# Nielsen and Chuang ex 2.73

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

$$p_i=\frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}$$

where $$p^{-1}$$ is defined to be the inverse of $$\rho$$, when $$\rho$$ is considered as an operator acting only on the support of $$\rho$$

$$\rho$$ 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...

---Update---

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 $$i\neq j$$.

Update

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 $$\rho_{\epsilon^\prime}$$.

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

• 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 – Sam Palmer Apr 17 at 21:22
• 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, – Sam Palmer Apr 17 at 22:23