For any orthonormal basis $B_1=\{|\psi_i\rangle\}_{i=1}^d$ and $B_2=\{|\phi_i\rangle\}_{i=1}^d$, there exists a unitary mapping $B_1$ to $B_2$ given by $U=\sum_{i=1}^d |\phi_i\rangle\langle\psi_i|$ such that $U|\psi_i\rangle=|\phi_i\rangle$.
How do I prove it from the definition of the change of basis matrix in the matrix form, $P_{B_1\to B_2}=\Big[[|\psi_1\rangle]_{B_2}\quad[|\psi_2\rangle]_{B_2}\quad\cdots\quad[|\psi_d\rangle]_{B_2}\Big]=B_2^{-1}B_1$ ?
What I know
The change of basis matrix from $B_1=\{|\psi_i\rangle\}$ to $B_2=\{|\phi_i\rangle\}$ is the matrix $P_{B_1\to B_2}$ whose columns are the elements of $B_1$ expressed in $B_2$, i.e., $P_{B_1\to B_2}=\Big[[|\psi_1\rangle]_{B_2}\quad[|\psi_2\rangle]_{B_2}\quad\cdots\quad[|\psi_d\rangle]_{B_2}\Big]=B_2^{-1}B_1$, where $B_1$ is the matrix with $|\psi_i\rangle$ as the columns and $B_2$ is the matrix with $|\phi_i\rangle$ as the columns.
Using this definition,
$$
P_{B_1\to B_2}=B_2^{-1}B_1=B_2^{\dagger}B_1=[|\phi_1\rangle\quad\cdots\quad |\phi_d\rangle]^\dagger[|\psi_1\rangle\quad\cdots\quad |\psi_d\rangle]\\
=\begin{bmatrix}\langle\phi_1|\\\vdots\\\langle\phi_d|\end{bmatrix}[|\psi_1\rangle\quad\cdots\quad |\psi_d\rangle]=\begin{bmatrix}\langle\phi_1|\psi_1\rangle&\cdots&\langle\phi_1|\psi_d\rangle\\\vdots&\ddots&\vdots\\\langle\phi_d|\psi_1\rangle&\cdots&\langle\phi_d|\psi_d\rangle\end{bmatrix}
$$
How come we equate this to $\sum_{i=1}^d |\phi_i\rangle\langle\psi_i|$ ? Or are both different ?
I may be missing something but it seems like $\sum_{i=1}^d |\phi_i\rangle\langle\psi_i|=B_2B_1^\dagger$.
Best Answer
Let $V$ be an $n$-dimensional complex inner product space (so $V \approx \mathbb{C}^n$). I will say a linear map $T \colon V \to V$ is unitary if $T^*T = I$, i.e., $T^* = T^{-1}$. One equivalent condition to $T^*T = I$ is that $T$ preserves inner products, i.e., $(Tv, Tw) = (v, w)$ for all $v, w \in V$. Another equivalent condition is that $T$ preserves norms, i.e., $\lVert Tv \rVert = \lVert v \rVert$ for all $v \in V$.
If $\{v_1, \dots, v_n\}$ is an orthonormal basis of $V$ and $\{w_1, \dots, w_n\}$ is an orthonormal basis of $V$, then the map $T \colon V \to V$ given by $Tv_i = w_i$ is unitary. This is because $(Tv, Tw) = (v, w)$ for all $v, w \in V$ (first for $v = v_i$, $w = w_j$, then for all $v, w \in V$ by linearity).
As for writing $T$ as a summation, we have $$Tv = \sum_{i = 1}^{n}(v, v_i)Tv_i = \sum_{i = 1}^{n}(v, v_i)w_i = \sum_{i = 1}^{n}(v_i^*v)w_i = \sum_{i = 1}^{n}w_iv_i^*v.$$ Hence $T = \sum_{i = 1}^{n}w_iv_i^*$.
I guess you are working in $\mathbb{C}^n$. The change of basis matrix, i.e. the identity transformation represented with respect to bases $B_1$, $B_2$ is $P = M_{B_1}^{B_2}(I) = B_2^{-1}B_1$. More fundamentally, the $j$th column of $M_{B_1}^{B_2}(I)$ has the coordinates of $Iv_j$ in the basis $B_2$. By orthonormality of $\{w_1, \dots, w_n\}$, we have $v_j = \sum_{i = 1}^{n}(v_j, w_i)w_i$. Thus $p_{ij} = (v_j, w_i)$. But $P \neq U$, so I don't know why you are computing it.
If you want to find the matrix representation of $U$, then you should be computing $M_{B_1}^{B_2}(U)$ instead. We have $Uv_j = w_j$, so $M_{B_1}^{B_2}(U)$ is the identity matrix, which is indeed unitary.