Consider two spin-$s_1$ and spin-$s_2$ particles coupled through a Hamiltonian of the form $$H = E_1 S_1^z + E_2 S_2^z + J S_1^x S_2^x,$$ for some $E_1,E_2,J\in \mathbb{R}$. Suppose we want to find the eigenstates/eigenvalues of this Hamiltonian. Now for a specific $s_1$ and $s_2$, a brute-force way of doing this would be to 1) find the matrix representation of the angular momentum operators in the joint eigenbasis of $S_1^z$ and $S_2^z$ (utilizing ladder operators for the transverse components). And then 2) plugging these matrix representations into $H$ and just finding the eigenvectors and eigenvalues of that matrix.

That works for specific $s_1$ and $s_2$ values, provided that they are small enough so that the matrix representation isn't too large. But I was wondering if there is a way to find the eigenstates for arbitrary spins instead. For Hamiltonians of the form $\mathbf{S}_1\cdot\mathbf{S}_2$, you can rewrite the dot product in terms of the total angular momentum of the two and then use Clebsch-Gordon coefficients. But that trick doesn't seem to be applicable here either.

  • $\begingroup$ For large $s_1$, $s_2$, a quasi-classical approximation can be used. $\endgroup$
    – Gec
    Sep 6, 2023 at 16:54

2 Answers 2


I'll rename your parameters in a more standard form: $$ H = -B_1S_1^z-B_2S_2^z-JS_1^xS_2^x $$ In general, I don't think that there is a simple closed form for arbitrary $s_2,s_2$. Intuitively, your Hamiltonian has no standard symmetries, so it'll be hard to directly diagonalise it using standard basis. I'll just propose various approximation schemes that can be useful to calculate the spectrum.

A first natural step is to treat $J$ perturbatively in the $J\to0$ limit. In this case, the natural basis is the simultaneous $S_1^z,S_2^z$ eigenbasis of $(2s_1+1)(2s_2+1)$ vectors: $|m_1,m_2\rangle_{|m_1|\leq s_1,|m_2|\leq s_2}$. They automatically form an energy eigenbasis: $$ H(J=0)|m_1,m_2\rangle = (-B_1m_1-B_2m_2)|m_1,m_2\rangle \\ E_{m_1,m_2}^{(0)} = -B_1m_1-B_2m_2 $$

For simplicity assuming that initially the energy spectrum is non degenerate ($B_1,B_2$ being incommensurable suffices), 1rst order perturbation theory gives a vanishing correction. At second order, you get the leading order correction. Using: $$ \langle m'|S_x|m\rangle = \frac{1}{2}(\delta_{m',m+1}+\delta_{m',m-1})\sqrt{(s+1)(m+m'-1)-mm'} $$ you get: $$ E_{m_1,m_2}^{(2)} =\frac{J^2}{4}\sum_{\epsilon_1,\epsilon_2\in\{\pm\}^2}\frac{[(s_1+1)(2m_1+\epsilon_1-1)-m_1(m_1+\epsilon_1)][(s_2+1)(2m_2+\epsilon_2-1)-m_2(m_2+\epsilon_2)]}{\epsilon_1B_1+\epsilon_2B_2} $$

Conversely, you could use perturbative theory in the $J\to\infty$ limit. This time, the natural basis is the simultaneous $S_1^x,S_2^x$ eigenbasis of $(2s_1+1)(2s_2+1)$ vectors: $|m_1,m_2\rangle_{|m_1|\leq s_1,|m_2|\leq s_2}$. They automatically form an energy eigenbasis: $$ H(J\to\infty)|m_1,m_2\rangle = -Jm_1m_2|m_1,m_2\rangle \\ E_{m_1,m_2}^{(0)} = -Jm_1m_2 $$ However, the energies are degenerate and you need to apply second order perturbation theory with lifted degeneracy which complicates matters.

