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Uuu in spin 1/2 config

  1. Feb 12, 2009 #1
    First, this is my 1st post on physicsforums.com, so I'm not too sure I'm posting this in the right place, but guts tell me this is a good place to start.

    Second I'm sure you've all heard this question a million times. I've seen it answered here a couple of times, but what I've seen is of no help. I'm trying to derive various hadronic wavefunctions, but I'm stuck on this one for now (and if I can't do this one, this means I don't really understand what I'm doing for the others even though they seem to work). It is my understanding that baryons are made of quarks, and have 4 degrees of freedom: Space, flavour, spin, and color. The spatial part of the wavefunction is assumed symmetric under particle exchange, and in the case of a uuu baryon, the flavour part is symmetric as well (which is trivial to prove). If this state of the uuu is to be seen, this leaves the spin+colour part to be globally antisymmetric, so the overall function is antisymmetric (per Pauli).

    Now everywhere, people invoke Pauli to exclude a spin 1/2 uuu baryon (or any spin 1/2 single-flavoured baryon). But it seems to me that if colour is chosen antisymmetric, then a spin 1/2 config is permitted.

    Let r, g, and b denote the three colours.

    You have 6 permutations of colours (rgb, rbg, grb, gbr, brg, bgr), so the wavefunction is

    |colour> = A |rgb> + B|rbg> + C|grb> + D|gbr> + E|brg> + F|bgr>

    Now under the constraints |psi 123> = -|psi 213> (A=-C, B=-E, D=-F) and |psi 123> = -|psi 321> (A=-F, B=-D, C=-E), a solution is possible, namely A=-B=-C=D=E=-F, which can be normalized to A=exp(i\phi)/sqrt{6}. And thus you get a colour wavefunction of

    |colour> = exp (i\phi)/sqrt{6} (|rgb> - |rbg> - |grb> + |gbr> + |brg> - |bgr>)

    I see this in books everywhere, so this tells me I'm doing something right here (although I'm perhaps seeing it in a different context than here). Now if I do the same for spins in a symmetric spin-1/2 config, I also get something.

    Let u and d denoted the spin up and spin down quark configs (not the up and down quarks).

    You have three permutations of spin 1/2 (uud, udu, duu), so the wavefunction is

    |spin> = A|uud> + B|udu> + C|duu>

    Now under the constraints |psi 123> = |psi 213> (A=A, B=C) and |psi 123> = |psi 321> (A=C, B=B), a solution is possible, namely A=B=C, which can be normalized to exp(i/phi)/sqrt{3}.

    |spin> = exp(i\phi)/sqrt{3} (|uud> + |udu> + |duu>)

    Now this puzzles me a great deal. I don't see anything wrong with what I did above, but yet this contradicts the explanation for the lack of uuu baryons in spin 1/2 config.

    I thought that I did something wrong above, because if colour is antisymmetric, and spin symmetric, then this is overall antisymmetric. Which books and reality tells me is impossible. So I thought that perhaps you have to deal with the "colourspin" wavefunction in order to get the proper result. However I still get something.

    You have three permutations of spin (uud, udu, duu) and six permutations of colour (rgb, rbg, grb, gbr, brg, bgr) so you have a wavefunction with 3 x 6 = 18 parts before symmetrisation:

    |psi> =
    + A|uud/rgb> + B|uud/rbg> +C|uud/grb> +D|uud/gbr> +E|uud/brg> +F|uud/bgr>
    + G|udu/rgb> + H|udu/rbg> +I|udu/grb> +J|udu/gbr> +K|udu/brg> +L|udu/bgr>
    + M|duu/rgb> + N|duu/rbg> +O|duu/grb> +P|duu/gbr> +Q|duu/brg> +R|duu/bgr>

    Under the |psi 123> = -|psi 213> constraint, you have
    A=-C, B=-E, D=-F, G=-O, H=-Q, I=-M, J=-R, K=-N, L=-P

    Under the |psi 123> = -|psi 321> constraint, you have the following conditions on the A-R constants:
    A=-R, B=-P, C=-Q, D=-N, E=-O, F=-M, G=-L, H=-J, I=-K

    Which is basically French for

    A=-C=Q=-H=J=-R;
    B=-E=O=-G=L=-P;
    D=-F=M=-I=K=-N

    So you get

    |psi> =
    +A ( |uud/rgb> - |uud/grb> + |duu/brg> - |udu/rbg> + |udu/gbr> - |duu/bgr> )
    +B ( |uud/rbg> - |uud/brg> + |duu/grb> - |udu/rgb> + |udu/bgr> - |duu/gbr> )
    +D ( |uud/gbr> - |uud/bgr> + |duu/rgb> - |udu/grb> + |udu/brg> - |duu/rbg> )

    with A, B, and D chosen so A*A + B*B +D*D = 18. I'm also kinda worried that I can't get a unique A:B:D ratio.

