2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $s_{tot}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.