Strange Hamiltonian of two particles on the surface of a sphere

[Salmone]

I have a problem with this Hamiltonian: two identical particles of mass $m$ and spin half are constrained to move on the surface of a sphere of radius $R$. Their Hamiltonian is $H=\frac{1}{2}mR^2(L_1^2+L_2^2+\frac{1}{2}L_1L_2+\frac{1}{2}S_1S_2)$. By introducing the two operators
$L=L_1+L_2$ and $S=S_1+S_2$ I was able to rewrite the Hamiltonian as: $H=\frac{1}{8}mR^2(3L_1^2+3L_2^2+L^2+S^2-\frac{3}{2}\hbar^2)$ this looks to me very strange since the Hamiltonian for two spinless particles on the surface of a sphere is $H=\frac{L_1^2+L_2^2}{2mR^2}$ so how can this be the Hamiltonian of two particles on the surface of a sphere?

And how can I find the eigenvalues of this Hamiltonian? For the resolution I thought I can separate the Hamiltonian into four parts: $H_1=\frac{3}{8}mR^2L_1^2$, $H_2=\frac{3}{8}mR^2L_2^2$, $H_3=\frac{3}{8}mR^2L^2$, $H_4=\frac{1}{8}mr^2S^2-\frac{3}{16}\hbar^2mR^2$ but still I don't know how to go on.

 

[DrClaude]

this looks to me very strange since the Hamiltonian for two spinless particles on the surface of a sphere is $H=\frac{L_1^2+L_2^2}{2mR^2}$ so how can this be the Hamiltonian of two particles on the surface of a sphere?

This Hamiltonian doesn't include any interaction between the particles. The other one clearly does.

And how can I find the eigenvalues of this Hamiltonian? For the resolution I thought I can separate the Hamiltonian into four parts: $H_1=\frac{3}{8}mR^2L_1^2$, $H_2=\frac{3}{8}mR^2L_2^2$, $H_3=\frac{3}{8}mR^2L^2$, $H_4=\frac{1}{8}mr^2S^2-\frac{3}{16}\hbar^2mR^2$ but still I don't know how to go on.

There are an infinite number of states, as $l_1$ and $l_2$ are unbounded. You can separate spin from orbital angular momentum, resulting in the usual singlet and triplet states. Then you can find orbital eigenstates for each value of $L$ starting at 0. Make sure that you only consider orbital + spin combinations that satisfy the Pauli principle.