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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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2 Answers 2

Starting with the first two identities you can actually show that the last expression is satisfied not just for the basis vectors but for any two vectors that are elements of the space spanned by the basis vectors, i.e. $\langle\eta|U^{\dagger}U|\theta\rangle = \langle\eta|\theta\rangle$. Therefore $U^{\dagger}U = I$.

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In finite dimensions,

$$\langle\psi_n|U^\dagger U|\psi_m\rangle$$

extracts the $(n,m)$ component of $U^\dagger U$ in basis $\{|\psi_n\rangle\}$. Since the result is $\delta_{nm}$, so $U^\dagger U = I$.

In infinite dimensions, you can apply double-integrals

$$\int_n \int_m |\psi_n\rangle(\cdot)\langle\psi_m|\; dm\; dn = I$$

on both sides, separate the integrals, and finally use

$$\int_p |\psi_p\rangle\langle\psi_p| \;dp = I$$

to "cancel" the vectors to get $U^\dagger U = I$.

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