# If angular momentum of an electron in $s$-orbital is zero, then is it true that the electron doesn't move?

If the electron does move, then what is the factor that leads angular momentum of the electron in s-orbital to be zero?

• An electron doesn't have a strictly well-defined position, so what do you mean by "move"? It sounds like you might be trying to ask a classical question about a quantum system. – Chiral Anomaly May 10 at 20:25
• Nothing "moves" in quantum mechanics because the notion of motion is ruled out by quantum mechanics. While an electron would never generically be in an eigenstate of its position operator, you can still measure its position. But, there would be absolutely no continuity in its positions as measured over infinitesimal intervals of time (positions measured over infinitesimal intervals of time can be finitely separated). So, among many other things, quantum mechanics also rules out "motion" as a physical concept. – Dvij Mankad May 11 at 5:30
• Then entropy is coming out to be the product of Boltzmann constant and the natural logarithm of the microstates of the electron, isn't it? – Coder Hashtag May 12 at 8:15
• Thanks for answering my question – Coder Hashtag May 12 at 8:15

Even for electrons with zero orbital angular momentum, having quantum number $$l=0$$, their average kinetic energy is still 1/2 of their ionisation energy.

Remember that an electron in rest would be a direct violation of the Heisenberg uncertainty principle.

In fact, s orbitals are of all high orbitals the most affected by relativistic effects, spending significant portion of time near the kernel at speed comparable to c.

E.g. 6s electrons of heavy atoms like gold or mercury must be treated by relativistic computations in relativistic quantum chemistry.

Involving relativistic effects on mercury 6s orbitals lead to decreased prediction error of mercury melting point from the value being 140 K off to just about 1 K off.

Zero orbital momentum means radial motion, smeared by Heisenberg uncertainty.

Orbitals s have spherical symmetry of probability of electron occurrence.

$$\psi_{n\ell m}(r,\vartheta,\varphi) =\\ \sqrt {{\left ( \frac{2}{n a^*_0} \right )}^3 \frac{(n-\ell-1)!}{2n(n+\ell)!}} e^{- \rho / 2} \rho^{\ell} L_{n-\ell-1}^{2\ell+1}(\rho) Y_{\ell}^{m}(\vartheta, \varphi )$$

where $$\rho = {2r \over {n a^*_0}}$$

$$a^*_0$$ is the reduced Bohr radius $$a^*_0={{4\pi\epsilon_0\hbar^2}\over{\mu e^2}}$$

$$L_{n-\ell-1}^{2\ell+1}(\rho)$$

is a generalized Laguerre polynomial of degree $${'n'' − ''ℓ'' − 1}$$ and $$Y_{\ell}^{m}(\vartheta, \varphi )$$

is a spherical harmonic function of degree ''ℓ'' and order ''m''.

• Please give me some plausible mathematical expressions – Coder Hashtag May 10 at 19:38
• There is nothing inplausible about these expressions – lux May 11 at 7:18
• @lux Expressions and reference in the answer came after the OP comment, so his comment was quite legitimate. :-) – Poutnik May 11 at 8:12
• I see - thank you for clarification – lux May 11 at 15:51

No.

In classical orbital mechanics, you can have an orbit with zero orbital angular momentum ($$L = 0$$), provided your source is a point source and your mass is a point mass.

The shape of an orbit is determined by two things: the orbital energy ($$E$$) and the orbital angular momentum ($$L$$). Higher energy orbits go farther out, while higher angular momentum orbits are more circular. If you hold the energy constant, and reduce the angular momentum, the orbit becomes more and more elliptical or eccentric, and in the limit of very tiny angular momenta, it effectively becomes a straight line (profoundly eccentric ellipse, i.e. $$e \approx \approx 1$$) and moves in and out.

The same thing is the case in quantum orbital mechanics, except that what we have is essentially an extremely "lossy" version of a classical orbit with very little information present defining it, as we are near the information limit. In particular, all that is available is that the electron is somewhere along the orbit, and that it is moving, but which direction and where, and moreover, which direction the orbit is oriented, are virtually if not completely undefined (maximal or near-maximal Shannon entropies on all these parameters). Thus, there is a sense of inward/outward motion, but only a very hazy sense, or subtle feeling, of it, that atmospheres the situation, is pretty much the best way to describe it, I'd think(*). Not only that, but the semi-major axis of the orbit is also only somewhat defined - that is why the orbital never goes to zero even very far from the atom.

(Now this is only an approximate description even here: in fact, due to quantum interference effects, the orbits cannot simply be considered as just a "lossy" version of a classical orbit: e.g. orbitals above $$\mathrm{1s}$$ have nodes in them, which make no sense classically because a classical particle never skips points in its orbit.)

It is important to point out that stasis of the positional probability distribution does not mean the absence of motion: "motion" is more generally defined by the denotation of certain parameters like kinetic energy and velocity, which in only some circumstances will manifest into anything mathematically moving in terms of positional or geometric parameters. Keep in mind that the mathematics is just a model of reality, and not "reality itself", so one should not reify it too hard.

In fact, it is this case in classical mechanics as well: a idealized perfect sphere can be in rotation but mathematically, nothing changes either because a sphere is fixed by rotational transformations about an axis. What "rotation" means here is that it can still possess positive angular momentum, and will set into rotation anything that touches it which does visibly rotate (or transfer AM to another sphere). Likewise with the description of the flow of an ideal fluid: nothing mathematically moves, yet you have a velocity field present throughout which will carry anything inserted into it.

Nonetheless, not only is the motion real, however "fuzzy" (Heisenberg's term) it may be, but it is real enough that it can still even be subject to relativistic effects in very heavy atoms (high charge on the nucleus) leading to orbital contraction, which is what gives Gold its beautiful color.

(Courtesy: "solarisgirl", https://www.flickr.com/photos/sanmitrakale/34953644670. cc-by-sa 2.0.)

(*) Mathematically, I believe you can actually make this idea a bit more precise by writing the wave function of the $$\mbox{1s}$$ orbital as a sum of ingoing and outgoing waves, but I haven't worked it out.