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

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 $$\rho$$.

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 $$\sum_{j}|\langle\alpha_j|\varphi_\mu\rangle|^2 = \langle\varphi_\mu|\varphi_\mu\rangle = 1$$.
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 $$\sqrt{q_k}|\varphi_k\rangle := \sum_i\sqrt{p_i} V_{k i} |\alpha_i\rangle$$.