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Particle octet (another try)

  1. Apr 24, 2008 #1
    I'm interested in comments on thishttp://camoo.freeshell.org/25.6.gif" [Broken].
    I tried to upload it to the forum but apparently that doesn't work well with .gifs.
    It's an exercise in a book asking you to explain the particle octet, why there aren't uuu, ddd and sss spin 1/2 particles, there are 1 each of uud, uss, etc. etc., and 2 types of uds particles. I'm wondering whether I got the wavefunctions for the uds particles right. I was just guessing, came up with something nice and symmetrical that looks right, but I don't know if there's a way to tell if it is right.
    Yes, I know that according to the color theory, the quarks aren't actually following boson statistics. This exercise was before that was explained in the book, all we were told was that the quarks are acting like bosons.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Apr 25, 2008 #2


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    I did not follow all your reasoning, I would just simply state that the only totaly symmetric state with respect to two particle interchange in 3quark system is if you have all three quark spins in the same direction, wich gives you S = 3/2

    Are you using "Particle Physics" by Martin & Shaw ?

    The wave function of uds will just be the total antisymmetric wave function when you couple 3 spin 1/2 particles. I.e you can first couple u & d to S = 1 or 0.
    Then you couple the s quark to S = 1 to get S_tot = 3/2 or 1/2.
    Or you couple the s quark to S = 0 to get S_tot = 1/2.
    i.e You'll get three such states for uds
  4. Apr 25, 2008 #3


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    Quarks are in a completely asymmetric color state, so what they follow, if color is not explicit, is effective Bose statistics. This could be called "quark statistics".
    For 6 of the octet, two quarks are the same, and must be in the symmetric spin one state. Coupling this spin one to spin 1/2 gives either 1/2 or 3/2.
    The decuplet arises because the 3 uuu type states and the uds can be included with the other 6. Since the spin 1/2 state is not completely symmetric, it can't be coupled with states like uuu.
    The 8 octet states include 2 uds states. That is because the ud pairing can be in either spin one or spin zero.
    The two uds spin fucntions are [tex][aba-baa]/\sqrt{2}[/tex] for the Lambda, and
    [tex][2aab-aba-baa]/\sqrt{6}[/tex] for the Sigma zero.
  5. Apr 26, 2008 #4
    It sounds like I got the basic idea of the spin states for the proton, neutron, etc. right, but my guess about the 2 possible uds particles was maybe too symmetric to be right. The two states I thought might be the particles are the complex conjugate of each other, at least if you're ignoring anything besides spin.
    I mean, you have this vector space of dimension 2 containing 3 possible spin 1/2 states for uds, and I wasn't sure how you decide which vectors are the "particles"? You make a Hamiltonian for the vector space and you find 2 eigenvectors, and they have different eigenvalues, and you go "Aha, those are the particles?"
    I think deciding which state vectors correspond to particles is something we don't know how to do yet in the book, there's been no explanation of flavor symmetry and that would be involved.
    I'm reading the Penrose book "The Road to Reality" supplemented by outside looking around when an exercise puzzles me.
    Last edited: Apr 26, 2008
  6. Apr 26, 2008 #5


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    The Sigma zero spin state should be the same as the Sigma plus.
  7. Apr 26, 2008 #6
    I see. So the particles that have the same letter have the same spin state, if symmetrized right? Then the [tex]\Lambda^0[/tex] is orthogonal to the [tex]\Sigma^0[/tex]. Seems like the spins are considered more important to what the particle "is" than the quark composition.
  8. Apr 27, 2008 #7


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    They play an equal role. If two particles have the same quark composition, then spin state distinguishes them.
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