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I was going through the following book "Alice and Bob meet Banach: The interface of Asymptotic Geometric Analysis and Quantum information theory" by Aubrun and Szarek. I found a problem on basic quantum theory (pure and mixed states) that looked deceptively simple but hard to prove. Before I state the problem, I will give a few definitions.

Consider $\mathbb{C}^d$, where $d \ge 1$ is the dimension. A vector $|v\rangle \in \mathbb{C}^d$ is called a pure state if $\langle v|v\rangle = 1$. Usually, a pure state is written in an operator form or as a density matrix as $|v\rangle\langle v|$. Now given all pure states, we take their convex hull to get a space of operators $\rho$, known as mixed states, of the form $\rho = \sum_{i=1}^N p_i|v_i\rangle\langle v_i| $, where the vectors $|v_i\rangle$ need not be orthonormal, $\sum_{i=1}^{N} p_i = 1$ for $p_i \ge 0$ and some $N$. Denote this space by $D(\mathbb{C}^d)$.

Note that according to the spectral theorem, for every state $\rho$, there exists an orthonormal basis $|e_i\rangle$ (called eigenvectors of $\rho$) and non-negative coefficients $\alpha_i$ (called eigenvalues of $\rho$) with $\sum_i \alpha_i = 1$ such that $\rho = \sum_{i=1}^d \alpha_i |e_i\rangle\langle e_i|$. I am also aware of a result that says pure states are extreme points of $D(\mathbb{C}^d)$.

The problem is as follows:

For $\rho \in D(\mathbb{C}^d)$, show that there exist $d$ pure states $|v_i\rangle$ not necessarily orthonormal such that $$\rho = \frac{1}{d}\sum_{i=1}^d |v_i\rangle \langle v_i|.$$

My Analysis: I observed that it is trivial for $d=1$. For $d=2$, suppose the state $\rho$ can be expanded as $\rho = p|e_1\rangle\langle e_1| + (1-p)|e_2\rangle\langle e_2|$ via spectral theorem, then setting $|v_1\rangle = |\sqrt{p}e_1 + \sqrt{1-p}e_2\rangle$ and $|v_2\rangle = |-\sqrt{p}e_1 + \sqrt{1-p}e_2\rangle$ gives us the desired form.

I figured a proof by induction would be in order. Indeed the book suggests using the following induction step. Use intermediate value theorem to show that $\rho - \frac{1}{d}|w\rangle\langle w| \in \partial PSD(\mathbb{C}^d)$ for some pure state $|w\rangle$. Here $\partial(.)$ means the boundary of the space. $PSD$ refers to the space of positive semidefinite matrices. A previous exercise was to show that the boundary of $D(\mathbb{C}^d)$ consisted of all those states which possessed at least one eigenvalue that was zero.

I got hopelessly stuck while trying to use the hint as well as without it. I'm not sure about which continuous function to use intermediate value theorem. I would be grateful if someone could assist me in this. I did not find a similar question on StackExchange.

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2 Answers 2

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So, if I understand correctly, you want to prove that any positive semidefinite matrix with trace $1$ is a simple average of $d$ endomorphisms of the form $\Pi_v: x \longmapsto (x \cdot v)v$ with $v$ a unit vectors.

In an orthonormal basis, a matrix represents some $\Pi_v$ iff it is hermitian, has rank and trace $1$.

In other words, you want to show that any non-negative diagonal matrix $M$ with trace $1$ is an average of $d$ hermitian matrices with rank and trace $1$.

Write $\alpha_i=M_{i,i}$. Then consider the matrices $N_0,\ldots,N_{d-1}$ where $(N_k)_{i,j}=\sqrt{\alpha_i\alpha_j}e^{2ik(j-i)\pi/d}$.

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  • $\begingroup$ Thanks for the wonderful approach. This almost looks like a Fourier transform. I will try this approach and if it works, I'll post my workout. $\endgroup$ Apr 8, 2020 at 13:07
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Thanks to Mindlack (the accepted answer) for the hint.

Essentially, we choose \begin{equation} |v_k\rangle = \sum_{l=1}^d \sqrt{\alpha_l}\exp\left(i\frac{2\pi kl}{d}\right)|e_l\rangle \end{equation} where $i = \sqrt{-1}$ above. I verified this construction and it works. I used $$\sum_{k=1}^d \exp\left(i\frac{2\pi k(l-j)}{d}\right) = d~\delta_{lj}$$ to simplify a step and show the equivalence.

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