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

[daudaudaudau]

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?

 

[daudaudaudau]

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 />
...

 

[saaskis]

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".

 

[daudaudaudau]

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 />