On existence of orthonormal basis for each subsystem in Separable state

Question

A separable state in $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$ is given by

$$\rho_{s}=\sum_{\alpha,\beta}p(\alpha,\beta)|\alpha\rangle\!\langle\alpha|\otimes|\beta\rangle\!\langle\beta|.$$

Now, my question is, is there a suitable choice of $\{|\alpha \rangle\}$ and $\{|\beta \rangle\}$ such that all of them are elements from a complete basis (possibly non-unique) in individual subsystem?

A reason I think the bases $\{|\alpha \rangle\}$ and $\{|\beta \rangle\}$ will form a complete basis is because separable state space is the convex hull of tesnor products of symmetric rank-$1$ projectors $|\alpha\rangle\!\langle \alpha|\otimes|\beta\rangle\!\langle \beta|$. The extreme points are orthonormal sets $\{|\alpha\rangle\!\langle \alpha|\}$ and $\{|\beta \rangle\!\langle \beta|\}$. Is it true? Any help is appreciated.

Answer 1

In general, no. It can be that the sets of states $\{|\alpha\rangle\}$ and $\{|\beta\rangle\}$ span their respective spaces. However, the size of each set can be larger than the dimension of the space, so the states are not all linearly independent and therefore not a basis.

For example, $\rho^{a/b}_1=p_1|0\rangle\langle 0|+(1-p_1)|1\rangle\langle 1|$ and $\rho^{a/b}_2=p_2|+\rangle\langle +|+(1-p_2)|-\rangle\langle -|$. So, $\{|\alpha\rangle\}=\{|0\rangle,|1\rangle,|+\rangle,|-\rangle\}$ and is not a basis.

Answer 2

As a trivial counterexample, let $\mathcal H$ be a Hilbert space with $\dim(\mathcal H)=299792458$, and let $\{|i\rangle\}_{i=1}^{299792458}$ be a basis for it. Then $\rho_s\equiv|1\rangle\!\langle1|\otimes|1\rangle\!\langle1|$ is separable and is already decomposed like you mention, but clearly $|1\rangle$ does not span the full space.