2
$\begingroup$

In ordinary probability theory the relation: $$E[XY]=E[X]E[Y]$$ holds when $X$ and $Y$ are uncorrelated (or independent). The analogous relation in quantum mechanics is: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\mean{\hat A \hat B}=\mean{\hat A} \mean{\hat B}$$ my question is when in general does this hold?

$\endgroup$
3
$\begingroup$

$$\langle \hat{A} \hat{B} \rangle = \langle \Psi | \hat{A} \hat{B} | \Psi \rangle \\ \langle \hat{A}\rangle \langle \hat{B} \rangle = \langle \Psi | \hat{A} | \Psi \rangle \langle \Psi | \hat{B} | \Psi \rangle $$ That means for the equation to hold, you need $$ \hat{A} | \Psi \rangle \langle \Psi |\hat{B} = \hat{A} \hat{B} $$

In general the equation holds if $|\Psi \rangle$ is an eigenstate of both of the operators.

$\endgroup$
  • 1
    $\begingroup$ It is sufficient. I edited my post. $\endgroup$ – Quantumwhisp Dec 9 '16 at 19:33

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.