I was wondering if it is important in Quantum Mechanics to deal with operators that have an orthonormal basis of eigenstates? Imagine that we would have an operator (finite-dimensional) acting on a spin system that has real eigenvalues, but its eigenvectors are not perpendicular to each other. Is there any reason why such an operator cannot describe an actual physical quantity?
Quantum Mechanics – Do We Need an Orthonormal Basis in Quantum Mechanics?
hilbert-spaceoperatorsquantum mechanics
Best Answer
If two states are orthogonal, this means that $\langle \psi | \phi \rangle = 0$. Physically this means that if a system is in state $|\psi\rangle$ then there is no possibility that we will find the system in state $|\phi\rangle$ on measurement, and vis versa. In other words the 2 states in some sense mutually exclusive. This is an important property for operators because it means that the results of a measurement are unambiguous. A state with a well defined momentum $p_1$, i.e. an eigenstate of the momentum operator, cannot also have a momentum $p_2 \ne p_1$. Observables having an orthogonal (and complete) set of eigenstates is therefore a requirement in order for the theory to make physical sense (or at least for repeated measurements to give consistent results, as is experimentally observed)