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Baryon Mass

  1. Mar 16, 2007 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive the mass of [itex]\Xi[/itex] using the formula:
    [tex]M\left(baryon\right)=m_1 + m_2 + m_3 + A' \left[\frac{S_1 \cdot S_2}{m_1 m_2} +\frac{S_1 \cdot S_3}{m_1 m_3} + \frac{S_2 \cdot S_3}{m_2 m_2\3}\right][/tex]


    2. Relevant equations
    We have:
    [tex] S_1 \cdot S_2 + S_1 \cdot S_3 + S_2 \cdot S_3 = \frac{\hbar^2}{2}\left[j\left(j+1\right)-2/4\right] = -3/4 \hbar[/tex] for octet

    and also:
    [tex]\left(S_u+S_d\right)^2 = S_u^2 + S_d^2 + 2S_u \cdot S_d [/tex]


    3. The attempt at a solution
    What I don't get it the last equation. In the case of [itex] \Sigma [/itex], is equal to [itex]2\hbar^2[/itex] because the isospin is 1 (and therefore [itex]S_u \cdot S_d = \hbar^2 /4[/itex]. Following the pattern, since the isospin of [itex]\Xi[/itex] is 1/2, I tried to figure out [itex]S_s \cdot S_s[/itex] which is needed, since the quark content for [itex]\Xi[/itex] is uss. I got [itex] S_s \cdot S_s = -3/8\hbar^2[/itex] which doesn't give the right answer.

    The right answer should be:
    [tex]M_\Xi = 2*m_s + m_u + \frac{\hbar^2}{4}A'\left(\frac{1}{m_s^2}-\frac{4}{m_u m_s}\right) [/tex]

    Can someone help me? thanks!
     
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 17, 2007 #2

    Meir Achuz

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    s1.s2=+1/4 for S=1, and -3/4 for S=0.
    For the s-s pair, this gives s1.s2=+1/4.
    for the two s-u pairs you have to recouple the spin states.
    The result is that each s-u pair is 1/4 spin 1, and 3/4 spin 0.
    Using those factors gives the right answer.
    This was first done in PR 154 (1967) 1608.
     
  4. Mar 17, 2007 #3
    Thanks for the reply.

    When you say for S=1 or S=0, what does that correspond to? I thought it was just an isospin of the particle (so 1/2 for [itex]\Xi[/itex] and 0 for [itex]\Lambda[/itex]), but I guess that's not the case. In the text (Griffiths), he says something about spins being "parallel" in the decuplet case. What does that mean exactly? What about in octet case?
     
  5. Mar 17, 2007 #4

    nrqed

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    in what section is that covered in griffiths? I think I have the book around and could look it up.

    The S's in the formula are the *spins* of the particles, not the Isospin. Spin and isospin are totally different concepts.
     
  6. Mar 17, 2007 #5
    This is from Griffiths "Introduction to elementary particles", chapter 5: bound states. pp 180 - 184.

    the spins of the quarks (u,d,s) are all 1/2, so [itex]\left(S_u + S_d \right)^2 [/itex] should aways be [itex] 2\hbar[/itex], but in the text it says it is equal to 0 for [itex]\Lambda[/itex] (it says it is equal to [itex]2\hbar[/itex] for [itex]\Sigma[/itex] though...)
     
  7. Mar 17, 2007 #6

    nrqed

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    They all have spin 1/2, that's true, but that does not mean that [itex]\left(S_u + S_d \right)^2 [/itex] is necessarily [itex] 2\hbar[/itex]! Two particles with a spin of 1/2 may either be in a total spin S=0 or a S=1 state. As Achuz said, if they are in the S=0 state, then [itex]\left(S_u + S_d \right)^2 = 0 [/itex] (in which case [itex]S_u \cdot S_d = -3/4 \hbar^2 [/itex]) whereas if they are in the S=1 state, [itex]\left(S_u + S_d \right)^2 = 2 \hbar^2 [/itex] (in which case [itex]S_u \cdot S_d = 1/4 \hbar^2 [/itex])
     
    Last edited: Mar 17, 2007
  8. Mar 17, 2007 #7
    So, how do you know if they are in S=0 state or S=1 state? Is this where isospin comes in (Griffiths mentioned this, but I didn't know what he was talking about)?
     
