I Strange Hamiltonian of two particles on the surface of a sphere

Salmone
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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.
 
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Salmone said:
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.

Salmone said:
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.
 
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