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Problem with a uuu particle with spin 1/2?

  1. Nov 24, 2006 #1


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    A baryon comprising uuu (delta++, spin 3/2) exists albeit briefly giving rise to evidence for colour as an additional quantum number, but what's the problem with a uuu particle with spin 1/2? Why can't the individual quarks be RGB, giving a colourless result with the spins being +,+,- to give +1/2 as a total spin?

    Secondly why can't you have hadrons comprising 4 quarks (colourlessness being retained by having qqbar pairs)?

    Thirdly, in state vectors, why do some components have positive signs and others negative?

  2. jcsd
  3. Nov 24, 2006 #2


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    Note that two fermions cannot be in a bound state and have the same quantum numbers. Work out your uuu puzzle with that fact in mind.
  4. Nov 24, 2006 #3


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  5. Nov 24, 2006 #4


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    So, let (u,d) a SU(2)_flavour doublet. Do not confuse it with a SU(2)_weak doublet. Let me call its "total flavour spin" F and its third component F3. So u is the F3=+1/2 and d the F3=-1/2 part of a F=1/2 doublet.

    lets tensor two doublets, we get a F=1 triplet and a F=1 singlet. The rules are spelled here:
    The triplet is
    [tex]1,+1 \equiv |u u>[/tex]
    [tex]1, 0 \equiv {1\over \sqrt 2} | u d> + {1 \over \sqrt 2} | d u>[/tex]
    [tex]1, -1 \equiv |d d>[/tex]
    And the singlet is
    [tex]0, 0 \equiv {1\over \sqrt 2} | u d> - {1 \over \sqrt 2} | d u>[/tex]

    Now lets add the third quark. For the singlet it is sort of trivial and it becomes again a doublet
    [tex]1/2, +1/2 \equiv {1\over \sqrt 2} | u d u > - {1 \over \sqrt 2} | d u u>[/tex]
    [tex]1/2, -1/2 \equiv {1\over \sqrt 2} | u d d> - {1 \over \sqrt 2} | d u d>[/tex]

    For the triplet we need to use again the tables. We are adding a F=1 to a F=1/2 sp we get a F=3/2 4-plet and a F=1/2 2-plet.

    let me do the first term of doublet with some detail:
    [tex]1/2, +1/2 \equiv \sqrt \frac 23 | 1, +1> | 1/2, -1/2> - \sqrt \frac 13 | 1, 0> |1/2, +1/2> =[/tex]
    [tex]= \sqrt \frac 23 |u u> |d> - \sqrt \frac 13 ({1\over \sqrt 2} | u d> + {1 \over \sqrt 2} | d u>) |u>
    =\sqrt \frac 23 |u u d> - \sqrt \frac 16 | u d u > - \sqrt \frac 16 | d u u >

    for the second term of the doublet I omit the F and use only F3, as the tables do:
    [tex]1/2, -1/2 \equiv \sqrt \frac 13 | 0> |-1/2> - \sqrt \frac 23 | -1> |+1/2> =[/tex]
    [tex]= \sqrt \frac 13 ({1\over \sqrt 2} | u d> + {1 \over \sqrt 2} | d u>) |d> - \sqrt \frac 23 |dd> |u>
    =\sqrt \frac 16 |u d d> + \sqrt \frac 16 | d u d > - \sqrt \frac 23 | d d u >

    At this moment you should be able to see where the signs come from and to note that we have got to decide what doublet is the one we wish to represent proton and neutron

    The 4-plet. Dont get enthusiastic, remember the F=3/2 is not spin but "flavspin".
    [tex]3/2, +3/2 \equiv |1> |+1/2> = |uu> |u> = |uuu>[/tex] that was easy.
    [tex]3/2, +1/2 \equiv \sqrt \frac 13 |+1> |-1/2> + \sqrt \frac 23 | 0> |+1/2> =[/tex]
    [tex]= \sqrt \frac 13 |u u> |d> + \sqrt \frac 23 ({1\over \sqrt 2} | u d> + {1 \over \sqrt 2} | d u>) |u>
    =\sqrt \frac 13 |u u d> + \sqrt \frac 13 | u d u > + \sqrt \frac 13 | d u u >
    Do you notice it is orthogonal to the one of the previous doublet? That is the way these things are built.
    [tex]3/2, -1/2 \equiv \sqrt \frac 23 | 0> |-1/2> + \sqrt \frac 13 | -1> |+1/2> =[/tex]
    [tex]= \sqrt \frac 23 ({1\over \sqrt 2} | u d> + {1 \over \sqrt 2} | d u>) |d> + \sqrt \frac 13 |dd> |u>
    =\sqrt \frac 13 |u d d> + \sqrt \frac 13 | d u d > + \sqrt \frac 13 | d d u >
    [tex]3/2, -3/2> \equiv |ddd> [/tex]

    (Note: I had never done this explicitly!)

    Note that we have learnt something about "flavspin" which must be useful now when considering spin: that the S=3/2 (or the F=3/2) composition of three S=1/2 (or F=1/2) particles is symmetric.

    Now the point is that we want the whole wavefunction to be antisymmetric. We need to see it as a product of the spatial wavefunction times spin times colour times flavspin. (OK, ok, I will call it Isospin. But at the age of isospin, there was not such quarks, they appeared to explain SU(3)_flavour).

    Someway, when the colour and spatial parts are sorted out, the rest is forced to be symmetric. This means that either both spin and isospin are in a symmetric representation, or both are in an antisymmetric representation.

    But |uuu> lives exclusively in a symmetric representation of flavspin (of isospin). So it must live also in a symmetric representation of spin. The only one available is S=3/2. Thus we have answered your question, or reduced it to:

    Why does the product of spatial and color parts of the eigenfunction happen to be fixed to be antisymmetric?

    And in the forest of doublets we have a secondary quest: what is the eigenfunction of the proton? It should be a product of antisymmetric isospin times antisymmetric spin
    Last edited: Nov 24, 2006
  6. Nov 26, 2006 #5


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    Fine, I killed the thread :-D
  7. Nov 26, 2006 #6


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    The posed question, I believe, is what happens about the state where we have three quarks [tex] |u, +1/2, R> \otimes |u, +1/2, G> \otimes |u, +1/2, B> [/tex], obviously such quarks do not have the same quantum numbers.
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