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Total spin quantum number

  1. Jul 21, 2013 #1
    I have a pair of non interacting, identical 1/2 spin fermions in a one dimensional infinite square well with walls at x=0 and x=L.
    One particle is in ground state, the other in first excited state.
    This two-particle system has total spin quantum number S=0

    I have normalized energy eigenfunctions for each and am trying to explain implication of this to symmetries of spin and spatial parts of the total wave function which I think is;

    ψ1.2(t) = ψ(x1, x2, t) Ims1, ms2>

    or for time t=0
    ψ1.2 = ψ(x1, x2) Ims1, ms2>

    So what exactly is the total spin quantum number please.


    I can see that the spin quantum number of a spin 1/2-particle is 1/2. So could the total spin quantum number be when adding the second part of the pair?

    Additionally, if this total spin quantum number becomes S=1 when they are both in the same eigenstate what does this relate to? Are both particles now either spin-up or both spin-down?
     
    Last edited: Jul 21, 2013
  2. jcsd
  3. Jul 22, 2013 #2

    tom.stoer

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    Science Advisor

    You have to write down the states with quantum numbers n,s for two possibilities a,b and two particles 1,2 as

    ##|\psi\rangle = |n_a,s_a\rangle_1\,|n_b,s_b\rangle_2 - |n_b,s_b\rangle_1\,|n_a,s_a\rangle_2##

    The total spin operator is

    ##S = S_1 \, \text{id}_2 + \text{id}_1 \, S_2##

    Now you can apply the spin operator using the fact that an operator with index 1 (2) acts on the ket with index 1 (2) and is the identity w.r.t. the ket with index 2 (1).

    The first of all four terms reads

    ##(S_1 \, \text{id}_2)(|n_a,s_a\rangle_1\,|n_b,s_b\rangle_2) = (S_1 |n_a,s_a\rangle_1)\,(\text{id}_2\,|n_b,s_b\rangle_2)##

    Now your state with one particle in ground state and one particle in the first excited state and total spin S=0 means

    ##n_a=1,\,n_b=2,\,s_a=+1/2,\,s_b=-1/2##
     
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