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I am trying to calculate the orbital magnetic moment, $\bar{\mu}$, for Sodium, which has an electron configuration of $1s^2 2s^2 2p^6 3s^1$. The full shells do not contribute to $\bar{L}$ and $\bar{S}$ so only the outer shell electron will contribute to them.

The $3s$ shell corresponds to $l=0$. And $\bar{L}=\hbar\sqrt{l(l+1)}$ which gives $\bar{L}=0$.

Is this possible? Doesn't that mean that the electron isn't "orbiting" the nucleus? I am thinking that $\bar{L}^2$ tells you if the electron is "orbiting" the nuclues, but then what is the significance of $\bar{L}$?

Extra question:

The z-component of the orbital magnetic moment is given by $\mu_z = -m_l\mu_b$. Since $m_l$ varies as $-l\le m_l\le l$ does that mean that there is more than one value for the z-component? Strange, no?

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    $\begingroup$ Electrons don't orbit the atom like planets orbiting a star. The electrons in atoms exist as a delocalised probability distribution. This distribution can have a zero or non-zero angular momentum, but the latter can't be simply interpreted as the electrons circling the nucleus. $\endgroup$
    – John Rennie
    Commented Feb 12, 2014 at 15:30
  • $\begingroup$ @JohnRennie You're right, I was being lazy. I added speech marks to make it a little bit more formal :) $\endgroup$
    – turnip
    Commented Feb 12, 2014 at 15:36
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    $\begingroup$ My point was that $L$ is just the angular momentum of the atomic orbital, and it can indeed be zero so the total orbital angular momentum of a sodium atom is zero. Why is it strange that $L_z$ can vary from $-L$ to $L$? $\endgroup$
    – John Rennie
    Commented Feb 12, 2014 at 15:41
  • $\begingroup$ Well I was more concerned with $\mu_z$. Since $m_l$ varies between $-l$ and $l$, I see why there can be a sign change but in cases where $l=2$ you can get four values (two for the positive "side" and two for the negative). What does that mean in physical terms for the electron? $\endgroup$
    – turnip
    Commented Feb 12, 2014 at 15:49
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    $\begingroup$ The magnetic dipole can take any of the $2l + 1$ angles (i.e. values of $l_z$) to the $z$ axis available to it. Indeed an NMR spectrometer works by measuring the energy of the transitions between different $l_z$ values. $\endgroup$
    – John Rennie
    Commented Feb 12, 2014 at 16:08

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Yes, the quantum angular momentum can be zero. But it seems you were trapped by your intuition from classical mechanics.

In classical mechanics an orbit with zero angular momentum means an orbit with a purely radial motion, i.e. kind of a degenerate ellipse with eccentricity $=1$. To show what I mean, here are elliptical orbits with eccentricity $0.99$ and $1$. (The nucleus is marked as a red dot)
enter image description here

But in quantum mechanics an orbital with $l=0$ (also called an $s$-orbital) is somewhat different. It is a wave function with spherical symmetry.
enter image description here
(image from LibreTexts Chemistry - 6.6: 3D Representation of orbitals)

To get a crude intuition, you may think of the $s$-orbital as a superposition of the classical degenerate ellipses from above, averaged over all possible radial directions.

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I read your question as: how can an electron orbital have zero angular momentum? For a classical orbit, zero AM means that the motion is strictly radial. If any motion can be associated with an s-orbital it should be strictly radial motion.

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Zero value to angular momentum doesn't mean that the electron is not revolving around the nucleus. It means that the motion of the electron is such that there is no preferred direction in space. For example, in a spherically symmetric orbital and that's exactly what s orbital is. Since s orbital is spherically symmetric there is no particular direction in space, which the angular momentum vector is pointing to and thus its average value is 0.

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    $\begingroup$ it's perfectly possible for a single electron to have $\ell=0$. In fact this is the case for the ground state of hydrogen. $\endgroup$
    – ZeroTheHero
    Commented Feb 27, 2022 at 19:30
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    $\begingroup$ This is not correct. The statement 'the angular momentum vectors cancel over time' has no meaning. $\endgroup$
    – my2cts
    Commented Feb 27, 2022 at 21:06
  • $\begingroup$ Quantum angular momentum cannot be discussed from the classical perspective that the electron is a localized particle. $\endgroup$
    – JGBM
    Commented Feb 28, 2022 at 0:33
  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – JGBM
    Commented Feb 28, 2022 at 0:34
  • $\begingroup$ Ok, I do see that I've treated electron as classical particle in my answer, but what "cancel over time" means is that the average value is zero. $\endgroup$
    – Shivam Singh Aswal
    Commented Feb 28, 2022 at 2:21

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