In statistics, given a probability distribution, the variance of a quantity $X$ is obtained by averaging $(X – \langle X\rangle)^2$ over the distribution. Here $\langle X\rangle$ is the average value of $X$. We can also write the variance as $\langle X^2\rangle – \langle X\rangle^2$. It can easily be verified that the two definitions are equivalent.
In quantum mechanics, given the wave function, we may define the variance of an operator $O$. One choice would be to evaluate the expectation value of $(O – A)^2$. Here $A$ is the expectation value of $O$. The other choice would be to take $O^2 – A^2$.
Unfortunately these two choices are not equivalent in QM. That is because the operator $O$ will (in general) not only act on the wave function, but also on $A$ since it need not be a constant but can be a function of space or momentum coordinates.
My question is: what is the preferred definition of the variance in QM ?
Best Answer
The two choices are equivalent in QM as well. Take a wavefunction $\psi$. Take an operator $Q$ with expectation value $\langle Q \rangle$ in state $\psi$.
$$(Q-\langle Q \rangle)^2=(Q-\langle Q \rangle)(Q-\langle Q \rangle)=Q^2-2\langle Q \rangle Q+\langle Q \rangle^2$$
Thus,
$$\langle \psi|(Q-\langle Q \rangle)^2|\psi \rangle=\langle\psi|Q^2|\psi\rangle-2\langle Q\rangle \langle \psi|Q|\psi \rangle+\langle Q \rangle^2$$
since $\langle Q \rangle$ is a constant. The above expression, as you can see is simply: $\langle Q^2 \rangle-\langle Q \rangle^2$
Where is the discrepancy? $\langle Q \rangle$ is a fixed number that does not depend on anything but time, in which case the operator Q will still not care about it since most quantum operators are time independent.