    Now it strikes me that between a scenario where I'm wrong because of a stupid mistake or me not seeing things through the correct glasses, and the scenario where all textbooks and web resources all being wrong on this, reality would choose the former over the latter.

    So what is it I'm not seeing or what is it I'm doing wrong? I'm greatly puzzled by this.
     
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 12, 2009 #2

    clem

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    Your post is quite long and involved and I don't have time for too much detail, but I can make a couple of suggestions.
    1. The simple reason why uuu can't be spin 1/2 is that space is symmetric, color is antisymmetric, flavor is symmetric. Therefor you need a symmetric spin function, which must be spin 3/2. Spin 1/2 is mixed symmetry, which can only combine with a mixed symmetry flavor state.
    2. If you look in detail at your functions, they will not have the combination symmetric flavor and mixed spin symmetry.
     
  4. Feb 12, 2009 #3

    malawi_glenn

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    In the quark model of baryons, two quarks of equal flavour must couple to Spin = 1, see e.g. martin - particle physics (wiley)
     
  5. Feb 12, 2009 #4
    Yeah, but I don't get a mixed symmetry spin, I get a symmetric spin. Namely

    |spin> = exp(i\phi)/sqrt{3} (|uud> + |udu> + |duu>)

    I'm deriving this from first principles, so there's something I'm doing wrong. I'm trying to pinpoint what exactly it is I'm doing wrong.
     
  6. Feb 12, 2009 #5
    You just posit that this state must be. You did not construct it from the representations of anything. In fact, what you need to do is just elementary "spin-like" addition from quantum-kindergarden. You should get that
    2x2x2=(3+1)x2=4+2+2
    where the last term is Symmetric+MixedSymmetric+MixedAntiSymmetric
    for SU(2), and there is no singlet here. The state that you posit is simply absent. An obvious problem with your state is that it will not be orthogonal with (for instance, unnormalized)
    udu+udd-2uud
    but this is the mixed-symmetric guy from above.

    I suggest you start over, with constructing first the SU(2) multiplet above, then the SU(3) flavor decuplet, two octets and singlet appearing in
    3x3x3=(6x3)+(3bx3)=10+8+8+1
     
  7. Feb 12, 2009 #6
    Thing is I don't get group theory, like at all. Which is actually the reason why I'm trying to derive these wavefunctions from first principles.

    I can get the two quarks spin wavefunctions correctly. From the same maths I used above (aka, build a general wavefunction with all permutations, then impose symmetry or antisymmetry), I get

    Symmetric:
    [tex]| S_z = 1 \rangle = | \uparrow \uparrow \rangle [/tex]
    [tex]| S_z = 0 \rangle = \frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right)[/tex]
    [tex]| S_z = 1 \rangle = | \downarrow \downarrow \rangle [/tex]

    Symmetric (12) Antisymmetric (13)
    [tex]| S_z = 1 \rangle = 0 [/tex]
    [tex]| S_z = 0 \rangle = 0 [/tex]
    [tex]| S_z = 1 \rangle = 0 [/tex]

    Antisymmetric (12) Symmetric (13)
    [tex]| S_z = 1 \rangle = 0 [/tex]
    [tex]| S_z = 0 \rangle = 0 [/tex]
    [tex]| S_z = 1 \rangle = 0 [/tex]

    Antisymmetric:
    [tex]| S_z = 1 \rangle = 0 [/tex]
    [tex]| S_z = 0 \rangle = \frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle - | \uparrow \downarrow \rangle \right)[/tex]
    [tex]| S_z = 1 \rangle = 0 [/tex]

    And the two S_z = 0 states are orthogonal to each other. I get a symmetric triplet, and an antisymmetric singlet. Which raises a green flag, because I see this result pretty much everywhere. And I suppose this matches the 4 = 3 + 1 decomposition whatever that's supposed to mean.

    Now on the three quark spin wavefunction, if I use the same maths (aka build a general wavefunction with all permutations, then imposing symmetry or antisymmetry), I get:

    Symmetric:
    [tex]| S_z = + \frac{3}{2} \rangle = | \uparrow \uparrow \uparrow \rangle [/tex]
    [tex]| S_z = + \frac{1}{2} \rangle = \frac{1}{\sqrt{3}} \left( | \uparrow \uparrow \downarrow \rangle + | \uparrow \downarrow \uparrow \rangle + | \downarrow \uparrow \uparrow \right)[/tex]
    [tex]| S_z = - \frac{1}{2} \rangle = \frac{1}{\sqrt{3}} \left( | \downarrow \downarrow \uparrow \rangle + | \downarrow \uparrow \downarrow \rangle + | \uparrow \downarrow \downarrow \right)[/tex]
    [tex]| S_z = - \frac{3}{2} \rangle = | \downarrow \downarrow \downarrow \rangle [/tex]

