I have three interacting spin-1/2 particles and I want to find the energy eigenvalues of H. The Hamiltonian for the system is $ H = \frac{J}{\hbar^2}(S_1\cdot S_2+S_2\cdot S_3+S_3\cdot S_1)$ (where J is positive and has energy units) and $S_{tot} = S_1+S_2+S_3$ and $S_{tot}^2 = S_{tot} \cdot S_{tot}$. I have expressed the Hamiltonian with $S_{tot}^2$ as

$H = \frac{J}{2\hbar^2}(S_{tot}^2-S_1^2-S_2^2-S_3^2)$ What I am struggeling with is using this new Hamiltonian on for example $H|\uparrow \downarrow \downarrow \rangle$ and all other spin combinations. I managed to do so with the first expression for the Hamiltonian, but not with the new one.

And $S_1 = S\otimes I \otimes I$, $S_2 = I \otimes S\otimes I$, $S_3 = I \otimes I\otimes S$


The energy eigenvalues are given by the spin quantum number as $S^2\mid\psi\rangle=s(s+1)\hbar^2\mid \psi\rangle$. For a system of three spins we have either $s_{tot}=\frac{3}{2}$ or $s_{tot}=\frac{1}{2}$, while $s=\pm\frac{1}{2}$. For a general three-spin system the energy eigenvalues of $H$ must be, for $s_{tot}=\frac{3}{2}, s=\frac{1}{2}$;

$\frac{J}{2\hbar^2}(S_{tot}^2-(S_1^2+S_2^2+S_3^2))\mid\psi\rangle = \frac{J}{2\hbar^2}\left(\frac{3}{2}(\frac{3}{2}+1)\hbar^2 - 3\frac{1}{2}(\frac{1}{2}+1)\hbar^2\right)\mid\psi\rangle = J\frac{3}{4}\mid\psi\rangle$.

The rest should be relatively straight-forward. By the way, you had a small error in the second Hamiltonian expression.

  • $\begingroup$ So when $s=1/2$, the energy eigenvalues would be negative, therefore we can disregard those states? $\endgroup$ – hbar-gal Oct 12 '16 at 20:17
  • $\begingroup$ $s_{tot} = 1/2$ I mean $\endgroup$ – hbar-gal Oct 12 '16 at 20:18
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    $\begingroup$ Not at all. That an energy eigenvalue is negative does not mean it is wrong and can be disregarded. I think that this Hamiltonian is an interaction Hamiltonian and therefore only part of the full Hamiltonian. Negative energy is OK as long as there is more to the problem. $\endgroup$ – Greg Winther Oct 12 '16 at 20:27
  • $\begingroup$ But $s_{tot}$ can be negative as well? $-3/2$ and $-1/2$? $\endgroup$ – hbar-gal Oct 12 '16 at 20:36
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    $\begingroup$ $s_{tot}$ can never be negative because its components, the spin angular momentum number $s_i$ is never negative. You may be mixing it up with the projection spin quantum number ($-s\leq m_s \leq s$). $\endgroup$ – Greg Winther Oct 12 '16 at 21:26

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