Unitary Transfomation from One Basis to Another 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? 
 A: 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 $\bar{U} = UP^\dagger$.
