I've been trying to solve exercise 2.73 (p.g 105), 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


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

$p_i = \frac{1}{\sum_{k} \frac{c_{i,k}^2}{\lambda_k}}$

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...


I think the answer of the first part can be corrected, as the density matrix for $\psi_i$ needs to be normalised, and therefore to normalise it

By the definition of $\rho$ is can be seen that to normalise the trace it is required that $p_i = \frac{1}{\sum_{k}\frac{c_{ik}^2}{\lambda_k}}$.

Hence $\rho_i= p_i|\psi_i\rangle \langle\psi_i| = p_i\sum_{k}\frac{c_{ik}^2}{\lambda_k}|\hat{k}\rangle \langle\hat{k}| = \sum_{k}|\hat{k}\rangle\langle\hat{k}|$. Which is our original definition of $\rho$.


This formulation of Ex 2.73 looks senseless to me (see the update of this answer).

In my edition of N&C the statement of Ex 2.73 is different. You are given density operator $\rho$ and linearly independent states $\{|\psi_i\rangle\}$ that span the support of $\rho$, and you have to prove that there are unique numbers $p_i$ such that $\rho = \sum p_i |\psi_i\rangle\langle \psi_i |$ (and $\sum p_i=1$ but this follows trivially from calculating the trace). Those numbers $p_i$ can be computed by $$p_i=\frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}.$$

Such formulation is sensible, but it's just wrong.

Let $\rho= \frac{1}{3}\big(|0\rangle\langle 0| + 2|1\rangle\langle 1|\big)$ and $|\psi_1\rangle=|+\rangle$, $|\psi_2\rangle=|-\rangle$. It's quite clear that there are no $p_1, p_2$ such that $\rho = p_1|+\rangle\langle +| + p_2|-\rangle\langle -|$, because $|+\rangle$ and $|-\rangle$ are orthogonal, so it has to be a spectral decomposition of $\rho$, but spectral decomposition is unique if eigenvalues are different.

The only correct formulation I can imagine is the following.

Let $\rho$ be a density operator and linearly independent states $\{|\psi_i\rangle\}$ span the support of $\rho$. Suppose we are given that $\rho = \sum p_i |\psi_i\rangle\langle \psi_i |$ for some $p_i>0$. Prove that $\sum p_i=1$ and $$p_i=\frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}.$$

And the proof is quite simple. Multiply that equation for $\rho$ by $\rho^{-1}|\psi_j\rangle$ from the right. We obtain $$ \rho\big(\rho^{-1}|\psi_j\rangle\big) = \sum_i p_i |\psi_i\rangle\langle \psi_i |\big(\rho^{-1}|\psi_j\rangle\big) $$ that is $$ |\psi_j\rangle = \sum_i \big(p_i \langle \psi_i| \rho^{-1}|\psi_j\rangle\big) \cdot |\psi_i\rangle $$

But $\{|\psi_i\rangle\}$ are linearly independent, so it must be $p_j \langle \psi_j| \rho^{-1}|\psi_j\rangle = 1$ and $p_i \langle \psi_i| \rho^{-1}|\psi_j\rangle = 0$ for $i\neq j$.


Another fact that can be proved is the following.

Let $\rho$ be a density operator which has support of dim $m$. Let $|\psi_0\rangle$ be some state from this support. Then there are states $|\psi_1\rangle, .., |\psi_{m-1}\rangle$ from this support such that $\rho = \sum p_i |\psi_i\rangle\langle \psi_i |$, $p_i>0$, $\sum_i p_i = 0$. And by the previous fact we can deduce that $p_i = \frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}$.

I guess it's what the editors actually meant.

The proof is also not hard. Consider $\rho_\epsilon = \rho - \epsilon |\psi_0\rangle\langle \psi_0 |$ for some small $\epsilon>0$. If $\epsilon$ is small enough then $\rho_\epsilon$ will be strictly positive on the support of $\rho$. But if we will raise $\epsilon$ then at some moment $\epsilon = \epsilon^\prime$ it will be $\rho_{\epsilon^\prime} \geq 0$ but not $\rho_{\epsilon^\prime} > 0$. This implies that $\rho_{\epsilon^\prime}$ has the support of dim $m-1$ and we can take $|\psi_1\rangle,..,|\psi_{m-1}\rangle$ as the corresponding eigenvectors of $\rho_{\epsilon^\prime}$.

  • 1
    $\begingroup$ I guess it seems even they didn't really know what they were looking for if the question is changing over editions! $\endgroup$ – Sam Palmer Apr 17 '20 at 22:30
  • $\begingroup$ @SamPalmer I think I've figured this out, see the update $\endgroup$ – Danylo Y Apr 19 '20 at 6:19

Define $p_i = \dfrac{1}{\sum_k \dfrac{|c_{ik}|^2}{\lambda_k} }$ and $q_{ik} = \dfrac{\sqrt{p_i}c_{ik}}{\sqrt{\lambda_k}}$ then

$$ \sum_k |q_{ik}|^2 = p_i \sum_k \dfrac{|c_{ik}|^2}{\lambda_k} = 1 $$

And also you have that

$$ \langle \psi_i| \rho^{-1}|\psi_i\rangle = \sum_k \dfrac{|c_{ik}|^2}{\lambda_k} $$

Note that I added $|c_{ik}|^2 $ to be mathematically accurate.

  • $\begingroup$ The issue I see is that my answer for part 2 doesn't match what i derived for $p_i$ in part 1, so one of them is wrong, it's missing the 1/ in part 1 $\endgroup$ – Sam Palmer Apr 17 '20 at 21:22
  • $\begingroup$ I've seen this answer before in the unofficial solution, idoc.pub/documents/…, however I feel like this answer was derived from working backwards, as it shows no workings of why we define $p_i$ as we do, $\endgroup$ – Sam Palmer Apr 17 '20 at 22:23

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