Why can we write $\rho=\sum_\mu q_\mu|\varphi_\mu\rangle\!\langle\varphi_\mu|$ iff $q\preceq \mathrm{spec}(\rho)$?

Question

Exercise 2.6 in Preskill's notes (chapter 2, around page 48, pdf available here) asks to prove that an arbitrary state $\rho=\sum_i p_i |\alpha_i\rangle\!\langle\alpha_i|$, where $p_i$ and $|\alpha_i\rangle$ are obtained from the spectral decomposition of $\rho$, can be also decomposed as $$\rho = \sum_\mu q_\mu |\varphi_\mu\rangle\!\langle \varphi_\mu|,$$ for some ensemble of pure states $|\varphi_\mu\rangle$, if and only if $q\preceq p$, that is, if and only if there is some doubly stochastic matrix $D$ such that $q_\mu=\sum_i D_{\mu i}p_i$.

As a hint, he remembers that the two decompositions above are simultaneously possible if and only if $$\sqrt{q_\mu}|\varphi_\mu\rangle = \sum_i \sqrt{p_i} V_{\mu i}|\alpha_i\rangle,$$ for some unitary $V$. He also says we can use Horn's lemma: if $q\preceq p$ then $q=Dp$ with $D_{\mu i}=|U_{\mu i}|^2$ for some unitary $U$.

I understand why the statements in the hints hold, but I'm not sure how to apply them to the given exercise. I tried using the above relation to isolate $q_\mu$, but then I get $$\sqrt{q_\mu} = \sum_i \sqrt{p_i} V_{\mu i}\langle \varphi_\mu|\alpha_i\rangle \implies q_\mu = \sum_{ij} \sqrt{p_i p_j} V_{\mu i}\bar V_{\mu j} \langle \varphi_\mu|\alpha_i\rangle \langle\alpha_j|\varphi_\mu\rangle,$$ which I'm not sure how to reframe as $q=Dp$. On the other hand, taking the expectation value over $|\alpha_i\rangle$, I get the relation $$p_i = \sum_\mu q_\mu |\langle\alpha_i|\varphi_\mu\rangle|^2,$$ but again, this amounts to $p=A q$ for some $A$ about which we only know the columns sum to one. Besides, even if $A$ was doubly stochastic, the relation would be in the wrong direction.

Shur-Horn's theorem (if $\rho$ is Hermitian then $\operatorname{diag}(\rho)\preceq \operatorname{spec}(\rho)$) seems to be also relevant, but the diagonal of $\rho$ in the above notation would have elements $q'_i = \sum_\mu q_\mu |\langle i|\varphi_\mu\rangle|^2$, so we'd get $q'\preceq p$, which is not quite the same as $q\preceq p$ as far as I can tell. With $\operatorname{spec}(\rho)$ I mean the vector whose elements are the eivanvalues of $\rho$.

Answer 1

We have $\sqrt{q_\mu}\langle\alpha_j|\varphi_\mu\rangle = \sum_i \sqrt{p_i} V_{\mu i}\langle\alpha_j|\alpha_i\rangle = \sqrt{p_j} V_{\mu j}$.

Thus $q_\mu |\langle\alpha_j|\varphi_\mu\rangle|^2 = p_j |V_{\mu j}|^2$ and $q_\mu \sum_{j}|\langle\alpha_j|\varphi_\mu\rangle|^2 = \sum_{j}p_j |V_{\mu j}|^2$.

Hence $q_\mu = \sum_{j}p_j |V_{\mu j}|^2$, since $\sum_{j}|\langle\alpha_j|\varphi_\mu\rangle|^2 = \langle\varphi_\mu|\varphi_\mu\rangle = 1$.

That is, it's the same unitary that appears in the HJW theorem and in Horn's lemma.

In the other way, suppose we are given distribution $q$ such that $q_\mu = \sum_{j}p_j |V_{\mu j}|^2$ for some unitary $V$. As in the HJW theorem we can consider purification $\sum_i \sqrt{p_i} |\alpha_i\rangle \otimes V|i\rangle = \sum_i \sqrt{p_i} |\alpha_i\rangle \otimes \sum_k V_{k i}|k\rangle = \sum_k \big(\sum_i \sqrt{p_i} V_{k i} |\alpha_i\rangle \big) \otimes |k\rangle$.
Then $\sqrt{q_k}|\varphi_k\rangle := \sum_i\sqrt{p_i} V_{k i} |\alpha_i\rangle$.