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Angular momentum of identical particles

  1. Nov 15, 2013 #1
    When combining N spin 's' states, is it always true that each multiplet has even or odd symmetry?

    I know that's the case for N=2 and s=1/2 or 1. For s=1/2, the triplet is symmetric and the singlet is antisymmetric. For s=1, the pentlet is symmetric, the triplet antisymmetric, and the singlet symmetric. But what about different s's and different N's?

    Also, I'm a bit confused about terminology. If you have a 2-neutron bound state whose total spin is zero, that means it's antisymmetric in the spins, so it must be symmetric in orbital angular momentum, i.e., [tex]Y^{m1}_{l1}(x_1)Y^{m2}_{l2}(x_2) +Y^{m1}_{l1}(x_2)Y^{m2}_{l2}(x_1)[/tex]. Under parity, due to a property of spherical harmonics, the orbital part changes by just a factor $$(-1)^{l_1+l_2}$$.

    However, I'm trying to make sense of the statement that if you have a 2-neutron bound state whose total spin is zero, that means it's antisymmetric in the spins, but must have an even total orbital angular momentum. Is that true? $$l_1+l_2$$ could be odd, but the total L could be even by taking the multiplet $$l_1+l_2-1$$ rather than highest multiplet as your total.
     
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  3. Nov 16, 2013 #2

    Bill_K

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    For two objects the only possible symmetries are "even" (symmetric) and "odd" (antisymmetric). But when N > 2 there are other possibilities. Each permutation of the N objects can be considered a group operation, and the group of all permutations on N objects is called the Symmetric Group SN. For example for N = 3 there are 6 possible permutations. In general there are N! When you talk about the symmetry possessed by a set of objects, you are talking about a representation of SN.

    SN always has a representation which is totally symmetric, in which for example ψ(1,2,3) = ψ(2,1,3) and so on. And for N >1 there's a representation which is totally antisymmetric, in which ψ(1,2,3) = - ψ(2,1,3). But for N > 2 there are other representations as well, generally called mixed symmetry, in which neither relationship holds, and ψ(1,2,3) and ψ(2,1,3) are linearly independent.

    This is discussed in many places, but usually in a rather abstract framework. Here's one paper which tries to be concrete. See especially his Example 3.2.
     
  4. Nov 16, 2013 #3
    Thanks. Does this mean that multiplets corresponding to mixed symmetry can never be states for identical particles, since they do not change by factors of -1 under exchange? So mixed symmetry only applies do distinct particles?
     
  5. Nov 16, 2013 #4

    Bill_K

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    Typically the particles' wavefunction will have several parts: a spin part, a space part, maybe other things like color, etc. For identical particles it's the complete wavefunction that must be symmetric or antisymmetric. So the spin part of the wavefunction can have mixed symmetry provided that the space part does also, and provided that their combination (tensor product) comes out right.
     
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