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Why no down/anti-down?

  1. Jul 25, 2004 #1
    I've searched all over for a down/anti-down (Dd) quark pair but the closest thing I can find is the neutral Pi meson, which is: (Uu + Dd)/Sqrt(2)

    I have two questions: 1) Can someone explain that equation - neither the numerator nor the denominator makes any sense to me. What does Uu+Dd equal? They are particles not numbers, right? And Sqrt(2)...WTF??? I thought mesons were made up of 2 quarks, not 2 quarks plus 2 others plus a mathematical expression. It makes as much sense as saying a proton is made up of UUD+5 !!! Can someone shed some light on this? 2) If all the other quarks can pair with their respective anti-quarks, then why does the Dd pair not exists alone?

  2. jcsd
  3. Jul 25, 2004 #2
    Uu refers to the wavefunctions of an U-quark combined with an u-quark. The combination in QM is implemented by multiplying the wavefunctions of each constituent quark. The fact that there is a squareroot 2 has to do with normalization, so we are sure that the probability-amplitude of the total pair is 1.

    In normal conditions (i mean the groundstate) all quarks have to be paired to form a meson or they form triplets which are the baryons. the reason for this is that the strong force gets "stronger" when the energy-scale of the quarks gets small. if you were to search for solo-quarks, they would have to be highly energetic. One can "find" such solo-quarks in accelerator-experiments. At high energies all known quarks can be found alone. This is called the asymptotic freedom of the strong force.
  4. Jul 25, 2004 #3


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    The neutral pi meson is a superposition that is equal parts Uu and Dd. The reason? A Uu meson and a Dd meson are identical in every respect to the outside world. There is no way to tell a Uu from a Dd, thus they are the same particle. The neutral pi meson is a superposition of both possible configurations.

    - Warren
  5. Jul 25, 2004 #4
    Thank you for your reply. But that still doesn't reallt clarify for me what the neutral Pi meson is composed of. Does it contain all 4 quarks: DdUu? If so, then I'm still searching for a meson that is just Dd. I've seen Cc, Ss, etc, and if the only difference between a Strange and a Down quark is the mass, then why can't a Dd meson be formed in the same way that a Ss meson can?
  6. Jul 25, 2004 #5


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    To the extent the eightfold way is a good symmetry, there are no u(ubar) or ddbar pseudoscalar mesons.

    The reason for this is how isospin adds (in much the same way as spin).

    Take two particles with isospin = 1/2. You get an isospin triplet and an isosinglet.

    the |1 0 > states are (uu - dd)/sqrt2
    and an isosinglet | 0 0 > = (uu + dd)/sqrt2

    The technicalities of this is part of group theory, and looking this up in Clebsch Gordon tables.

    I like to think of these things as fields that are waving in such a way that they are not pure states. If you could decompose the neutral pi meson somehow, half the time you'd get uubar, the other half ddbar.
    Last edited: Jul 25, 2004
  7. Jul 25, 2004 #6
    Ohhh... Ok, now I'm getting it.

    Chroot - I replied before I saw your post. I see what your saying.

    Now my questions is, how come there is no way to distinguish between the Dd and Uu, but we can distinguish between the Ss, right? If the D and the U have unique quantum numbers, do we not get unique answers when computing those numbers?
  8. Jul 27, 2004 #7
    We can distinguish between the Ss,they have opposite strangeness.
    We can not say UUbar is identical to DDbar. We can say : in the (ud) plane, the physical (observable) vectors are NOT the basis u and d. They are another basis, which is the original u and d basis rotated with an angle pi/4.
    Last edited: Jul 27, 2004
  9. Jul 28, 2004 #8


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    Isn't the muon a quark-anti quark pair?
  10. Jul 28, 2004 #9
    No ! It is a lepton, such as the electron.
  11. Jul 28, 2004 #10


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    Mk, half a century ago, the muon was mistaken by the pion, but it took only a few years to notice the error. And it was years before the quark theory.
  12. Jul 29, 2004 #11


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    hmmm... ok, thanks
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