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Total nuclear spin of deuteron

  1. Feb 24, 2012 #1
    Hello all, I am having trouble understanding how this works. In Krane there arises a discussion on total angular momentum I of the deuteron. While it is true it has three components, namely the individual spins of the neutron and proton, but also the orbital angular momentum l of the nucleons as they move about their common center of mass. This total angular momentum can be denoted by

    [itex]I=s_{p} + s_{n} + l[/itex]

    He continues on to talk about the different ways to couple these contributions and states there are only four possibilities. I can see the first two possibilities for total angular momentum I=1, but the other two make no sense. These are the possibilities:

    (1) [itex]s_{n}[/itex] and [itex]s_{p}[/itex] are parallel with [itex]l=0[/itex]
    (2) [itex]s_{n}[/itex] and [itex]s_{p}[/itex] are antiparallel with [itex]l=1[/itex]
    (3) [itex]s_{n}[/itex] and [itex]s_{p}[/itex] are parallel with [itex]l=1[/itex]
    (4) [itex]s_{n}[/itex] and [itex]s_{p}[/itex] are parallel with [itex]l=2[/itex]

    One can see why (1) and (2) hold by inspection but (3) and (4) make my brain hurt. Perhaps I am just not seeing the correct orientation. Any suggestions?
     
    Last edited: Feb 24, 2012
  2. jcsd
  3. Feb 25, 2012 #2

    Bill_K

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    matumich26, Do you know how to couple two angular momenta? You don't just add them. When you couple J1 to J2 the combined system can have any value of J in the range from their sum J1 + J2 to their difference |J1 - J2|. You must include all of these possibilities. For example when you coupled the two spins together, 1/2 ⊗ 1/2 = 1 ⊕ 0. That's the way we write it, and it means the coupled system can have either S = 1 (parallel) or S = 0 (antiparallel).

    Ok, you wanted to look at the cases with S = 1. When you further couple S with L,

    1 ⊗ 0 = 1
    1 ⊗ 1 = 2 ⊕ 1 ⊕ 0
    1 ⊗ 2 = 3 ⊕ 2 ⊕ 1

    You can see in all three of these cases, I = 1 is one of the possibilities.
     
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