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Homework Help: Quantum Numbers J,S and L in p+n and n+n system

  1. Feb 24, 2010 #1
    I have some problems concerning one of the exercises I got last week

    1. The problem statement, all variables and given/known data
    List the quantum numbers J,S,L (in spectroscopic notation [tex]^{2S+1}L_J[/tex] for partial waves of the n + p systemup to total angular momentum J=2
    The quantum numbers J, S and L represent the total angular momentum, the total spin, and the relativ orbital momentum, repectively.
    b) Repeat for the n+n system
    c) Use the above information, together with the known spectroscopy of the deuteron, to explain why there are no bound states of two neutrons

    3. The attempt at a solution

    Ok first of all we can have S=0 or S=1. Then we go through the different L states.
    With L=0: S L=1 : P L=2: D
    L=0: [tex]^1S_0[/tex]
    L=1: [tex]^1P_1[/tex]
    L=2: [tex]^1D_2[/tex]

    Now I have some problems with the S=1 states. Lets assume we have L=1 and S=1. What values can J have. Is it the same as I learned for the electron orbits [tex]J=|L-S|..L+S[/tex]? So can J be in the example be 0,1 and 2?
    When yes then I get 7 additional cases
    L=0: [tex]^3S_1[/tex]
    L=1: [tex]^3P_0 ^3P_1 ^3P_2[/tex]
    L=2: [tex]^3D_1 ^3D_2[/tex]
    L=3: [tex]^3F_2[/tex]

    I have really no idea if this is correct or not. But when this is true I am not sure about the second part. Now we have a n+n system, this means we have to consider the Pauli principle. Which of the combination is than not allowed any longer? Of course the [tex]^3S_1[/tex] one, because then both have same spin. But are all S=1 states forbidden, even when they have a relative angular momentum?

    In the last part the question about a bound state of 2 neutrons. I know that the ground state in the deuteron is the [tex]^3S_1[/tex] state. This one is not allowed in the 2 neutron system, but why not the next higher state is the ground state for the 2 neutrons?(e.g. the first excited state of the deuteron)

    Would be really nice if you could help me
    Best regards
  2. jcsd
  3. Feb 24, 2010 #2


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    Looks good.
    The symmetry of the entire state needs to be antisymmetric for the two-neutron system. The S=1 spin state is symmetric, so the orbital part of the state needs to be antisymmetric so that the state as a whole is antisymmetric.
  4. Feb 24, 2010 #3
    Thx alot!
    Do you mean with an antisymmetric orbital part that the L has to be 1 or 3 or how do I know what symmetry the orbital part has?

  5. Feb 24, 2010 #4


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    Yes, that's what I meant. I'm not sure how you'd subsequently rule those states out for the n-n system though.
  6. Feb 25, 2010 #5
    hmm maybe it is possible to argue with the parity, the relative orbital momentum gives [tex](-1)^l[/tex] so in the n+n system and S=1 l has to be an odd number and in the S=0 and even number?
    Or am I thinking totally wrong?
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