quantum mechanics - Proving the unitary relation of ensemble decompositions

In my class it was told that ensemble decompositions of a density operator $\rho$ are not unique, but that the ones that exist are related by a unitary operator. I'm trying to prove this, but I get stuck somewhere along the way.

Lets begin by assuming two different decompositions of density operator $\rho$: $\rho = \sum_{j=1}^n{p_j|\psi_j\rangle\langle\psi_j|} = \sum_{k=1}^m{q_k|\phi_k\rangle\langle\phi_k|}$

Now, these two decompositions live in a Hilbert space $\mathcal A$. We can then define a purification of both, using a system described by a Hilbert space $\mathcal B$ of dimension $k=\max (n,m)$, so that we get $|\Psi_1\rangle_{\mathcal A\mathcal B} = \sum_{j=1}^n{\sqrt{p_j}|\psi_j\rangle \otimes |b_j\rangle}$ and $|\Psi_2\rangle_{\mathcal A\mathcal B} = \sum_{k=1}^m{\sqrt{q_k}|\phi_k\rangle \otimes |b_k\rangle}$.

Now, here we can use that as these pure states are purifications of the same density operator, there must be a unitary $U$ connecting them: $(1_A \otimes U_B)|\Psi_1\rangle_{\mathcal A\mathcal B} = |\Psi_2\rangle_{\mathcal A\mathcal B}$.

Here is where I get stuck. I should be able to use this to prove the unitary relation between the $\psi$ and the $\phi$, but it is not obvious to me how I should do this.

Update: after reviewing the comments to the first question, I should have written that the $\psi$ and $\phi$ states do NOT have to be orthonormal, per se.

Answer

We will prove the following Theorem:

Let $\rho=\sum_{i=1}^Np_i\lvert\phi_i\rangle\langle\phi_i\rvert$ be a eigenvalue decomposition and $\sigma=\sum_{i=1}^Mq_i\lvert\psi_i\rangle\langle\psi_i\rvert$ ($M\ge N$). Then, $\rho=\sigma$ if and only if $$\begin{equation} \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle\ , \end{equation}$$ with $\sum_j v_{ij}v_{i'j}^*=\sum_k \delta_{ii'}$, i.e., $V\equiv(v_{ij})$ is an isometry.

Proof:

The "if" direction is straightforward: $$\begin{align} \rho&=\sum_{j}q_j\lvert\psi_j\rangle\langle\psi_j\rvert\\ &=\sum_{i,i',j} v_{ij} v_{i'j}^* \sqrt{p_i}\sqrt{p_{i'}} \lvert\phi_i\rangle \langle\phi_{i'}\rvert \\ &=\sum_i p_i\lvert\phi_i\rangle \langle\phi_{i}\rvert = \sigma\ , \end{align}$$ where in the last step we have used that $\sum_j v_{ij}v_{i'j}^*=\delta_{ii'}$.

To prove the converse, let $$v_{ij} := \langle\phi_i\lvert\psi_j\rangle\,\sqrt{q_j/p_i}\ .$$ Then, $$\sum_i v_{ij} \sqrt{p_i}\lvert\phi_i\rangle = \sum_i \lvert\phi_i\rangle\langle \phi_i\lvert\psi_j\rangle\sqrt{q_j} = \sqrt{q_j}\lvert\psi_j\rangle\ ,$$ i.e., $v_{ij}$ is the desired basis transformation. Further, $$\sum_{ii'}\underbrace{\delta_{ii'}\sqrt{p_ip_{i'}}}_{=:a_{ii'}}\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_ip_i\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_jq_j\lvert\psi_j\rangle\langle\psi_j\rvert = \sum_{ii'}\underbrace{\sum_jv_{ij}v^*_{i'j}\sqrt{p_ip_i'}}_{=:b_{ii'}}\lvert\phi_i\rangle\langle\phi_{i'}\rvert\ .$$ Now, since the $\lvert\phi_i\rangle$ are orthogonal (as they form an eigenbasis), the $\lvert\phi_i\rangle\langle\phi_{i'}\rvert$ are linearly independent, and thus, $a_{ii'}=b_{ii'}$, which implies $\sum_jv_{ij}v^*_{i'j}=\delta_{ii'}$.

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If the $\lvert\phi_i\rangle$ do not form an orthonormal basis, we can generalize the theorem by going through an orthonormal basis $\rho=\sum r_k\lvert\chi_k\rangle\langle\chi_k\rvert$: Then, $$\begin{align} \sqrt{p_i}\lvert\phi_i\rangle&=\sum_k u_{ki}\sqrt{r_k}\lvert\chi_k\rangle\\ \sqrt{q_j}\lvert\psi_j\rangle&=\sum_k w_{kj}\sqrt{r_k}\lvert\chi_k\rangle \end{align}$$ with $\sum_i u_{ki}u_{k'i}^*=\delta_{kk'}$ and $\sum_j w_{kj}w_{k'j}=\delta_{kk'}$. The second equation then yields $\sqrt{r_k}\lvert\chi_k\rangle= \sum_j w_{kj}^*\sqrt{q_j}\lvert\psi_j\rangle$. After inserting this in the first equation, we obtain $$\sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle$$ with $v_{ij} = \sum_k u_{ki}w_{kj}^*$, i.e., $V=UW^\dagger$ is a partial isometry.