Finally, there is a third general approach outlined by Gec. Just as the low $s_1,s_2$ limit is easy to compute, the high $s_1,s_2$ limit is also easier. You just need to think classically since large $s$ is equivalent to vanishing $\hbar$. $S_1,S_2$ are now classical vectors of respective norm $s_1,s_2$. Assuming that they are both uniformly (using the standard rotation invariant measure), independently distributed on their respective spheres, this gives the energy density: $$ \rho(E) = \frac{(2s_1+1)(2s_2+1)}{(4\pi)^2}\int \delta(E+B_1s_1\cos\theta_1+B_2s_2\cos\theta_2+Js_1s_2\sin\theta_1\sin\theta_2\cos\phi_1\cos\phi_2)\sin\theta_2\sin\theta_2d\phi_1d\theta_1d\phi_2d\theta_2 $$ Once again, I don't think that you can calculate this analytically in general, but simulating it using Monte Carlo methods is very easy.

Hope this helps.


LPZ's answer inspired me to write more about the quasi-classical approach. I agree that it is hardly possible to find an exact formula for the eigenvalues of the original Hamiltonian. Nevertheless, modern quasi-classical methods provide a powerful tool for finding the asymptotics of the spectra of wide classes of quantum Hamiltonians with a small (or large) parameter. And asymptotic formulas often work surprisingly well at moderate parameter values.

To demonstrate the power of quasi-classics, let's try to find the spectra of states with $S^z_1\approx s_1$, $S^z_2\approx s_2$. In the case when $J, E_1, E_2 \sim 1$, $s_1, s_2\gg 1$, for the arbitrary direction of spins, it is easy to see that the terms $E_1 S^z_1+E_2 S^z_2 \sim s_{1,2}$ are small in comparison with the term $JS^x_1S^x_2\sim s_1s_2$. In this case, the perturbation theory mentioned by LPZ will work. To consider a more interesting example, let's further assume that $E_1 = s_2 h_1$, $E_2 = s_1 h_2$, where $J, h_1, h_2, s1/s2\sim 1$, $s_1, s_2 \gg 1$.

I want to find main term and first corrections in the energies of states with $S^z_1\approx s_1$, $S^z_2\approx s_2$. Therefore, it is convenient to use the following approximate representation of spin operators $S^\alpha_i$ in terms of bosonic operators $a^\pm_i$: $$ S_i^z \approx s_i - a^+_ia^-_i,\quad S_i^x \approx \sqrt{\frac{s_i}2}(a^+_i + a^-_i), \quad S_i^y \approx i\sqrt{\frac{s_i}2}(a^+_i - a^-_i),\quad (*) $$ $$ [a^-_i,a^+_{i'}] = \delta_{ii'}. $$ (don't confuse the index $i$ with $i = \sqrt{-1}$) For our purpose, there is no need to write precise expressions for $S$ through $a$. It is easy to check the approximate validity of spin commutation relations.

Substituting (*) into the original Hamiltonian gives: $$ H = s_1s_2(h_1 + h_2) + $$ $$ +\frac12J\sqrt{s_1s_2}\left((a_1^+ + a_1^-) (a^+_2 + a_2^-) - \frac{2h_1}{J}\sqrt{\frac{s_2}{s_1}} a^+_1 a^-_1 - \frac{2h_2}{J}\sqrt{\frac{s_1}{s_2}} a^+_2 a^-_2 \right) + O(s_{1,2}^0) $$ In this last expression, the first term is the quasi-classical energy $E_1s_1 + E_2s_2$ of the order $s_1s_2$ of states with $S^z_1\approx s_1$, $S^z_2\approx s_2$. And the second term is an exactly solvable Hamiltonian of two coupled oscillators whose eigenvalues give quantum correction of the order $\sqrt{s_1s_2}$ to the quasi-classical energy.

To find quantum corrections to the quasi-classical energy in case of arbitrary direction of quasi-classical spins $S_i^\alpha$ it is sufficient to express initial spin operators in terms of appropriate rotated spins $S'^\alpha_i$ and use the same technique.


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