The total angular momentum of the system is obtained by coupling l_{1} =1 and l_{2} =1: \hat{\vec{L}}=\hat{\vec {L}} _{1}+ \hat{\vec{L}} _{2}. This leads to \hat{\vec {L}} _{1} .\hat {\vec{L}} _{2}=\frac{1}{2} (\hat{L}^{2}-\hat{L}^{2} _{1}-\hat{L}^{2} _{2} ), and when this is inserted into the system’s Hamiltonian it yields
\hat{H} =\frac{\varepsilon _{1} }{\hbar ^{2} } (\hat{\vec{L}} _{1} .\hat{\vec{L}} _{2}+\hat{L}^{2}_{2})+ \frac{\varepsilon _{2} }{\hbar ^{2} } \hat{L}^{2}_{z} =\frac{\varepsilon _{1} }{2\hbar ^{2} } (\hat {L}^{2} -\hat{L}^{2}_{1}+\hat{L}^{2}_{2} ) + \frac{\varepsilon _{2} }{\hbar ^{2} } \hat{L}^{2}_{z} . (7.436)
Notice that the operators \hat{H} ,\hat{L}^{2}_{1},\hat {L}^{2} _ {2},\hat{L}^{2}, and \hat{L}_{z} mutually commute; we denote their joint eigenstates by |l,m 〉. The energy levels of (7.436) are thus given by
E_{lm} =\frac{\varepsilon _{1} }{2} \left[l(l+1)-l_{1}(l_{1}+1) +l_{2}(l_{2}+1)\right] +\varepsilon _{2}m^{2} = \frac{\varepsilon _{1} }{2}l(l+1)+\varepsilon _{2}m^{2} , (7.437)
since l_{1} =l_{2} =1.
The calculation of |l,m 〉 in terms of the states |l_{1 },m_{1} 〉|l_{2},m_{2} 〉=|l_{1},l_{2};m_{1},m_{2} 〉 was carried out in Problem 7.3, page 436; the states corresponding to a total angular momentum of l = 2 are given by
|2,\pm 2 〉=|1,1;\pm 1,\pm 1 〉, |2,\pm 1 〉=\frac{1}{\sqrt{2} } \left(|1,1;\pm 1,0 〉+|1,1;0,\pm 1 〉\right) , (7.438)
|2,0 〉=\frac{1}{\sqrt{6} } \left(|1,1;1,-1 〉+2|1,1;0,0 〉+|1,1;-1,1 〉\right). (7.439)
From (7.437) we see that the energy corresponding to l = 2 and m=\pm 2 is doubly degenerate, because the states |2,\pm 2 〉 have the same energy E_{2,\pm 2}= 3\varepsilon _{1}+4\varepsilon _{2}. The two states |2,\pm 1 〉 are also degenerate, for they correspond to the same energy E_{ 2,\pm 1}= 3\varepsilon _{1}+\varepsilon _{2}. The energy corresponding to |2,0 〉 is not degenerate: E_{20}= 3\varepsilon _{1}.