Von Neumann entropy in terms of the mutual overlap? I have $N$ pure, but nonorthogonal, states $|\psi_n\rangle$ with density matrix $\rho_n=|\psi_n\rangle\langle\psi_n|$.
Say we call the the total density matrix $\rho=\frac{1}{N}\sum_n \rho_n$. 
Are there any formulas to calculate $S_{vn}[\rho]=-\mathrm{Tr}[\rho\log{\rho}]$ solely from the overlaps $\langle\psi_n|\psi_m\rangle$ (and the fact that all individual states are pure)? Gram-Schmidt orthogonalization might be a possibility but perhaps there is an easier way/existing result?
Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?
 A: Let us denote the overlap matrix by $O$, this is, $O_{nm}=\langle\psi_n\vert\psi_m\rangle$.  Then, 
$$
S_{\mathrm{vN}}(\rho)=S_\mathrm{vN}(O)\ .
$$
More generally, $O$ has the same non-zero eigenvalues as $\rho$, so any function of the eigenvalues (which is insensitive to zero eigenvalues) can be evaluated on $O$ instead of $\rho$.
This can be seen by defining a matrix $X$ whose columns are the states $\vert\psi_n\rangle$, i.e. $X_{kn}=\langle k\vert\psi_n\rangle$.  Then, $\rho=XX^\dagger$, while $O=X^\dagger X$.  However, two matrices $AB$ and $BA$ always have the same non-zero eigenvalues, and thus, the non-zero eigenvalues of $\rho$ and $O$ are the same.
A: In such situations it is common to calculate the linear entropy instead of the full VN entropy. The linear entropy comes from the first term approximation in the expansion of $\log(\rho)$ as Taylor series around $\rho= I$, which is $I-\rho$ (I might be wrong about the sign here). Then the first term approximation to the VN entropy is 
$$Tr[\rho\log\rho]\longmapsto Tr[\rho (I-\rho)]=1-Tr[ \rho^2]$$
where $Tr[ \rho^2]$ is commonly referred to as purity. The purity is 1 for pure states and $\frac{1}{d}$ for completely mixed states  (here $d$ is the dimension of the Hilbert space). 
Back to your problem. The purity of $\rho$ is 
$$Tr[\rho^2]=Tr[(\frac{1}{N}\sum_n \rho_n)^2]=\frac{1}{N^2}\sum_{n,m}Tr[\rho_n\rho_m] $$
which, as you have anticipated, depends on the overlaps. 
A: Since part of the questions is

Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?

let me provide an explicit counterexample:
Let
$$\newcommand{\ket}[1]{\vert#1\rangle}\ket{\psi_1}=\frac{\ket0+\ket1}{\sqrt{2}}, \ket{\psi_2}=\frac{\ket1+\ket2}{\sqrt{2}}, \ket{\psi_3}=\frac{\ket0+\ket2}{\sqrt{2}}\ .
$$
The mixing entropy of $
\newcommand{\kb}[1]{\vert#1\rangle\langle#1\vert}
\rho=\frac13(\kb{\psi_1}+\kb{\psi_2}+\kb{\psi_3})$ is then 
$S(\rho)=\log(3)-\tfrac13\log(2)\approx0.87$.
On the other hand, if we change $\ket{\psi_3}$ to $$\ket{\psi_3}=\frac{\ket0-\ket2}{\sqrt{2}}\ ,$$
we obtain $S(\rho)=\log(2)\approx0.69$.
