2

    ZapperZ

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    Wow! You don't ask simple questions, do you? :)

    When I talked to the high energy physicists in my division a few years ago, after a long conversation, they pointed to me these three articles of varying degree of difficulties. They were helpful (to a varying degree also), and they contain the description of the weak decay of both the K_short and K_long and how they compare with the SM prediction. So who knows, these things may be helpful to you too.

    http://xxx.lanl.gov/abs/hep-ph/0109240
    http://xxx.lanl.gov/abs/hep-ph/0101336
    http://xxx.lanl.gov/abs/hep-ex/0002039

    Zz.
     
  4. May 4, 2005 #3

    Meir Achuz

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    1. Most books refer to K_1 and K_2 as the CP estates. Then K_S and K_L are the mixed states you describe. If you saw a book that said K_S and K_L were CP estates, it must have been written in the 50's or early 60's.
    2. Before CP violation was seen, the argument went like this:
    The original K in p-->K^0+Lambda is an even mix of CP + and CP - states, as you say.
    The CP+ can decay relatively quickly into the CP+ state of 2 pions.
    The CP- cannot decay into two pions by CP conservation.
    Therefore, it will have a much longer lifetime and travel a long distance before decay into 3 pions. The observation of these long lived K's into 3 pi's was originally a triumph for CP conservation. There should have been no 2 pi decays at long distance, so their later observation by Cronin and Fitch demonstrated CP violation.
    3. I am not sure what you are saying in your third point. It might be that the key element you have missed is that the CP+ state can decay quickly into two pions, so no CP+ should be left in the long lived beam.
    4. There is an old book by P. K. Kabir (maybe titled "the CP Puzzle") that gives probably the best discussion of this.
     
  5. May 4, 2005 #4
    Actually, I was somewhat confused about the fact that CP violation seems to be correlated to phase space arguments. But, it seems to be clear to me now, that the different dynamics because of the different phase spaces just influence the way we detect the CP violation in this special case -- it has nothing to do with CP violation directly. In other words: if the two CP eigenstates would have the same phase space, then we would not observe CP violation, but it would still be there -- just hidden by the fact that the neutral Kaon has the same "amount" of CP+ as it has CP-...

    Is that perception correct?

    Btw: if a Kaon does not decay into a 2pi within the first few inches, where does the CP+ part of the wave function go? Isn't it still there? Why doesn't it decay any more?
     
    Last edited: May 4, 2005
  6. May 4, 2005 #5
    This book is really good, but really expensive.

    http://www.oup.com/us/catalog/gener...lecularOpticalphysics/?view=usa&ci=0198503997

    Also - search for CP violation at

    http://www.arxiv.org/find

    There are hundreds of articles.
     
    Last edited: May 4, 2005
  7. May 4, 2005 #6

    Meir Achuz

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    Yes, it is the small phase space for three pions that give the K_L its long life.
    The K_S lifetime is about 10^-10 sec, so they are all gone in a short distance.
    I am afraid the CP papers on the arxiv would be too technical, and not much good for basic understanding.
     
  8. May 4, 2005 #7
    There should be a few out there which are pitched at more general audiences.
     
  9. May 5, 2005 #8
    Thx. I'll have a look at it, tomorrow.
     
  10. May 6, 2005 #9
    Unfortunately, the suggested book treats the indirect CP violation in the neutral Kaon system as superficial as most others do.
     
  11. May 17, 2005 #10
    Ok, here is a second try:

    Is it possible to explain using the standard model that if there is no K_S decay within 10^-10 seconds, then there won't be any K_S decays after, for example, 10^-8 seconds? The neutral Kaon is a balanced superposition of amplitudes for CP=1 and CP=-1, correct? So if the Kaon does not decay as a K_S within 10^-10, there is no reason to assume that its amplitude has disappeared somehow just because some time has elapsed, correct? The only possible way _I_ can explain this, is that the Kaon decides immediately after its creation to be a K_S or a K_L -- but wouldn't that require another, additional interaction which itself violates CP symmetry maximally, ie. much more than the weak force?

