Orbital angular momentum of an electron in s-orbital

Question

I'm a beginner in quantum mechanics, and I'm pretty much confused with the orbital angular momentum of electron in s-orbital of hydrogen atom. I've read that average angular momentum of an electron in $1s$ orbital of $\ce{H}$-atom is $0 \pu{kgm^2s^-1}$, but when I took the wave function of $1s$ orbital of $\ce{H}$-atom and solving by the following way is giving me instantaneous angular momentum is $0\pu{kgm^2s^-1}$.

My way of solving:

The momentum operator is $\hat p = -i\hbar\nabla$ and the $\psi_{1s}= k \times e^{\frac{-r}{a_0}}$, where $k$ is normalizing constant

Applying momentum operator for $1s$ orbital,

$$\hat p \psi_{1s} = p \psi_{1s}$$ I'm getting some thing like $\hat p \psi_{1s} = -(x \hat i+ y\hat j+z\hat k)k^{'}e^\frac{-r}{a_0}$, which implies that the momentum $\overrightarrow p$ of electron in $1s$ orbital is $$\overrightarrow p = k^{''}(x \hat i + y \hat j + z \hat k)$$

So, angular momentum can now be given by $$\overrightarrow L = \overrightarrow r \times \overrightarrow p$$ The position vector $\overrightarrow r = x\hat i + y\hat j + z\hat k$

As the position vector, $\overrightarrow r$ and the momentum vector, $\overrightarrow p$ are in anti-parallel directors at every point in space, I have concluded that $\overrightarrow L = \overrightarrow 0$ at every instant, which contradicts to what I've read.

(NOTE: The source from where I've read is not very much reliable, but I didn't find any thing related to this any where else)

Answer 1

Electron orbitals in atoms are eigenstates of both energy and total orbital angular momentum. These are therefore constant in any orbital unless the orbital is changed, as it would be for instance when forming a bond to another atom. So whatever "average" you have for these variables applies at every instant as well.