Can a correlation matrix be regarded as a quantum density matrix? Consider a set of random variables $\mathbf{X}=\{x_1(t),\ldots, x_n(t) \}$. We can compute the Pearson correlation matrix of these variables with elements
$C_{ij} = \dfrac{\mathbb{E}[(x_i-\mu_{x_i})(x_j-\mu_{x_j})])}{\sigma_{x_i}\sigma_{x_j}}$.
The correlation matrix $C$ is then positive semi-definite and symmetric with 1 on the diagonal. If I divide each element of the correlation matrix by the number of random variables $n$ I have a matrix that has unitary trace, positive eigenvalues and is symmetric.
Can this be regarded as a quantum density matrix? I see that quantum density matrices have exactly the same properties.
 A: Yes indeed, there are at least some instances where correlation matrices are the same as density matrices. Let me demonstrate this in the opposite direction.
The density operator for a general mixed state can be expressed as
$$ \hat{\rho} = \sum_n |\psi_n\rangle P_n \langle\psi_n| , $$
where $P_n$ are probabilities, such that $\sum_n P_n = 1$. Already from this expression one can interpret this as an expectation value: $\hat{\rho}={\cal E}\{|\psi_n\rangle \langle\psi_n|\}$.
Let's make this more explicit. We consider a particular complete orthogonal basis $|p\rangle$, such that
$$ \sum_p |p\rangle \langle p| = 1 $$
Then we can expand the states $|\psi_n\rangle$ in terms of this basis
$$ |\psi_n\rangle= \sum_p |p\rangle \psi_{np} , $$
where $ \psi_{np} = \langle p|\psi_n\rangle $. Note also that since $\langle\psi_n|\psi_n\rangle=1$, it follows that
$$ \sum_p |\psi_{np}|^2 =1 $$
If we now extract the density matrix elements from the density operator we get
$$ \rho_{pq} = \langle p| \hat{\rho} |q\rangle = \sum_n \psi_{np} \psi_{nq}^* P_n = {\cal E}\{ \psi_{p} \psi_{q}^* \} . $$
Since $P_n$ is a probability, one can express this as an expectation value of the product of the two vectors. Alternatively one can see this as the ensemble average, where $n$ is an index for the ensemble elements. So then $\rho_{pq}$ represents the correlation matrix.
Just to check that this gives a valid density matrix, let's compute the trace
$$ \sum_p \rho_{pp} = \sum_{np} |\psi_{np}|^2 P_n = \sum_n P_n = 1. $$
So, yes, it works.
