One can read in Wikipedia that quantum correlations are

the expectation value of the product of the outcomes on the two sides

which indicates that $QC(a,b)=\langle( \vec\sigma \cdot \vec a)^{(1)}(\vec\sigma \cdot \vec b)^{(2)} \rangle$ or in classical mechanics represented by hidden variables $\int d\lambda \rho(\lambda)A(a,\lambda)B(b,\lambda)$.

However, in statistics, correlation is strictly defined as $$corr(X,Y)=\frac{cov(X,Y)}{\sigma_X\sigma_Y}= \frac{E(X-\mu_X)E(Y-\mu_Y)}{\sigma_X\sigma_Y} $$

My question:

Is there any relation between the quantum correlation and correlation in statistics? Or are they independent? Sometimes physics definition use the name in math or statistics but refer to different objects which make people confused.


1 Answer 1


Correlation is a general term that can be applied to many situations where two or more quantities are not independent on each other.

Even in statistics correlation means more than the Pearson correlation coefficient quoted in the OP, which is only a measure of a specific type of correlation - linear correlation between two variables. There exist many other correlation coefficients, such as Spearman's rank correlation coefficient, Kendall's tau and others.

Still, they remain correlation coefficients - mathematical metrics used to measure correlation, but not the phenomenon itself. Perhaps, the most general way of defining correlation between two variables $A$ and $B$ in statistics is by saying that that their joint probability distribution does not decompose into a product of individual probability distributions: $$P(A,B)\neq P(A)P(B)$$ or, equivalently, $$P(A|B)\neq P(A),P(B|A)\neq P(B).$$

Quantum mechanics
If we now take a quantum system of two or more particles, described by a joint wave function, then it may well be that this wave function cannot be decomposed into a product of the wave functions of individual particles. The wave function of two non-interacting but identical particles, $$ \psi_{a,b}(x_1,x_2)=\frac{1}{\sqrt{2}}\left[\psi_a(x_1)\psi_b(x_2)\pm \psi_a(x_2)\psi_b(x_1)\right]$$ is the simplest example, but many other examples could be given. Thus, such a correlation is very much a correlation in statistical sense. (Note that correlation here is nearly synonymous with entanglement, although I would not claim that the two are the same.)

Quantum statistical phsyics
The reason why we talk about quantum correlations is to distinguish them from the correlations and statistical phenomena that have nothing to do with quantum nature of particles. For example, a quantum gas exhibits randomness that is both due to the quantum nature of its constituents, as well as due to the uncertainty in particle positions or velocities that is already present in classical statistical phsyics.

Phase transitions (in what is aptly called strongly correlated systems) may be due to classical, as well as quantum interactions - e.g., coupling between spins in XY-model can be simple dipolar coupling or exchange mediated.

  • $\begingroup$ The proof of CHSH uses the renaming of dummy indexes $\int A(a,l)B(b,l)dl+\int A(a,k)B(b',k)dk=\int A(a,k)(B(b,k)+B(b',k))dk$ which doesnt care about the change of variable $l=f(k)$ and induces a link. Namely one could maybe go the reverse way to obtain a classical statistical violation ? $\endgroup$
    – Cretin2
    Apr 19 at 14:07
  • $\begingroup$ I mean like : $S=-A+C+AB+AC+2BC-ABC$. When considering possible local values $\{-3,-1,3,5\}$, versus the average considering the classical covariance $cov(a,b)=-1+\frac{2}{\pi}\arccos(\cos(b-a))$ gives $-4$. Qm violates both values but I'm not sure if it attains the algebraic minimum of $-7$ $\endgroup$
    – Cretin2
    Apr 24 at 16:59

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