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Person nifty with young tableaux sought!

  1. May 9, 2012 #1
    Many years ago I was taught Young tableau but I can't for the life of me remember how to do it, but now I seem out of the blue to need a Clebsch decomposition! Does anyone who is nifty at these things know what the decomposition of 3x3x3x3x3* is in SU(3)?

    (FWIW, the reason I'm asking this is because I'm reading one of Gell-Mann's first papers on the quark model, and he contemplates whether we could consider baryons to be made up of four quarks and an anti-quark. Now I'm wondering how the baryonic world would be different if they were!)

    Thanks guys!
     
  2. jcsd
  3. May 9, 2012 #2
    I haven't read through this carefully but this seems to cover what you're looking for:

    http://www.isv.uu.se/~rathsman/grouptheory/Beckman-Loffler-report.pdf [Broken]

    Basically it is about "putting the boxes" where they are allowed and see what you end up with. Then you will of course have to know which diagram corresponds to which representation and so on.

    The "baryon" decomposition 3x3x3 is well known and is equal to 10+8+8+1. So all you have to do is take the product of this with a 3x3*. This in turn is the "meson" decomposition, equal to 3x3* = 8+1. Last step is then to take the product (10+8+8+1)x(8+1), which basically involves taking the products 8x8 and 10x8. And this I leave to you ;)
     
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  4. May 9, 2012 #3

    Vanadium 50

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    8 x 8 is easy. It's 27 + 10 + 10bar + 8 + 8 + 1.

    You're on your own for 10 x 8.
     
  5. May 9, 2012 #4

    fzero

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  6. May 9, 2012 #5

    Vanadium 50

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    That's easy to see, because of the 10bar.

    The 8x8 is also easy to "cheat" on. I know I need 64 elements. I need an asymmetric guy, the 27, and exactly one fully symmetric guy, the 1. So I have 36 left. I know from the 3-gluon vertex that there is an 8, leaving me 28.

    If I want a 10, I need a 10bar, which leaves me another octet.

    If I don't want a 10, I can't make 28 with combinations of 3's and 6's, which, like 10's must occur in pairs. If I add another 8, I get 20, and it still doesn't work. If I add two 8's, I have 12 left, which cannot be 3+3+3bar+3bar because there is only one fundamental representation, so the only two options are 27 + 10 + 10bar + 8 + 8 + 1 and 27 + 6 + 6bar + 8 + 8 + 8 + 1.

    If you remember that there is a baryon decuplet, you know it's the first option and not the second.
     
  7. May 10, 2012 #6
    One would use the Littlewood-Richardson rule, and there's a nice explanation of it in

    Particle Data Group > Reviews, Tables, Plots > Mathematical Tools > SU(n) multiplets and Young diagrams
    38.4. Coupling multiplets together (PDF page 3)

    I used my Lie-algebra code on it, and I found all the results that everybody else here mentioned. For the 4q1q* case, I found

    (3)4 * (3*) -> 35 + 3(27) + 4(10) + 2(10*) + 8(8) + 3(1)

    Yes, 3 color singlets.
     
  8. May 10, 2012 #7
    All of you are completely awesome, thanks so much for taking the time to help me out! ;)
     
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