Is the Product of Expectation Values Always True in Quantum Mechanics?

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In quantum mechanics, when is this true
<br /> \langle\psi|AB|\psi\rangle=\langle\psi | A|\psi\rangle\langle\psi |B|\psi\rangle<br />
? In probability theory, when the two variables are independent, the mean value of the product is the product of the mean values. What about QM?
 
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It is true if |\psi\rangle is a normalized eigenstate of both A and B because then
<br /> \langle \psi|AB| \psi\rangle=a_\psi b_\psi<br />
...
 
The equation
<br /> \langle \psi|AB|\psi\rangle = \langle \psi|A|\psi\rangle \langle \psi|B|\psi\rangle \ \forall \psi<br />
also leads to [A,B]=0. But assuming that [A,B]=0 and that both are usual Hermitian observables does not seem to imply the above equation for general states, even though any state can be expanded as a linear combination of common eigenstates: this would require
<br /> \sum_i a_ib_i|c_i|^2 = \sum_{i,j} a_i b_j |c_i|^2 |c_j|^2<br />
for
<br /> |\psi\rangle = \sum_i c_i |\psi_i \rangle.<br />
So I don't have an answer to your question but I wrote anyway :) But
<br /> \langle AB \rangle = \langle A \rangle \langle B \rangle<br />
of course means that in the particular configuration, the operators are uncorrelated and there is e.g. no Heisenberg uncertainty in measuring both observables "simultaneously".
 
Yeah, it is not enough for the operators to commute, because A commutes with A, but
<br /> \langle\psi|A^2|\psi\rangle\neq\langle\psi | A|\psi\rangle\langle\psi |A|\psi\rangle<br />
 
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