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Flavor Spin Wavefunctions

  1. Aug 19, 2009 #1


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    I'm having a little trouble recreating some things from a paper and it is due to my lack of knowledge of working with Flavor-Spin wavefunctions.
    I'm trying to show that :
    \left\langle \Lambda \left|b_s^{\dagger }b_b\right|\Lambda _b\right\rangle =\frac{1}{\sqrt{3}}
    \left\langle p \left|b_u^{\dagger }b_c\right|\Lambda _c^+\right\rangle =\frac{1}{\sqrt{2}}

    These two papers both take their operation from the same text, but none of it explicitly shows how they get it or what wavefunctions they use. I assume this is because it is ELEMENTARY but while searching around, many books/articles give different wavefunctions using different approaches.

    I'm wondering where I should start. I feel like this should be a simple multiplication but everytime I try it I don't get their answers.

    From you guys, do any of you know offhand know the flavor-spin states for a lambda and p (both in s=1/2)? One paper gives both octet proton flavor wavefunctions as :

    Which is right? Or is it that only in a linear combination with some spin states to make the total flavor-spin function symmetric :
    [tex]\left.|56,S=\frac{1}{2},8\right\rangle =\frac{1}{\sqrt{2}}(p'\chi '+p\text{''}\chi \text{''})[/tex]

    where the spinors are pretty much the same as the protons function but with spin arrows up and down instead of "u u d".

    Am I on the right track?
  2. jcsd
  3. Aug 19, 2009 #2


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    Also, if I get something like:

    [tex]u^{\dagger } u^{\dagger } d^{\dagger } u c^{\dagger } d u c[/tex]

    and I believe its the trace of this (since i summed over spin states?)
    Is there a way to simplify? I know I can cyclically permutate traces to simplify, but that doesn't help this one.
    Since I'm summing over spin, and this is merely the flavor representation, can I interchange any two operators freely? Or it would add the (-1) because its not symmetric as just the flavor representation in s=1/2?

    If thats the case, then this would simplify to:
    [tex]- u^{\dagger} u u^{\dagger} u d^{\dagger} d c^{\dagger} c[/tex]

    Which is just -1?

    If I follow that method, I get something that isn't what they have. Or am I way off?
  4. Aug 24, 2009 #3


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    Ok, so no responses. Let me clean up my questions then.

    From Appendix A (Page 25&26) of http://arxiv.org/PS_cache/hep-ph/pdf/9304/9304286v1.pdf

    I have, in their choice of wavefunctions for a proton and a lambda_c:

    [tex]p=\frac{1}{\sqrt{3}}\left[u u d \chi _s +(13)+(23)\right][/tex]
    [tex]\Lambda _c^+=\frac{-1}{\sqrt{6}}\left[(u d c- d u c) \chi _A +(13)+(23)\right][/tex]
    [tex]\chi _s=\frac{1}{\sqrt{6}}\left[a^{\uparrow } b^{\uparrow } c^{\downarrow }-a^{\uparrow } b^{\downarrow } c^{\uparrow }-a^{\downarrow } b^{\uparrow } c^{\uparrow }\right][/tex]
    [tex]\chi _A=\frac{1}{\sqrt{2}}\left[a^{\uparrow } b^{\downarrow } c^{\uparrow }-a^{\downarrow } b^{\uparrow } c^{\uparrow }\right][/tex]

    Where (13) is just the permutation.

    I'm trying to show that:

    [tex]N_{\text{fi}}=\, _{\text{flavor} \text{spin}}\left\langle \Lambda \left|b_s^{\dagger }b_c\right|\Lambda _c^+\right\rangle {}_{\text{flavor} \text{spin}}=\frac{1}{\sqrt{3}}[/tex] (in http://arxiv.org/PS_cache/hep-ph/pdf/9502/9502391v3.pdf, bottom of page 13, same author)

    [tex]N_{\text{fi}}=\, _{\text{flavor} \text{spin}}\left\langle p\left|b_u^{\dagger }b_c\right|\Lambda _c^+\right\rangle {}_{\text{flavor} \text{spin}}=\frac{1}{\sqrt{2}}[/tex] (page 18, same article)

    I've tried working it out by hand, I don't htink I'm doing it right because I keep getting 3/Sqrt[6] for the proton one, and I get 1 for the lambda_c to lambda.

    I know I shouldn't get one, I think I'm doing the operator wrong.

    What would I do for the 2 quark flavor spin operator here? Does it replace all c's with u's? Or does it affect spin in any way? Etc. I know its killing off the c, and creating a u, but how does that individually affect "u c d" parts of the wavefunction?

    Please help guide me, I just can't seem to do it correctly. ANY help will be amazing.

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