Now, an even operator A ^ ^ is one that satisfies, A ~ ~ = P A ^ ^ P [ P , A ^ ^ ] = 0 while an odd operator B ^ ^ is one that satisfies P B ^ ^ + B ^ ^ P = 0 Since the square of the momentum operator p ^ ^ 2 is even, if the potential V(r) is even, the Hamiltonian H ^ ^ is said to be an even operator. In that case, if each of its eigenvalues are non-degenerate, each eigenvector is necessarily an eigenstate of P, and therefore it is possible to look for the eigenstates of H ^ ^ among even and odd states. However, if one of the energy eigenstates has no definite parity, it can be asserted that the corresponding eigenvalue is degenerate, and P | ψ ψ ⟩ ⟩ is an eigenvector of H ^ ^ with the same eigenvalue as | ψ ψ ⟩ ⟩ .