    Symmetric (12) Antisymmetric (13)
    [tex]| S_z = + \frac{3}{2} \rangle = 0 [/tex]
    [tex]| S_z = + \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{3}{2} \rangle = 0 [/tex]

    Antisymmetric (12) Symmetric (13)
    [tex]| S_z = + \frac{3}{2} \rangle = 0 [/tex]
    [tex]| S_z = + \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{3}{2} \rangle = 0 [/tex]

    Antisymmetric:
    [tex]| S_z = + \frac{3}{2} \rangle = 0 [/tex]
    [tex]| S_z = + \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{1}{2} \rangle = 0 [/tex]
    [tex]| S_z = - \frac{3}{2} \rangle = 0 [/tex]

    So by analogy with the above 4 = 3 + 1 decomposition, I get a 4 = 4 + 0 + 0 + 0 decomposition this time. Which apparently should be 8 = 4 + 2 + 2 + 0.

    Now this approach of mine is found in R. Shankar's Principles of Quantum Mechanics (1994, 2 ed.), p.271-273 (section called "System of N Identical Particles". The relevant passage is the following:

    Now if we plug in [tex]n_1 = \uparrow [/tex] , [tex] n_2 = \uparrow [/tex], [tex]n_3 = \downarrow[/tex], in the symmetric case, then we get what I get, aka a symmetric spin-1/2 state.

    So again, what I am doing wrong? There's something I'm not seeing.
     
  8. Feb 12, 2009 #7

    clem

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    Now you've got me confused. Does your u above stand for an up quark or for the spin projection +1/2?
    If it does sstand for the spin projection, then
    the spin state you have there is the m=1/2 state of total spin 3/2.
    The spin 1/2 state would be [2uud-udu-duu]/sqrt{6}, having mixed symnmetry.
    This spin state cannot combine with 3 u quarks.
     
  9. Feb 12, 2009 #8
    The u denotes spin projection +1/2. I didn't know the forum supported TeX at the time of my original post.

    I'm not quite sure you understand what my problem is. I'm applying first principles as I understand them (see the quotation from R. Shankar above), and the only thing I'm getting is a symmetric wavefunction, namely:

    [tex] \frac{1}{\sqrt{3}} \left[ | \uparrow \uparrow \downarrow \rangle + | \uparrow \downarrow \uparrow \rangle + | \downarrow \uparrow \uparrow \rangle \right] \ .[/tex]

    So what is it I'm doing wrong? What is it I'm not getting?

    I know it should be obtaining the mixed symmetry (amongst other possibilities)
    [tex] \frac{1}{\sqrt{6}} \left[ |\uparrow \downarrow \uparrow \rangle + |\downarrow \uparrow \uparrow \rangle - 2 |\uparrow \uparrow \downarrow \right] \ . [/tex]

    But this is not what I'm obtaining, what I'm obtaining is

    [tex] \frac{1}{\sqrt{3}} \left[ | \uparrow \uparrow \downarrow \rangle + | \uparrow \downarrow \uparrow \rangle + | \downarrow \uparrow \uparrow \rangle \right] \ .[/tex]

    So A) What am I doing wrong? Why is it wrong? B) What is the correct way to obtain the wavefunctions, if not for the "Shankar approach" quoted above?
     
  10. Feb 12, 2009 #9
    Mmmm... I've reread your post (clem), and I now see that this would the be the [tex]\Delta^{++}[/tex]'s S=3/2 Sz=1/2 spin config. Which is why the math checks out. I'm simply not working with a S = 1/2 config like I was under the impression. Also I now noticed that humanino made a small mistake that confused me (said the state was udu+udd-2uud, when it should've been udu+duu-2uud).

    Thanks for pointing that out, it's been bumming me for a couple of days now.

    Now onto obtaining the mixed symmetry spin states. If you got tips, go right ahead.

    It seems that imposing both a symmetry and an antisymmetry on the same space leads to a dead end. Would I be crazy to impose the 1<-->2 symmetry on spin, then to say absolutely nothing about the 1<-->3 exchange then try to combine it with a color wavefunction on which I've imposed a condition on 1<-->3, but none on 1<-->2 and see where that leads?
     
    Last edited: Feb 12, 2009
  11. Feb 13, 2009 #10

    clem

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    The color wave function only enters in giving effective Bose statistics for spin, flavor, space. After that, just ignore color.
     
  12. Feb 20, 2009 #11
    Yeah, but I also want to derive the possibilities with symmetric colour/space.

    Which shouldn't be too hard if I substitute quark flavour and quark colour.
     
  13. Feb 20, 2009 #12

    clem

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    Do you mean that you want to assume a symmetric color state?
    If so, just apply effective Fermi statistics to the space-spin-flavor states.
     
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