I've seen the uncertainty of $A$ written as $$ (\Delta A)^2 = \langle (A - \langle A \rangle)^2 \rangle. $$ But what does this even mean since $ A $ is an operator and $ \langle A \rangle $ is a scalar. How can we compute the difference of the two?


2 Answers 2


For any vector $v$ and scalar $\alpha$, we have $\alpha v = \alpha I v$, so multiplication by a scalar $\alpha$ behaves the same way as the linear operator $\alpha I$. Therefore, we interpret $A - \langle A \rangle$ as $A - \langle A \rangle I$.


Note that

$$\big(Z - \langle Z \rangle \big)^2 = Z^2 - 2Z \langle Z \rangle + \langle Z \rangle^2 $$

And therefore $$ \langle \big(Z - \langle Z \rangle \big)^2 \rangle = \langle Z^2 - 2Z \langle Z \rangle + \langle Z \rangle^2 \rangle $$

Then by linearity, we have

\begin{align} \langle Z^2 - 2Z \langle Z \rangle + \langle Z \rangle^2 \rangle &= \langle Z^2 \rangle - \langle 2Z \langle Z \rangle \rangle + \langle \langle Z \rangle^2 \rangle \\ &= \langle Z^2 \rangle - (2\langle Z \rangle) \langle Z \rangle + \langle Z \rangle^2 \\ &= \langle Z^2 \rangle - 2 \langle Z \rangle^2 + \langle Z\rangle^2 \\ &= \langle Z^2 \rangle - \langle Z \rangle^2 \end{align}

So therefore,

$$ (\Delta Z)^2 = \big(Z - \langle Z \rangle \big)^2 = \langle Z^2 \rangle - \langle Z \rangle^2 $$

Which is what you see if you search up Variance


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.