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So we have two orthonormal linearly independent basis $\{ |\phi_1 \rangle, \dots, |\phi_n \rangle \}$ and $\{ |\psi_1 \rangle, \dots, |\psi_n \rangle \}$. We can express the basis vectors of the first set in terms of the second set by using the closure relation: $$ | \phi_i \rangle = \sum_{j=1}^n | \psi_j \rangle \langle \psi_j |\phi_i \rangle = \sum_{j=1}^{n} U_{ji}| \psi_j \rangle, \quad U_{ji} = \langle \psi_j |\phi_i \rangle $$

$U_{ji}$ is the matrix element of a unitary operator $U$. The thing that is confusing me is I am seeing this unitary transformation written as $$ |\phi_i \rangle = U | \psi_i \rangle $$

Why is this correct?

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closed as unclear what you're asking by Bill N, user36790, Kyle Kanos, ACuriousMind, HDE 226868 Oct 18 '15 at 16:30

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  • $\begingroup$ you need to use the definition of orthonormality $\endgroup$ – diffeomorphism Oct 18 '15 at 1:43
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    $\begingroup$ What makes you doubt it? $U$ is the $n\times n$ transformation matrix. I don't understand why you have a question. $\endgroup$ – Bill N Oct 18 '15 at 2:49
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It's very simple actually. For each $|\phi_i\rangle$ write the expansion coefficients $$ U_{ji} = \langle \psi_j | \phi_i \rangle $$ as explicit matrix elements in the basis $\{|\psi_j\rangle\}_j$ of a transformation $U$ that acts on $|\psi_i\rangle$: $$ U_{ji} = \langle \psi_j |U|\psi_i\rangle $$ The expression for $|\phi_i\rangle $ becomes (trivially) $$ |\phi_i\rangle = \sum_j {\langle \psi_j |U|\psi_i\rangle |\psi_j\rangle} = U|\psi_i\rangle $$ The $U$ so defined is a unitary application, $UU^\dagger = U^\dagger U = I$: $$\langle \psi_j| UU^\dagger |\psi_i\rangle = \sum_{i,j}{\langle \psi_j |U|\psi_i\rangle\langle \psi_j |U|\psi_i\rangle^*} = \sum_{i,j}{\langle \psi_j | \phi_i \rangle \langle \phi_i | \psi_j \rangle} = \delta_{ij}, \;\;\text{etc.} $$

Note that in writing the overlap $U_{ji}$ as a matrix element, the target vector $|\psi_i\rangle$ could be replaced by any other member of the $\{|\psi_j\rangle\}$ basis, say $|\psi_{J(i)}\rangle$, with $J(i)$ some permutation of the set ${1,2,...,n}$. In this case we would define $$ U_{ji} = \langle \psi_j |\bar{U}|\psi_{J(i)}\rangle $$ such that $$ |\phi_i\rangle = \bar{U}|\psi_{J(i)}\rangle $$ The resulting $\bar{U}$ is again a unitary application that differs from the previous $U$ by a permutation application $P$, $U = \bar{U}P$. Here $P$ is defined by $P|\psi_i\rangle = |\psi_{J(i)}\rangle$ and is itself a unitary application, $PP^\dagger = P^\dagger P = I$, so we have $\bar{U} = UP^\dagger$.

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