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Isospin question

  1. May 15, 2007 #1


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    Hi you guys!

    I am a little bit confused about nuclear isospin.

    u and d quarks have I = ½

    u has I_3 = ½ , d has I_3 = -½ (third component)

    right so far?

    Now why does the proton have I = ½ and I_3 = 1/2, and
    neutron I = ½ and I_3 = -1/2 ??

    but when you add upp the I on the quark level , you might get things like 3/2 etc, has that something to do with the strong intreaction?

    And how do one assign I and I_3 to p-bar and n-bar? [anti-proton, anti-neutron]

    please help dear Nuclear and Particle lovers :)
  2. jcsd
  3. May 15, 2007 #2
    AHH, this is a classic one. The answer goes like this :

    Angular momentum doesn't add up linearly, as you were suggesting to do. That is, a proton contains three quarks, but its spin is not 3/2, but 1/2. When you add angular momentum, you have a range of possible results, which goes from the maximum (the arithmetic sum) down to the minimum (the larger minus the smaller) with possible values at unit steps in between.

    If you add two spin-1/2 fermions, you get either spin 0 or spin 1 (a boson). If you add 5 units of angular momentum to 2 units, you get 3, 4, 5, 6, or 7 units of angular momentum. If you add 3/2 units and 2 units, you get 1/2, 3/2, 5/2, or 7/2 units. A proton has spin 1/2. As you should expect, there's another particle with the same quark content as the proton, but spin 3/2, the Δ+.

    In QM, one can calculate all of this using the Clebsch Gordon coefficients. Do you know this technique ? If not, THIS is what you need to be studying then.

  4. May 15, 2007 #3


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    aha ok its "just" vector summation?

    This is overcourse right now :) I will have courses in advanced QM in the autuum =) thanks!!
  5. May 15, 2007 #4
    No, this is slightly more elaborated :smile:
    Something like "Decomposition in a direct sum of a tensor product of irreductible representations of Lie groups"
    It is actually not as complicated as it might seem. You can check wikipedia about Clebsch-Gordan coefficients
  6. May 15, 2007 #5


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    hehe ****! =)
  7. May 16, 2007 #6


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    So does the anti-neutron have I = 1/2 and I_3 = 1/2
    and anti-proton have I= 1/2 and I_3 = -1/2 ??
  8. Jun 26, 2010 #7


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    Sorry to raise a dead thread but I have the same question, I think that the answer is yes, but not sure. Can anyone confirm or disconfirm?

    EDIT: Nevermind, it's yes.
    Last edited: Jun 26, 2010
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