I've been trying to solve exercise 2.73 (p.g 105) in Nielsen Chuang, and I'm not sure if i'v been overthinking it and the answer is as simple as i've described below or if I am missing something, or i'm just wrong!
Ex 2.73:
Let $\rho$ be a density operator. A minimal ensemble for $\rho$ is an ensemble $\{p_i,|\psi_i\rangle\}$ containing a number of elements equal to the rank of $\rho$. Let $|\psi\rangle$ be any state in the support of $\rho$. Show that there is a minimal ensemble for $\rho$ that contains $|\psi\rangle$, and moreover that in any such ensemble $|\psi\rangle$ must appear with probability
$p_i=\frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}$
where $p^{-1}$ is defined to be the inverse of $\rho$, when $\rho$ is considered as an operator acting only on the support of $\rho$
My answer so far is:
$\rho$ is positive the therefore has a spectral decomposition $\rho=\sum_k\lambda_k|k\rangle\langle k|$.
The density operator cann be defined as $\rho=\sum_kp_k|k\rangle\langle k| = \sum_k|\hat{k}\rangle\langle \hat{k}|$, where $|\hat{k}\rangle=\sqrt{\lambda_k}|k\rangle$, and therefore $|k\rangle = \frac{|\hat{k}\rangle}{\sqrt{\lambda_k}} $.
For any $|\psi_i\rangle = \sum_k c_{ik}|k\rangle$, using the above definition of $|k\rangle$:
$|\psi_i\rangle = \sum_k \frac{c_{ik}}{\sqrt{\lambda_k}}|\hat{k}\rangle$
The density operator is given by $\rho=\sum_i|\psi_i\rangle\langle\psi_i|$, therefore
$\rho = \sum_{i}\sum_{k}\frac{c_{ik}^2}{\lambda_k}|\hat{k}\rangle \langle\hat{k}|$.
By the definition of $\rho$ is can be seen that $p_i = \sum_{k}\frac{c_{ik}^2}{\lambda_k}$.
— reading this back i'm not sure this is correct at all 🙁
For the second part working backwards a bit:
$\langle \psi_i|\rho^{-1}|\psi_i\rangle = \langle \psi_i|\sum_k \left( \frac{1}{\lambda_k}|k\rangle\langle k| \right) |\psi_i\rangle = \sum_k \frac{1}{\lambda_k}\langle \psi_i|k\rangle\langle k |\psi_i\rangle = \sum_{i,k} \frac{1}{\lambda_k}c_{i,k}^2\langle i|k\rangle \langle k |i\rangle $
Given that $|i\rangle$ is of basis $|k \rangle$, $\langle k |i\rangle = \langle i |k\rangle = 1 $ if $i=k$, therefore
$\langle \psi_i|\rho^{-1}|\psi_i\rangle = \sum_{k} \frac{c_{i,k}^2}{\lambda_k}$, so it then follows that
$p_i = \frac{1}{\sum_{k} \frac{c_{i,k}^2}{\lambda_k}}$
However the above result does not match with the result I got for $p_i$ in the first part, so one of them is wrong…
Best Answer
Sketch of proof: We need the following fact:
This can be proven with the spectral theorem. However, this is the only thing that we need the spectral theorem for.
Recall that an operator $\rho$ is positive semidefinite iff for all vectors $|\phi \rangle$, we have $\langle \phi |\rho | \phi \rangle > 0$. For part one, use the fact that $\langle \psi|\rho|\psi \rangle$ to show that there exists an $\alpha$ with $0<\alpha<1$ for which $\sigma = \rho - \alpha |\psi \rangle \langle \psi |$ is positive semdefinite. Now, take any ensemble $\{p_i,|\psi_i\rangle \}$ for $\hat \sigma = \frac{\sigma}{1-\alpha}$, and show that $\{(1-\alpha)p_i,|\psi_i\rangle\} \cup \{\alpha, |\psi\rangle\}$ is an ensemble for $\rho$.
I'm not quite sure what they're asking for part $2$, but here are my thoughts. Let $r$ denote the rank of $\rho$, and let $\alpha_* = \frac{1}{\langle \psi |\rho^{-1}|\psi\rangle}$. It suffices to note/show that $\sigma = \rho - \alpha |\psi \rangle \langle \psi |$ will fail to be positive semidefinite for $\alpha > \alpha_*$, and that $\sigma$ will have rank $r$ (instead of $r-1$) when $\alpha < \alpha_*$.
Proof that this is the correct value for $\alpha_*$: With the Schur complement, we see that $\rho - \alpha xx^\dagger$ is positive semidefinite iff the matrix $$ M = \pmatrix{\rho & x\\x^\dagger & \alpha^{-1}} $$ is positive semidefinite. By taking the Schur complement relative to $\rho$, we find that $M$ is positive semidefinite iff $\alpha^{-1} - x^\dagger\rho^{-1}x \geq 0$, which is to say that $\alpha \leq \alpha_* = \frac{1}{x^\dagger\rho^{-1}x}$, as was desired.
A matrix version of the proof given on the QIT SE site:
Let $D = \operatorname{diag}(p_1,\dots,p_r)$, and let $a_1,\dots,a_r$ be the linearly independent vectors (corresponding to $\hat \psi_i = \sqrt{p_i}\psi_i$. Let $A$ be the matrix with columns $A$; we have $\rho = AA^\dagger$. Note that $$ A = \rho \rho^{-1} A = AA^\dagger \rho^{-1} A= A[A^\dagger\rho^{-1}A]. $$ $A$ has linearly independent columns and is therefore left-cancellable. Conclude that $A^\dagger \rho^{-1} A = I_{r}$. If we consider the $j,j$ entry, we have $$ 1 = a_j^\dagger\rho^{-1}a_j \leadsto 1 = \langle \hat \psi_j | \rho^{-1} | \hat \psi_j \rangle = p_j \cdot \langle \psi_j | \rho^{-1} | \psi_j \rangle. $$
Second version of the matrix proof: $\rho - \alpha xx^\dagger$ is positive semidefinite iff $\rho^{-1/2}[\rho - \alpha xx^\dagger] \rho^{-1/2} = I - \alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger$ is postive semidefinite. It's easy to see that the lowest eigenvalue of this matrix is $1-\lambda$, where $\lambda$ is the largest eigenvalue of $\alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger$. Because this matrix has rank $1$, we see that $$ \lambda = \operatorname{Tr}(\alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger) = \alpha (\rho^{-1/2}x)^\dagger) (\rho^{-1/2}x) = \alpha x^\dagger \rho^{-1} x. $$ We reach a threshold at $\lambda = 1$, i.e. $\alpha = \frac{1}{x^\dagger \rho^{-1} x}$.