Because H 11 = H 22 as well as H 12 = H 21 , the linear combination of | 1 ⟩ and | 2 ⟩ that diagonalizes the Hamiltonian is | ± ± ⟩ = 1 2 ± ± 2 ⟨ 1 | 2 ⟩ ( | 1 ⟩ ± ± | 2 ⟩ ) (after normalization). Now as [ H , i ] = 0 for H 2 + , the states | ± ± ⟩ are also eigenstates of i. It turns out that | + ⟩ and | − − ⟩ are the eigenstates of i with eigenvalues +1 and -1 (in other words, the wave functions ⟨ r | + ⟩ and ⟨ r | − − ⟩ are gerade (symmetric) and ungerade (unsymmetric), respectively). The corresponding expectation value of the energies are E ± ± = 1 1 ± ± ⟨ 1 | 2 ⟩ ( H 11 ± ± H 12 ) .