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I Direct and indirect CP violation

  1. Feb 9, 2018 #1

    Gene Naden

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    The Wikipedia article on CP violation says there is indirect violation in Kaon decay but does not give details. What is the reaction that demonstrates CP violation?
     
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  3. Feb 9, 2018 #2

    Gene Naden

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    The Wikipedia article on CP violation says direct violation was observed in kaon decay but does not give details. What is the reaction and why is it considered direct cp violation?
     
  4. Feb 9, 2018 #3

    ChrisVer

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  5. Feb 10, 2018 #4

    Gene Naden

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    I am not sure if I should be posting to this thread or the other one. They both say I should have made just one thread.

    I looked at the reference, chapter 2, and got stuck right away. It said, that the action of CP is to convert the kaon into an antikaon with a 180 degree phase shift (in the footnote it says that since flavor is conserved by strong interactions, you can use phase-shifted states. Could someone give me a hint about this, or where to read about it? I guess the states are eigenstates of the strong interaction Hamiltonian but cannot get further in my reasoning.
     
  6. Feb 10, 2018 #5

    Gene Naden

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    I found a source that says kaons have odd intrinsic parity. I guess that is why the action of CP transforms a kaon into minus the antikaon. Still don't understand the footnote but maybe that is not important...
     
  7. Feb 10, 2018 #6

    Gene Naden

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    The reference, chapter 2, page 16, says "the physical KL were not purely CP eigenstates but the result of a mixing between both CP odd K2 and CP even K1, or that these transitions directly violated CP since an odd state decayed into an even state." So, which is it, direct or indirect? I realize I am zeroing in on the details when I need to get the big picture.
     
  8. Feb 11, 2018 #7

    ChrisVer

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    Well, first of all, do you understand that the decays to three pions is CP-odd and to two pions is CP-even? I mean why you need the K1 (CP-even) and K2 (CP-odd). If CP is conserved, then the K1 and K2 are the Klong and Kshort.

    If CP was not violated, then that would be always the case. However, in experiment we observed that the K2 (long) would decay into 2 pions (Fitch and Cronin)! You do that by exploiting that the Kshort lives for a shorter time; so if you have a beam of neutral Kaons and allow it to travel some distance you'll be dominantly left with K-longs. CP-violation.

    Then the explanation comes from 2 components:
    1. CP-indirect violation. As the name suggests you introduce CP-violation by doing something indirectly... in which case the Klong and Kshort are not exactly the CP-eigenstates, but a mixture of those)... So the Klong has both a component of K1 and K2 which would allow it to decay into 2 or 3 pions with probabilities depending on the CP-state mixture. This is parametrized by the parameter ##\epsilon## which is the mixing phase:
    [itex] K_L = \frac{1}{\sqrt{1-|\epsilon|^2}} ( K_2 + \epsilon K_1 ) [/itex]
    So the ##K_L## decays to 3 pions via the ##K_2## and to 2 pions via the ##K_1## (which is weighted by the ##\epsilon##)

    2. CP-direct violation, again as the name suggests, here you brute-heartedly violate the symmetry. In which case you have the direct violation of CP : ##K_L = K_2 \rightarrow \pi \pi## which is parametrized by the parameter ##\epsilon'##.

    Both mechanisms contribute in the Kaon CP-Violation, but the 2nd not as strongly...

    In fact you can go further into this and start thinking of the actual physics/mechanisms in those cases, so you would start breaking the phases into real parts (eg. the ##Re(\epsilon)##, the ##Re(\epsilon ')## and the imaginary parts: ##Im(\epsilon) , Im(\epsilon ')## measure different things in CP-Violation)
     
    Last edited: Feb 11, 2018
  9. Feb 12, 2018 #8

    Gene Naden

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    Thanks for your reply. That was helpful. It sounds like ##K_L \rightarrow \pi \pi## includes both direct and indirect CP violation. I wonder how the experimenters distinguish between the two.
     
  10. Feb 12, 2018 #9

    ChrisVer

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    You target different measurable quantities/predictions of each case...

    The indirect CP violation if I recall well is measured precisely using the transitional probabilities (As you have neutral Kaon oscillations of the form [itex]K^0 \rightarrow \bar{K}^0 \rightarrow K^0 \rightarrow ... [/itex]).

    The direct CP violation in the Kaon system was done by NA48 Experiment measuring the ratio:
    [itex]R = \Bigg|\dfrac{\frac{A(K_L \rightarrow 2\pi^0)}{A(K_S \rightarrow 2 \pi^0)}}{\frac{A(K_L \rightarrow \pi^+\pi^-)}{A(K_S \rightarrow \pi^+ \pi^-)}}\Bigg|^2[/itex]
    That quantity is also proportional to ##\epsilon'## by ##R \approx 1 - 6Re \Big( \frac{\epsilon'}{\epsilon}\Big)## . If there's no direct CP-Violation, then that fraction will be 1 (##\epsilon' =0##).

    So all you need to know is the rate the KL decays to 2 neutral pions over that of KS, and the similar ratio for the decay to 2 charged pions.
    What they did was to measure those fractions and they resulted to:
    [itex] Re(\epsilon'/\epsilon ) = (1.47 \pm 0.22) \times 10^{-3}[/itex]
    This is non-zero and thus the direct-CP Violation was proved/observed. Also it gives a feeling to what I said above- that the indirect CP-Violation is playing a more dominant role.
     
  11. Feb 12, 2018 #10

    Gene Naden

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    Thanks.

    Well that is interesting and I am ready to dive into what you have written. But I have a question about "A". Would that be the reaction rate? I don't think it could be amplitude as you cannot measure amplitudes directly, right?
     
  12. Feb 12, 2018 #11

    Gene Naden

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    So I looked up a Cornell University report on the NA48 experiment and it says that the A's are indeed amplitudes. So I am wondering how you measure amplitudes. (I have ordered Griffiths' book on Particle Physics, that may help, but it won't arrive for another week or so). It talks about ##\Gamma##. Maybe that is what is measured?
     
  13. Feb 12, 2018 #12

    Gene Naden

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    So it seems that ##\Gamma## is the decay width... now to look at the theory. Oh joy.
     
  14. Feb 13, 2018 #13

    ChrisVer

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    It's absolute value squared, which is basically a probability density and proportional to the widths ##\Gamma## [I could have written just the width ratios]
     
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