Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B and Anti-B mesons

  1. Aug 9, 2004 #1
  2. jcsd
  3. Aug 9, 2004 #2
    Is this kind of direct CP violation easily accountable for in the Standard Model ? :confused:
    Maybe one will have to add a new angle for the [tex]B\bar{B}[/tex] system proper states, analogous to the Cabibbo angle ? :uhh:
     
  4. Aug 14, 2004 #3
    Could someone explain the difference between ordinary CP violation and direct CP violation. I guess I am not the only one to be confused here. :uhh: :confused:
     
  5. Aug 15, 2004 #4
    Two modes of possible CP violation:
    1) Indirect: particles exist that are not CP eigenstates. For example, the neutral K mesons.
    2) Direct: CP is violated during a decay process.
    This may not be 100% clear, so let me explain. The neutral kaon system is a messy thing, because kaons are usually produced by strong interactions, in quark eigenstates, eg [itex] d\bar{s} [/itex]. But the weak interaction is not diagonal wrt quark flavors, so a higher order process can convert [itex] d\bar{s} [/itex] to [itex] s\bar{d} [/itex]. Hence the free particle states are not K and K-bar, but combinations thereof. (I know, I know. Only in QM!)
    Now, you can build CP eigenstates like this:

    [tex]K_1 = \frac{1}{ \sqrt{2} }(K^0 + \bar{K^0}) [/tex]

    [tex]K_2 = \frac{1}{ \sqrt{2} }(K^0 - \bar{K^0}) [/tex]

    Direct CP violation would imply that one of these states would decay to a final state that has the opposite CP number. This usually means the number of pions it decays into. This is also easy to experiment on, because an even(?) CP state decays to two
    pions, and an odd(?) state to three pions. The three pion decay takes longer so one state is longer lived.

    Indirect CP violation means that the particles observed in the lab are something like:

    [tex]K_S = \frac{K_1 + \epsilon K_2}{ \sqrt{1 + \epsilon^2 } } [/tex]

    [tex]K_L = \frac{\epsilon K_1 + K_2}{ \sqrt{1 + \epsilon^2 } } [/tex]
    These states are not pure CP states. The notation means S(hort-lived) and L(ong-lived); the idea is that the long-lived particle will largely prefer a three pion decay but on occasion will go to two pions. (The mixing parameter is small, a few parts per thousand). The source of the violation is the fact that the states are mixes of the CP eigenstates - we could see further CP violation (the direct type) if the separate components were allowed to violate CP in their decays.

    There are other interesting questions raised by this system. For example, what should we call a particle? The strong interaction states do not have a definite mass and lifetime, yet without a doubt they're what get produced. The L and S states do have definite masses and lifetimes. It's sorta like passing a linearly polarized photon beam through a medium that preferrentially absorbs a circular polarization; only this time the medium is the vacuum!

    Anyway, I know for a fact that is possible to distinguish experimentally between the two kinds of CP violation. See http://www.slac.stanford.edu/slac/media-info/20040802/ and the links therein for a more detailed explanation; the second one from the top is most relevant.
     
    Last edited: Aug 16, 2004
  6. Aug 16, 2004 #5
    thank you zefram_c ! It is much clearer to me now. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: B and Anti-B mesons
Loading...