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.

  • $\begingroup$ It's not exactly clear to me what you mean by "regarding" this matrix as a density matrix. Yes, it has exactly the same properties as a density matrix, but the space it lives on is not a Hilbert space of states and I don't think it's eigenvalues represent probabilites to be in a particular state. $\endgroup$
    – ACuriousMind
    Sep 28, 2016 at 12:59

1 Answer 1


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.

  • $\begingroup$ I don't understand though if it is like a correlation or a covariance matrix, as a correlation matrix has 1 on the diagonal and since the trace of a density matrix is 1 this is not possible, right? What normalization must be done exactly? $\endgroup$
    – linello
    Dec 16, 2016 at 14:31
  • $\begingroup$ The correlation matrix has $\sum_n|\psi_{np}|^2 P_n$ on the diagonal. So it sums to one when one computes the trace. The covariance matrix requires one to subtract the means, and I don't see where that would come from. So I'd say it is the correlation matrix and not the covariance matrix. $\endgroup$ Aug 11, 2017 at 7:02
  • $\begingroup$ What I want to say, re: the connection between correlation $C$ and the density matrix $\rho$, is that $C_{pq} = \dfrac{\rho_{pq}}{\sqrt{\rho_{pp}\rho_{qq}}}$ in which case $C_{pp} = 1$ identically. This amounts, I think, to normalizing by setting the variance of each observable to one. $\endgroup$ Mar 30, 2018 at 16:15

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