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Three spin-1/2 particles

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the energies for a 3 spin-1/2 particles with the Hamiltonean:
    [tex]H=\frac{E_0}{\hbar^2}(\vec{S_1}.\vec{S_3}+\vec{S_2}.\vec{S_3})[/tex]

    3. The attempt at a solution

    From the Clebsh-Gordon table one gets all the spin functions:
    [tex]|\frac{3}{2},\frac{3}{2}>...|\frac{1}{2},\frac{1}{2}>...|\frac{3}{2},-\frac{3}{2}>[/tex] (6 states in total)

    So, to get the matrix elements for the Hamiltonian I tried developing the dot product so I could work directly with the operators i.e.:

    [tex]\vec{S_1}.\vec{S_3}=S_{1x}S_{3x}+S_{1y}S_{3y}+S_{1z}S_{3z}[/tex]

    Now the problem is that the states as they are defined represent only one particle and the spin operators act on each particle individually and there's no CG table for a 1/2x1/2x1/2 spin addition.

    What's the best way to approach this?
     
  2. jcsd
  3. Jun 8, 2012 #2

    vela

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    Hint: You can write the Hamiltonian as
    $$\hat{H} = \frac{E_0}{\hbar^2}(\vec{S}_1+\vec{S}_2)\cdot\vec{S}_3.$$
     
  4. Jun 8, 2012 #3
    Yes but my question is how do these operators act on these states? The basis I'm using only shows the m numbers of two particles because I've already summed the spins of the first two...
     
  5. Jun 8, 2012 #4
    The usual trick here is to write

    [itex]S_{1}\cdot S_{2}=\frac{1}{2}((S_1+S_2)^2-S_1^2-S_2^2)[/itex]

    Those operators you DO know how they act.

    [itex]S_i^2\mid s,m\rangle = s(s+1)\mid s,m\rangle=\frac{3}{4}\mid s,m\rangle[/itex]

    The same goes for [itex](S_1+S_2)^2[/itex], since you now its possible values from Clebsch-Gordan coefficients
     
    Last edited: Jun 8, 2012
  6. Dec 6, 2012 #5
    Sorry to revive this thread but I am also trying to work out 3 spin 1/2 particle addition and this comment confused me. How can you find [itex](S_1+S_2)^2[/itex] in the |S,m> basis using C.G. coefficients when S represents the total spin [itex]S_1+S_2+S_3[/itex]?
     
  7. Dec 6, 2012 #6
    I think I partly answered my own question: I guess you can treat [itex](S_1+S_2) + S_3[/itex] as an addition of two spin systems? If that is the case then [itex](S_1+S_2)^2|S,m>=h^2(S_1+S_2)(S_1+S_2+1)|S,m>[/itex]. But then how do we treat [itex]S_1[/itex] and [itex]S_2[/itex]? And I don't seem to need any C.G. coefficients using this method?
     
    Last edited: Dec 6, 2012
  8. Dec 6, 2012 #7

    TSny

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    You can plow through it by using the 8 "product states" as basis states:

    ##|+++\rangle\equiv |\uparrow\rangle_1|\uparrow\rangle_2|\uparrow \rangle_3 ##

    ##|-++\rangle\equiv |\downarrow\rangle_1|\uparrow\rangle_2|\uparrow \rangle_3 ##

    etc. where ##|\uparrow\rangle_1## is spin up along z for particle 1.

    Find the matrix representation of H in this basis and then find the eigenvalues of H. It's not too tedius. Most matrix elements are zero and H will be in block diagonal form. Not very elegant, but it works.
     
  9. Dec 6, 2012 #8
    I agree that is probably the most straight forward way of going about this, I wanted to try to avoid explicit construction of the Hamiltonian in matrix form though with the hopes of possibly being able to generalize the process. Any hints on the route I was working on the previous post?

    Thanks!
     
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