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I have three Spin 1/2 Particles and a Hamiltonian given by $$H=A(S_1\cdot S_2)+B(S_2\cdot S_3+S_1\cdot S_3)$$ In order to find the energy spectrum, I want to diagonalize H in terms of $(S_1+S_2+S_3)^2$ and a coupling of 2 of them (I already have spectrum of $H=A(S_1\cdot S_2)$). However, every combination I have tried if terms of coefficients and couplings have not been successful, so I was hoping someone could give me a hint as to how to couple these guys. Thanks.

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1 Answer 1

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Hints:

  1. Notice that if we define $\mathbf S_{123} = \mathbf S_1 + \mathbf S_2 + \mathbf S_3$ and $\mathbf S_{12} = \mathbf S_1 + \mathbf S_2$, then we have \begin{align} \mathbf S_{123}^2 = \mathbf S_{12}^2 + \mathbf S_3^2 + 2\mathbf S_{12}\cdot\mathbf S_3 \end{align}

  2. Notice that your hamiltonian can be written as follows: \begin{align} H = \frac{A}{2}(\mathbf S_{12}^2- \mathbf S_1^2 - \mathbf S_2^2) + B(\mathbf S_{12}\cdot\mathbf S_3) \end{align}

  3. Combine the last two hints.

  4. Recall that the representation theory of the angular momentum algebra (aka addition of angular momentum) tells us that a tensor product of three spin-$1/2$ reprsentations splits into a direction sum as follows: \begin{align} \tfrac{1}{2}\otimes\tfrac{1}{2}\otimes\tfrac{1}{2} &= (1 \oplus 0)\otimes\tfrac{1}{2} \\ &= (1\otimes \tfrac{1}{2}) \oplus (0\otimes \tfrac{1}{2}) \\ &= \underbrace{(\tfrac{3}{2}\oplus\tfrac{1}{2})}_{s_{12}=1, s_3 = \frac{1}{2}}\oplus \underbrace{\tfrac{1}{2}}_{s_{12}=0, s_3=\frac{1}{2}} \end{align}

  5. (Addendum) To clarify some discussion in the comments, the last line in number 4 means that the Hilbert space admits an orthonormal basis of states which I'll label $|s_{123}, s_{12}\rangle$ for which \begin{align} \mathbf S_{123}^2|s_{123}, m_{123},s_{12}\rangle &= \hbar^2 s_{123}(s_{123}+1)|s_{123}, m_{123},s_{12}\rangle \\ S_{123}^z |s_{123}, m_{123},s_{12}\rangle &= \hbar m_{123}|s_{123}, m_{123},s_{12}\rangle\\ \mathbf S_{12}^2|s_{123}, m_{123},s_{12}\rangle &= \hbar^2 s_{12}(s_{12}+1)|s_{123}, m_{123},s_{12}\rangle \\ \mathbf S_{1}^2|s_{123}, m_{123},s_{12}\rangle &= \hbar^2 \tfrac{1}{2}(\tfrac{1}{2}+1)|s_{123}, m_{123},s_{12}\rangle \\ \mathbf S_{2}^2|s_{123}, m_{123},s_{12}\rangle &= \hbar^2 \tfrac{1}{2}(\tfrac{1}{2}+1)|s_{123}, m_{123},s_{12}\rangle \\ \mathbf S_{3}^2|s_{123}, m_{123},s_{12}\rangle &= \hbar^2 \tfrac{1}{2}(\tfrac{1}{2}+1)|s_{123}, m_{123},s_{12}\rangle \end{align} and explicitly, the basis is as follows: \begin{align} \left.\begin{array}{l} |\tfrac{3}{2}, \tfrac{3}{2},1\rangle \\ |\tfrac{3}{2}, \tfrac{1}{2},1\rangle \\ |\tfrac{3}{2}, -\tfrac{1}{2},1\rangle \\ |\tfrac{3}{2}, -\tfrac{3}{2},1\rangle \end{array}\right\} s_{123} = \tfrac{3}{2}, s_{12} = 1, \dim = 4\\ \left.\begin{array}{l} |\tfrac{1}{2}, \tfrac{1}{2},1\rangle \\ |\tfrac{1}{2}, -\tfrac{1}{2},1\rangle \end{array}\right\} s_{123} = \tfrac{1}{2}, s_{12} = 1, \dim = 2\\ \left.\begin{array}{l} |\tfrac{1}{2}, \tfrac{1}{2},0\rangle \\ |\tfrac{1}{2}, -\tfrac{1}{2},0\rangle \end{array}\right\} s_{123} = \tfrac{1}{2}, s_{12} = 0, \dim = 2 \end{align} This is a basis for the entire three spin-$1/2$ particle Hilbert space; it is eight-dimensional. There are therefore a total of 8 orthonormal eigenvectors for the hamiltonian.

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You are actually the best. As a followup, from yesterday, I know the spectrum of $S_{12}$ and then $S_{123}$ can be 3/2 or 1/2? –  yankeefan11 Mar 28 at 2:34
    
Haha ;). Yup, I think you got it. Basically the last line of 4 is saying that the Hilbert space decomposes into orthogonal subspaces, two of which have total spin quantum number $s_{123} = \tfrac{1}{2}$ and one of which has $s_{123} = \tfrac{3}{2}$, and within each of these subspaces, $s_{12}$ has a certain value based on where that subspace came from when we first combined the first and second spins. –  joshphysics Mar 28 at 2:39
    
For degeneracies, I know that I can have 4 $S_{123}=3/2$ states, and that would be coupled with 3 $S_{12}=1$ states and just one $S_{12}=0$ state. Same goes for $S_{123}=1/2$, except I only have two possible states for that to coupled with $S_{12}$? –  yankeefan11 Mar 28 at 3:26
    
If I do that, I only get 24 states. Shouldn't I have 27? –  yankeefan11 Mar 28 at 3:51
    
@yankeefan11 I don't quite follow. The spin-$\frac{1}{2}$ Hilbert space is $2$-dimensional, so its threefold tensor product has dimension $2^3 = 8$. This matches the dimension of the direct sum since $\dim(\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{1}{2}) = 4 + 2 + 2 = 8$. –  joshphysics Mar 28 at 5:11

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