# In calculations with uncertainty principle why could you equate the uncertainty in momentum with the actual momentum of the system

This website is trying to calculate the confinement energy of a electron starting from the uncertainty principle, but it does this: $\Delta p=p$. Why is this valid?

The electron is stuck to the atom, which isn't going anywhere. This means that $\langle p \rangle = 0$. The uncertainty $\Delta p$ measures the RMS fluctuation of the momentum:

$$\Delta p^2 = \langle p^2 - \langle p \rangle^2 \rangle = \langle p^2 \rangle$$

Since $E = p^2 / 2m$, this means that

$$\langle E \rangle = \frac{\Delta p^2}{2m}$$

The article you linked to is being fast and loose in going back and forth between operators and their expectation values. It's correct to say that

$$\Delta p = \sqrt{\langle p^2 \rangle}$$

but this is not equal to $\langle |p| \rangle$ in general:

$$\langle p^2 \rangle \neq \langle |p| \rangle^2$$

The article gets away with it since it is only after a calculation of $\langle E \rangle$, which involves $\langle p \rangle^2$ and not $\langle |p| \rangle$.