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)

  • $\begingroup$ So you've read that it should be $0$ and also found it to be $0$. What's the problem? $\endgroup$ Sep 19, 2021 at 7:43
  • $\begingroup$ @IvanNeretin I've read average angular momentum is zero but I got instantaneous also zero as well. Is it correct? $\endgroup$
    – Adithya
    Sep 19, 2021 at 7:45
  • $\begingroup$ Does it make the average different from $0$? No. So, there is no contradiction. $\endgroup$ Sep 19, 2021 at 7:47
  • $\begingroup$ @IvanNeretin It doesn't make any thing different, but is the instantaneous value also $0$? $\endgroup$
    – Adithya
    Sep 19, 2021 at 7:48
  • $\begingroup$ Yes.$\mathstrut$ $\endgroup$ Sep 19, 2021 at 7:49

1 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.


Your Answer

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

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