# Product of expectation values

1. Oct 31, 2010

### daudaudaudau

In quantum mechanics, when is this true
$$\langle\psi|AB|\psi\rangle=\langle\psi | A|\psi\rangle\langle\psi |B|\psi\rangle$$
? 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?

2. Oct 31, 2010

### daudaudaudau

It is true if $|\psi\rangle$ is a normalized eigenstate of both A and B because then
$$\langle \psi|AB| \psi\rangle=a_\psi b_\psi$$
...

3. Oct 31, 2010

The equation
$$\langle \psi|AB|\psi\rangle = \langle \psi|A|\psi\rangle \langle \psi|B|\psi\rangle \ \forall \psi$$
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
$$\sum_i a_ib_i|c_i|^2 = \sum_{i,j} a_i b_j |c_i|^2 |c_j|^2$$
for
$$|\psi\rangle = \sum_i c_i |\psi_i \rangle.$$
So I don't have an answer to your question but I wrote anyway :) But
$$\langle AB \rangle = \langle A \rangle \langle B \rangle$$
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".

4. Oct 31, 2010

### daudaudaudau

Yeah, it is not enough for the operators to commute, because A commutes with A, but
$$\langle\psi|A^2|\psi\rangle\neq\langle\psi | A|\psi\rangle\langle\psi |A|\psi\rangle$$