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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 $\sum_{i=1}^{n}\alpha_i=1$?

Thinkpad
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  • $(1+\varepsilon)D_1-\varepsilon D_2$ will be a density matrix for sufficiently small but positive $\varepsilon$. – Michal Adamaszek Jun 16 '23 at 08:42
  • Is this a counterexample? $$2\begin{bmatrix} 0.5 & 0 & 0\ 0 & 0.5 & 0\ 0& 0 & 0 \end{bmatrix}-\begin{bmatrix} 0 & 0 & 0\ 0 & 1 & 0\ 0& 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\ 0 & 0 & 0\ 0& 0 & 0 \end{bmatrix}$$ – Thinkpad Jun 16 '23 at 09:31
  • Does the underlying state space have finite dimension $n$? Because if not then this is clearly false by the answer of Luftbahnfahrer. – jd27 Jun 16 '23 at 11:27
  • @jd27 Yeah, I should have mentioned that the underlying state space have finite dimension, maybe say $N\geq n$. – Thinkpad Jun 16 '23 at 12:08
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    Your question basically splits into 2 steps: (i) suppose the $D_i$ are of the form $\mathbf e_i\mathbf e_i^T$ [standard basis vectors].... can I argue that convex combinations generate all the density matrices in their span? The answer is yes, though it contradicts the comment from @MichalAdamaszek . (ii) What if I choose as "vertices" some matrices in the proper convex hull of (i) -- can I argue that the convex hulls are the same? And it should be geometrically obvious that this is not true... you would need some not convex affine combinations here. – user8675309 Jun 16 '23 at 16:11
  • @user8675309 This viewpoint is really helpful! Now I have a better understanding. – Thinkpad Jun 16 '23 at 23:45

2 Answers2

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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 $S$.

Luftbahnfahrer
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  • You can not choose $n^2$ linearly independent density matrices, since they have to be self-adjoint. – jd27 Jun 16 '23 at 18:14
  • @jd27 Luftbahnfahrer's answer should be true, see this link – Thinkpad Jun 16 '23 at 23:43
  • @Thinkpad interesting thanks. I think this only works if the base field is $\mathbb{C}$ though, because then the unitary matrices have the same dimension as $M_n$, but in the real case the symmetric ones only have dimension $1/2n(n-1)$. But i guess for density matrices only the case $\mathbb{C}$ is of real interest anyways. – jd27 Jun 17 '23 at 04:17
  • @jd27 Thanks. This is a good catch anyway, I totally agree with you. – Thinkpad Jun 17 '23 at 10:49
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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.

jd27
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