Why do minimal ensemble decompositions for $\rho$ contain $|\psi⟩\in{\rm supp}(\rho)$ with probability $1/\langle\psi|\rho^{-1}|\psi⟩?$

Question

I came across the following exercise (2.73) in Nielsen & Chuang and am trying to understand it intuitively.

Here is my reasoning of what is going on:

The purpose of this exercise:

Let’s say we are given a density matrix, $\rho$. We are finding a way to decompose it into an ensemble of pure states. One way would be to decompose it into its eigenvectors, and the corresponding probabilities are the eigenvalues. However, in this exercise, we are finding a more general approach to decomposing the density matrix into any ensemble of pure states, but the ensemble of pure states must span the vector space that the eigenvectors span.

How?

1). We take an arbitrary state vector, $|\psi\rangle$, with the condition that it must be in the support of $\rho$. The support of $\rho$ is the vector space that $\rho$ lives in.

2). We take the inverse of the density matrix, which inverts the eigenvalues (or probabilities) for each eigenvector of the density matrix. This inversion leaves us with low probabilities for eigenvectors that account for most of $\rho$, and high probabilities for eigenvectors that don’t contribute to $\rho$ as much

3). Now, we take the expectation of our arbitrary state, $|\psi\rangle$, with the inverted density matrix. To understand this expectation, let’s think about the expectation value if we left out all the eigenvalues. We would calculate the inner product of $|\psi\rangle$ with each eigenvector, and the result should be one because the eigenvectors span the vector space, so a linear combination of them should return exactly $|\psi\rangle$. Now, if we include the eigenvalues (not inverted), the result will tell us which direction (combination of eigenvector) $|\psi\rangle$ most closely aligns with, by adding together eigenvalues for all the different directions weighted by the inner product with that eigenvector. This just tells how much of our system represented by our density matrix is in the state $|\psi\rangle$, or the probability that our system is in the state $|\psi\rangle$.

Where is the flaw in my reasoning? And what is the correct explanation of what is going on in this exercise?

Answer 1

I suspect that the tangle you're getting into is that you're implicitly expecting things to use an orthonormal basis. To get this untangled, it's probably worth looking at an example: $$ \rho=\begin{bmatrix} \frac34 & 0 \\ 0 & \frac14 \end{bmatrix}. $$ The eigenvectors clearly span the whole space, so for $|\psi\rangle$, I can pick any vector I want. For example, $|\psi\rangle=|+\rangle$.

Now, if I went for the apparently obvious way of calculating the weight, I'd get $$ \langle\psi|\rho|\psi\rangle=\frac12. $$ But then, how do I complete the ensemble? $$ \rho-\frac12|+\rangle\langle+|=\frac14\begin{bmatrix}2 & -1 \\ -1 & -1\end{bmatrix}. $$ You'll notice that this matrix has a negative determinant. It cannot be represented by a density matrix, let alone something that's rank 1, as required for the minimal decomposition.

Instead, let's compute $$ p=\frac{1}{\langle\psi|\rho^{-1}|\psi\rangle}=\frac38. $$ From here, you can show that $$ \rho=p|\psi\rangle\langle\psi|+(1-p)|\phi\rangle\langle\phi| $$ where $$ |\phi\rangle=\frac{1}{\sqrt{10}}(3|0\rangle+|1\rangle). $$

What's going on here is that the $\langle\psi|\rho|\psi\rangle$ has two terms contributing to it: $p|\psi\rangle\langle\psi|$ is the obvious one, but since $\langle\phi|\psi\rangle\neq 0$, the other term also contributes something.

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To add a bit more towards the actual proof, consider the strategy that if it's going to be a minimal ensemble, $$ \rho-p|\psi\rangle\langle\psi| $$ should have a rank that's 1 less than the rank of $\rho$. Let's assume we're working just with a subspace on which $\rho$ is full rank, dimension $d$. So, we require $$ \text{rk}(\rho(I-p\rho^{-1}|\psi\rangle\langle\psi|))=d-1 $$ Since the $\rho$ has full rank, this is the same as requiring $$ \text{rk}(I-p\rho^{-1}|\psi\rangle\langle\psi|)=d-1. $$ Clearly, the $I$ is full rank, and has trace $d$. If the whole thing has rank $d-1$, then the only way that can happen is of $$ \text{Tr}(p\rho^{-1}|\psi\rangle\langle\psi|)=1. $$ This is exactly your condition. Now, you can run this argument recursively to progressively select states to complete the decomposition. (It is easy to show that the density matrix remains a density matrix has you remove each term.)

Answer 2

The goal is, given $\rho$ and $|\psi_1\rangle\in\operatorname{supp}(\rho)$, to find a decomposition $$\rho=\sum_{k=1}^{\operatorname{rank}(\rho)} p_k |\psi_k\rangle\!\langle\psi_k|$$ for some set of $0<p_k\le1$ such that $\sum_k p_k=1$. Note that here $|\psi_k\rangle$ are not in general eigenvectors of $\rho$, nor are they orthogonal (we still assume them normalised though).

By construction, this means that if you remove the first term (or any other term, really), you're left with a singular matrix. That is, you must have $$\det(\rho-p_1 |\psi_1\rangle\!\langle\psi_1|) =0.$$ If $\rho$ itself is not singular (that is, if $\operatorname{supp}(\rho)$ is the full space), then you can use the matrix determinant lemma to write $$\det(\rho-p_1 |\psi_1\rangle\!\langle\psi_1|) = \det(\rho) (1- p_1 \langle\psi_1|\rho^{-1} |\psi_1\rangle) =0,$$ which in turn amounts to $p_1=\langle\psi_1|\rho^{-1}|\psi_1\rangle^{-1}$. Nothing really changes if $\rho$ isn't invertible: you can simply redefine the ambient space to be $\operatorname{supp}(\rho)$, and go back to the case of $\rho$ invertible. Practically speaking, amounts to observing you can block decompose the state as $\rho=\tilde\rho\oplus 0$ in some basis, with $\tilde\rho$ the (invertible) restriction of $\rho$ onto its support. Then follow the reasoning above on the block containing $\tilde\rho$.

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As a toy example, take $\rho=\begin{pmatrix}1/3&0\\0&2/3\end{pmatrix}$ and $|\psi_1\rangle=|+\rangle\equiv\frac{|0\rangle+|1\rangle}{\sqrt2}$. Then if $$p_1= \langle +|\rho^{-1}|+\rangle^{-1} = \left(\frac12(3+3/2)\right)^{-1}=\frac49,$$ you'll find that $$\rho - \frac49 |+\rangle\!\langle +| = \frac19\begin{pmatrix}1&-2\\-2&4\end{pmatrix} = \frac59\mathbb{P}\left(\frac{|0\rangle+2|1\rangle}{\sqrt5}\right), $$ where I used the shorthand notation $\mathbb{P}(|\psi\rangle)\equiv |\psi\rangle\!\langle\psi|$. In other words, the minimal decomposition containing $|+\rangle$ is $$\begin{pmatrix}1/3&0\\0&2/3\end{pmatrix} = \frac49 \mathbb{P}\left(\frac{|0\rangle+|1\rangle}{\sqrt2}\right) + \frac59 \mathbb{P}\left(\frac{|0\rangle+2|1\rangle}{\sqrt5}\right).$$