Here, the system is not spherically symmetric. So, [ H , L 2 ] ≠ ≠ 0 , and the state cannot be depicted in terms of l as an eigenstate of the Hamiltonian is not an eigenstate of L 2 anymore (in contrast to the atomic term symbol, where the states were written as 2 S + 1 L J ). But, as [ H , L z ] = 0 , the eigenvalues corresponding to L z can still be used. If,