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 ?


2 Answers 2


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 $\psi$.

$$(Q-\langle Q \rangle)^2=(Q-\langle Q \rangle)(Q-\langle Q \rangle)=Q^2-2\langle Q \rangle Q+\langle Q \rangle^2$$


$$\langle \psi|(Q-\langle Q \rangle)^2|\psi \rangle=\langle\psi|Q^2|\psi\rangle-2\langle Q\rangle \langle \psi|Q|\psi \rangle+\langle Q \rangle^2$$

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.

  • $\begingroup$ In case the physical quantity is the kinetic energy its operator acts as the Laplacian on the wave function. The resulting average kinetic energy will be position dependent. Thus the operator acting on the average value is non-zero. $\endgroup$
    – M. Wind
    Jul 19, 2021 at 18:51
  • $\begingroup$ The average of an operator is defined as: $\int d^3r\ \psi^*(\vec r,t)\ \hat Q (\psi(\vec r,t))$ which is only time dependent since space coordinates are dummy variables. $\endgroup$
    – Physiker
    Jul 19, 2021 at 18:52

Hint: use carets like $\hat X$ to denote operators.

Then $x:=\langle \hat X\rangle$ is a number since average values are numbers, not operators. Using this:
\begin{align} \langle (\hat X-x)^2\rangle =\langle \hat X^2 - 2x\hat X + x^2\rangle &= \langle \hat X^2\rangle - 2x \langle \hat X\rangle + x^2 \, , \tag{1}\\ &= \langle \hat X^2\rangle - x^2 = \langle \hat X^2\rangle - \langle \hat X\rangle ^2 \end{align} where $\langle \hat X\rangle=x$ and $\langle x\rangle=x\, , \langle x^2\rangle = x^2$ (since $x$ is a number) have been used.

Thus the two definitions are identical.


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