    Are there any theories about the construction of the weak force out of two simpler theories: one that conserves CP and one that violates CP maximally?
     
  12. May 17, 2005 #11
    I'm don't really understand all of your question. Maybe it's about the nature of measurement in QM?

    In the Branco book that I mentioned, there is the formalism for mixing mentioned on p.60.

    You start off with the flavor eigenstates, K0, K0bar. As the book mentions - with no weak interaction, these two eigenstates would be stable and not mix. When you switch on the weak interaction (which I guess is essentially immediately), these states mix and decay, so we are interested in the mixing, (or mass) eigenstates, K_L and K_S.

    On p91, they mention K_S regeneration, which occurs when you let the beam go through some material. This material scatters the kaons via the strong interaction, and thus, you select the K0 and K0bar, thus regenerating the K_S component.

    Also - when you detect semileptonic neutral kaon decays, you detect the flavor eigenstate.
     
  13. May 18, 2005 #12
    This is almost touching the part I find very obscure, but again: the authors talk of an entire beam, of a statistical entity, instead of just one process.

    Ok, let me state a question that should help to state my problem clearly:

    Assume the production of a neutral Kaon. Assume we are able to detect if a singly generated Kaon has not decayed after a specific lifetime, eg. 10^-7 seconds. What is the quantum mechanical or field theoretical description of that single Kaon at that point in time?
     
  14. May 18, 2005 #13
    Assume we start with a single K0. The only interpretation that would make sense to me is to say that the presence of the weak interaction selects either the K_L or K_S state with 50% probability, and the time evolution of the K_L or K_S is then given by eqns. 6.38 in the book - the standard exponential evolution laws with well-defined masses and decay widths (and which I don't have time right now to retype).

    That's how event generators for particle accelerator experiments do it. Any K0 in the vacuum is automatically "decayed" to a K_L or K_S with a 50% probability for each, then the K_L or K_S is decayed according to its lifetime.
     
  15. May 18, 2005 #14
    Exactly. That is my very problem: how does the weak (or any othertype of) interaction do this selection? You must keep in mind that the neutral Kaon mixing due to the weak force is described by a rather small parameter -- so it seems very unlikely to me that the weak force can do such an extreme thing in respect to CP eigenstates... for selecting a K_S or a K_L, it is necessary to convert (practically) instantaneously the whole CP + or - fraction into the opposite CP eigenstate.
     
  16. May 18, 2005 #15
    That's the way quantum mechanics works, I believe. As long as you can write down the mixing eigenstates due to the presence of the weak force, that is what ANY presence of the weak force will select.

    I mean, how are the ideas that much different from those demonstrated by the Stern-Gerlach experiment?
     
    Last edited: May 18, 2005
  17. May 18, 2005 #16
    Think about it: start with a spin +1/2 electron in the z direction. If you measure its spin in the x direction, it IMMEDIATELY becomes a superposition of spin +1/2 and spin -1/2 states in the z direction.
     
  18. May 27, 2005 #17
    I don't think that has much to do with the weak force. I came to the conclusion that "measuring that the Kaon has not decayed" is the same as we have measured the Kaon itself: therefore it is an indirect quantum-mechanical measurement that has the same effect as regeneration.

    Agreed?
     
  19. May 27, 2005 #18
    No - it has 100% to do with the weak force, since without the weak force, there would be no decay to start with.
     
  20. May 27, 2005 #19
    Also - pretty much everything I've said in the previous posts in this thread deal with weak interaction phenomena which would still occur in the absence of CP violation, i.e. a complex phase in the standard model CKM matrix. So when trying to understand this stuff, just throw CP violation out the window. It's irrevelant. Just assume that CP eigenstates are also mass eigenstates. The formalism still holds.
     
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