  9. Mar 18, 2007 #8

    Meir Achuz

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    The Ispin is irrelevant here. I trhink that PR article would be easier clearer on this than Griffiths. Each pair of quarks adds their spin to 12+1/2=0 or 1.
    Two identical quarks (like s-s) must add to spin one, for which s1.s2=+1/4.
    The difficulty comes for two different quarks (like s-d).
    Then their spin addition can give 1 or 0. For spin zero, s1.s2=-3/4.
    YOu can show that the s-d pair has probability 1/4 of being in the spin one state and probabilty 3/4 of being in the spin zero state. This should be enough info to get all the octet masses.
    The decuplet has spin 3/2, so each quark pair is in the spin one state.
    Griffiths can't be followed because he tries to simplify, and leaves out the physics.
     
  10. Jul 11, 2007 #9
    Achuz said, "The difficulty comes for two different quarks (like s-d).
    Then their spin addition can give 1 or 0. For spin zero, s1.s2=-3/4.
    YOu can show that the s-d pair has probability 1/4 of being in the spin one state and probabilty 3/4 of being in the spin zero state."

    Why do you say that?

    I thought the probability of being in a state with spin 1 is higher and should be 3/4 instead?

    Thanks ^_^
     
  11. Jul 12, 2007 #10

    Meir Achuz

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    The Xi total spin 1/2 wave function is (ssu)[2aab-aba-baa]/sqrt{6},
    where a is for spin up and b for spin down. This has the first two (ss) quarks in a spin one state. If you recouple the spin states, this combination can be written as (ssu){[sqrt{3}/2][a(ab-ba)/sqrt{2}]-[1/2][2baa-aab-aba]/sqrt{6}}.
    The first term has the 2nd and 3rd (su) quarks in a spin zero state and the second term has them in a spin one state.
    You can read the probabilities as 3/4 and 1/4.
     
  12. Jul 12, 2007 #11
    Hi Meir, thx for the response again....I kind of know what you mean...

    One more question though, when you said "The Xi total spin 1/2 wave function is (ssu)[2aab-aba-baa]/sqrt{6}", how did you obtaint his result?

    Here's what I think:
    Xi is a spin 1/2 particle and one should use EQ 5.113. However applying that equation doesnt seem to allow me to get what u got. Any comments?

    thx
    joe
     
  13. Jul 12, 2007 #12
    Oh Hi Meir again, sorry to bother you again ...

    you've mentioned that the original paper that talks about baryon mass estimation is found in Physics review. I tried finding it yesterday and failed. I'd like to take a look at it. Do you have it? Would you mind sending it to me?

    My email : joechien0218@yahoo.com

    Thank you so much for your help! You are a good person!
     
  14. Jul 13, 2007 #13

    Meir Achuz

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    I don't have Griffiths, but my guess is that he oversymmetrizes the wave function, which can make it quite complicated.
    The first two quarks in ssu are identical, so they must be in a spin 1 state.
    This gives the spin addition 1/2 + 1/2 +1/2 =[1] + 1/2 = 1/2, which is the spin wave function I gave. This will be clearer in a PR 172(1968)1807 paper I will send you. I find it amusing that the quark model was better understood 40 years ago than it is today.
     
  15. Jul 13, 2007 #14

    Meir Achuz

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    I will email the pdf of another PR paper I just got off PROLA on the web.
    The first paper I mentioned does not go as much into details as this one:
    PR 171(1968)1807. The paper is too long to upload here.
    Actually the first paper to use the s_1.s_2 interaction for hadrons was by Sakharov in 1966, then a 1975 paper by deRujula, Glashow, and Georgi deduced the 1/m_1m_2 from QCD..
     
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