I'm using the Dirac notation for vectors here, since this is a quantum mechanics question.

Suppose the complete orthonormal bases $\{|\psi_n\rangle\}$ and $\{|\psi{'}_n\rangle\}$ are related by the transformation matrix $U$:

$$ |\psi{'}_n\rangle = U|\psi_n\rangle \\ \langle\psi{'}_n| = \langle\psi_n|U^{\dagger} $$

We want to prove that $U$ is a unitary operator, i.e. it satisfies the relationship $U^{\dagger}U = I$.

By operating the LHS of the second on the LHS of the first, and similarly for the RHS:

$$ \langle\psi{'}_n|\psi{'}_m\rangle = \langle\psi_n|U^{\dagger}U|\psi_m\rangle $$

Due to orthonormality $$ \langle\psi{'}_n|\psi{'}_m\rangle = \delta_{nm} = \langle\psi_n|\psi_m\rangle $$

Hence $$ \langle\psi_n|U^{\dagger}U|\psi_m\rangle = \langle\psi_n|\psi_m\rangle $$

My question is that is there any other relationship other than $U^{\dagger}U = I$ that satisfies this equation? i.e. is this a sufficient proof that $U$ is unitary?


What you have proved is $$ \langle \psi_n | U^\dagger U -I | \psi_m\rangle = \delta_{m,n}. $$ Since the $\psi_n$ are a basis this shows $U^\dagger U -I=0$, or equivalently $U^\dagger U =I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.