We know that in the case of $O$ being an operator, $\langle O^2\rangle-\langle O\rangle^2$ equals to uncertainty as long as $\langle\rangle$ means the mean value (expectation value). if we have $A$ and $B$ as two operators what does this parameter mean? what physics is hidden on below difference?

$$ \langle AB\rangle-\langle A\rangle\langle B\rangle $$


That's the covariance of X and Y.
It tells you, in a way, "how much" X and Y are correlated, with covariance 0 meaning uncorrelated (Not to be confused with independent).
Because $\langle A\rangle$ and $\langle B\rangle$ can be quite large, it is customary to define the correlation coefficient: ($cov(A,B)$ is the covariance of A and B)
Which is then bounded between -1 and 1, and tells you how correlated they are. Either positively correlated (Positive values far from 0), not correlated (Values close to 0), or negatively correlated (Negative values far from 0).

I've yet to see any physicist use this, at least in theoretical calculations.
It can be useful in data analysis, trying to found out if two parameters are correlated, but other than that, I can't think of other uses in physics.

| cite | improve this answer | |
  • 2
    $\begingroup$ The product $AB$ doesn't yield an observable in general, so I wonder if one can really interpret the OP's formula as a covariance. Perhaps one can consider the Jordan product $A\circ B=\frac12(AB+BA)$ instead and set $\operatorname{cov}(A,B) = \langle A\circ B\rangle - \langle A\rangle\langle B\rangle$? $\endgroup$ – Phoenix87 Nov 28 '15 at 11:19
  • $\begingroup$ @Phoenix87 It's true that it's not always an observable, but that just means that IF it's defined properly, it's covariance. $\endgroup$ – Omry Nov 28 '15 at 11:53
  • $\begingroup$ @Omry, besides so much thanks of your answer, but I should say I have not seen the word "Covariance" in quantum literature. Although I have seen in other categories but I can't understand what is the properties of a "Covariance" operator in Quantum realm? is there any references about that? $\endgroup$ – Inzo Babaria Nov 28 '15 at 13:13
  • $\begingroup$ @InzoBabaria As I've mentioned, I've yet to see any theoretical usage of this. This is mostly a statistical concept, which is therefore used more in experiments. $\endgroup$ – Omry Nov 29 '15 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.