# All states are simple averages of pure states

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.

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 $$(N_k)_{i,j}=\sqrt{\alpha_i\alpha_j}e^{2ik(j-i)\pi/d}$$.

• 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. Apr 8, 2020 at 13:07

Thanks to Mindlack (the accepted answer) for the hint.

Essentially, we choose $$$$|v_k\rangle = \sum_{l=1}^d \sqrt{\alpha_l}\exp\left(i\frac{2\pi kl}{d}\right)|e_l\rangle$$$$ where $$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.