About density matrices as both the vertices of a convex hull and the basis of a matrix space

Question

Given $n$ density matrices $D_1, \dots, D_n$, that is, $D_i$ is positive semi-definite and $\operatorname{Tr}(D_i)=1$ for all $1\leq i\leq n$. Suppose that $D_1, \dots, D_n$ are linearly independent.

Denote by $\mathcal{S}:=\operatorname{span}\{D_1, \dots, D_n\}$ the linear space spanned by the given $n$ density matrices. Denote by $\mathcal{E}$ the set of all density matrices in $\mathcal{S}$.

Question: is $\mathcal{E}$ equal to the convex hull formed by the vertex set $\{D_1, \dots, D_n\}$? In other words, take an arbitrary element $A\in\mathcal{E}$, do we always have that $A=\sum_{i=1}^{n}\alpha_i D_i$ where $\alpha_i$'s are all non-negative and $\sum_{i=1}^{n}\alpha_i=1$?

Answer 1

No. The extreme points of the set of density matrices are the pure states of the form $|\psi\rangle\langle \psi|$.

If you have any collection of density matrices, $S=\{D_1, \dots, D_n\}$, any pure state $|\psi\rangle\langle \psi|$ not in $S$ will not be in the convex hull $\mathcal{E}=\operatorname{conv}(S)$.

Edit: I realized after more carefully reading your question that I didn't directly answer it. To address your question a bit better, let $\operatorname{D}_n$ denote the set of $n\times n$ density matrices in $\operatorname{M}_n$, the space of $n\times n$ matrices, which has dimension $n^2$ as a vector space. You can choose $n^2$ linearly independent density matrices $$S=\{D_1, \dots, D_{n^2}\}\subset \mathrm{D}_n$$ that span the whole space, i.e. $\mathrm{span}(S)=\operatorname{M}_n$. Again, any pure state $|\psi\rangle\langle \psi|$ not in $S$ will not be in the convex hull $\mathcal{E}=\operatorname{conv}(S)$, although it must be in the span of $S$.

Answer 2

A simple counter example is the following (which is just the one you gave in the comments i realize now): Let the base space have dimension $n=2$. Let $$D_1 = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} $$ and $$D_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$ Clearly they are density matrices and linearly independent. Let $$X = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ Clearly $X$ is a density matrix. Since $X = 2 D_1 -D_2$ it is also contained in $\mathrm{span} \{D_1,D_2 \}$. But there does not exist a $\lambda \in [0,1]$ so that $X = \lambda D_1 +(1-\lambda ) D_2$. Because to obtain the one at the index $(0,0)$ of $X$ we need that $\lambda = 2$. Which is clearly not allowed.