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Homework Help: Particle decays

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data
    My textbook takes a look at the [itex]\Delta(1232)[/itex] particle
    It says that
    [tex]\left|\pi p;\frac{3}{2},\frac{3}{2}\right>=\left|\pi;1,1\right>|N;\frac{1}{2},\frac{1}{2}\right>[/tex]

    where N stands for a nucleon and pi could be any of the three flavours of pion.
    They then go on by applying ladder operators (not explicitly, this is not at that level yet) to give
    [tex] \left|\pi p;\frac{3}{2},\frac{3}{2}\right>=-\sqrt{\frac{1}{3}}\left|\pi^+ n\right>+\sqrt{\frac{2}{3}}\left|\pi^0 p\right>[/tex]

    My question is how they came up with that

    2. Relevant equations
    Clebsch Gordon coefficients
    For now use the wikipedia source but if you can suggest a better source please suggest it

    3. The attempt at a solution
    Is the isospin of the pi+ is 1 and the isospin of the proton is 1/2?

    in either case how did they come up with the coefficients of -root 1/3 and root 2/3??
    the two spin values are j=1 and j=1/2. so we see two possiblities,
    first is m=3/2
    why is this possibility rejected?

    the other possibility is where m=1/2
    there are two possible j values. Look at the m1 values i could tell which woul the pi+/-/0 possibility. But the only way i would know if there was a neutron or proton would be to deduce it from the pion's spin and charge? Is that correct?

    Also while reading the CG coefficients, is the j,m of the decaying particle, and then j1, and j2 of the products?

    Thank you for all your help and advice!
    Last edited: Apr 8, 2008
  2. jcsd
  3. Apr 8, 2008 #2


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    Yes. To be more precise, the value of I for any of the pions is 1 (I is the analogue of l for orbital angular momentum). The value of [tex] I_z [/tex] is [tex]0, \pm 1 [/tex] depending on which pion you are considering (the analogue of [tex] I_z[/tex] is [tex] m_l[/tex]). The isospin of a nucleon (proton or neutron) is I=1/2. [tex] I_z [/tex] is [tex] \pm 1/2 [/tex] depending on whether you are considering the neutron or the proton.
    I am confused. are you absolutely sure they wrote 3/2, 3/2 for the total state? For that state the CG table does not give the coefficient you give. Are you sure it's not 3/2,1/2 or 1/2,1/2?
  4. Apr 9, 2008 #3
    Yes it is my mistake they had 3/2 1/2

    \left|\pi p;\frac{3}{2},\frac{1}{2}\right>=-\sqrt{\frac{1}{3}}\left|\pi^+ n\right>+\sqrt{\frac{2}{3}}\left|\pi^0 p\right>
    How did they come up with the coefficients then

    I dont know how to read the table...

    since spin is 3/2 and 1/2 the CG coeffs should be 1/2 or root3/4 but none of those appear??
    Last edited: Apr 9, 2008
  5. Apr 9, 2008 #4


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    I use a table in Griffiths' book and I get different coefficients: I get both coefficients to be positive.

    In any case, just look at the table for combining 1/2 and 1. Now look at the state 3/2 1/2 You find

    |3/2, 1/2 > = srqt(1/3) |1,1> |1/2,-1/2> + sqrt(2/3) |1,0> |1/2,1/2>

    |1,1> is a pi+
    |1,0> is a pi^0
    |1/2,1/2> is a proton
    |1/2,-1/2> is a neutron
  6. Jun 22, 2008 #5
    The Wiki table has a disclaimer about negative values, I think... read the fine print and it might apply to this situation.
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