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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?

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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 $U^\dagger U =I$.

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