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Addition of Angular Momentum for identical particles

  1. Jun 20, 2017 #1
    This is the problem I'm trying to understand:

    Consider two particles with spin 1 without orbital angular momentum. If they are distinguishable, from the rule of addition of angular momentum applied to spin, we'll have states of total spin [itex]j=0,1,2[/itex]. If we have, however, identical particles which are the possible states?

    In textbooks, the addition of angular momentum is never treated in terms of distinguishable and identical particles, at least I don't recall it. The way I would approach this problem is to acknowledge that the possible total spin would be [itex]j=0,1,2[/itex] and then, from the state [itex]|-1\rangle |-1\rangle[/itex], I would use the ladder operator to build all the other four states compatible with [itex]j=2[/itex].


    How do I build the states for [itex]j=0,1[/itex]? The state [itex]|j=1,m_{j}=-1\rangle[/itex] must be built from the same states as [itex]|j=2,m_{j}=-1\rangle[/itex], that is, [itex]|-1\rangle |0\rangle[/itex] and [itex]|0\rangle |-1\rangle[/itex]. So how are they any different?

    Thank you very much.


    3. The attempt at a solution
     
  2. jcsd
  3. Jun 20, 2017 #2

    Orodruin

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    What is important for distinguishable particles is that the total state is (anti-)symmetric (depending on Bose/Fermi statistics). You need to take the spin state into account as well as the spatial wave function.
     
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