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Sp - Sn in deuteron is zero?

  1. Feb 5, 2014 #1
    Hi, apologies if this is simple. I'm a bit confused with a piece of text from Introductory Nuclear Physics by Wong. It's talking about finding the expectation value of the magnetic moment of the deuteron. In the deuteron it is known the total spin quantum number is S = 1. In deriving the total [itex] \mu [/itex] we have a term

    [itex] < S_p - S_n> [/itex] (note: this is meant to be operators).

    Quoting from the text: "Since the operator [itex] S_p - S_n [/itex] acts on proton and neutron spins with opposite signs, it can only connect between two states, one with S = 1, and the other with S = 0, and as a result, cannot contribute to the expectation value of interest to us here".

    I'm at a bit of a loss as to what it's saying here, I know

    [itex] < S_p - S_n> = \int \phi^*S_p\phi dV - \int \phi^*S_n \phi dV[/itex]

    where [itex] \phi [/itex] is the total deuteron wavefunction. Since the deuteron is in an S = 1 state the proton and neutron either have the same spin z-component or opposite. Since really when we talk about the magnetic moment we're talking about the z-component, any state where the z-component is the same for the proton and neutron cancel out in this [itex] S_p - S_n [/itex] expectation value....but what about if the spins are opposite? I don't really comprehend what the text is saying.
  2. jcsd
  3. Feb 5, 2014 #2


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    No, that's incorrect. The S = 1 states are the three states that are symmetric:
    The state with Sz = +1 is |↑↑>,
    The state with Sz = -1 is |↓↓>,
    And the state with Sz = 0 is (|↑↓> + |↓↑>)/√2
    The remaining state, the antisymmetric state, (|↑↓> - |↓↑>)/√2, has S = 0 and Sz = 0

    Note that
    (Sp - Sn)|↑↑> = 0,
    Sp - Sn)|↓↓> = 0,
    and (Sp - Sn)(|↑↓> + |↓↑>)/√2 = (|↑↓> - |↓↑>)/√2.
    So acting on the S = 1 states it either gives you zero, or else it gives you the S = 0 state.
  4. Feb 5, 2014 #3
    Thanks for the reply.

    Was I incorrect because I should've said the proton and neutron have the same z-spin or are in a superposition of states with opposite z-spins?

    But then how does obtaining the S = 0 state create a zero expectation value? If calculating the expectation value then:

    [itex] \int \phi^*\chi_{S=1}^*<S_p-S_n>\phi\chi_{S=1} dV = \int \phi^*\phi\chi_{S=1}^*\chi_{S=0} dV = \chi_{S=1}^*\chi_{S=0}[/itex]

    assuming the space parts are normalised. Is this final product supposed to be zero?
  5. Feb 5, 2014 #4


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    Spin states with different values of S are orthogonal. The S = 0 state is orthogonal to the S = 1 states, so the inner product is zero.
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