6
$\begingroup$

I have read multiple answers on StackExchange about this question, but I wasn't able to find a concrete answer. Like other questions, the reason I ask about the $s$-orbital is because it has a zero orbital angular momentum. But, the implication of having a zero angular momentum is unclear? Some answers discuss probability distributions, but the question is how can there be multiple places for an electron, if it can't move. I read another answer that says that the electron passes through the nucleus or curve around it as a wave. I would appreciate it if anyone could provide a resolution to this question.

$\endgroup$
  • $\begingroup$ "Move" means time evolution of the state, which is given by $| \psi(t) \rangle = e^{-iHt} | \psi(0) \rangle = e^{-iE_s t} | \psi(0) \rangle $. $\endgroup$ – user26143 May 31 '14 at 9:03
  • $\begingroup$ Electrons can exist at more than one place at the same time. $\endgroup$ – Schrödinger's Cat May 31 '14 at 9:28
  • $\begingroup$ +1 to the answer by anna v. The word "particle" means something entirely different inside a physics classroom than outside. (Same is true for "work" and "heat" and "strangeness", etc). You'll have to put aside your everyday mental images. $\endgroup$ – garyp May 31 '14 at 11:36
  • 5
    $\begingroup$ It doesn't move around in that orbital, it exists in that orbital. When you probe it repeatedly, it can be found in different places without "moving" between them but by still remaining in the exact same spatial distribution. $\endgroup$ – Peteris May 31 '14 at 14:49
11
$\begingroup$

For me the physical implication of zero angular momentum is that the electron's probability distribution is spherically symmetric. At the deepest level, the angular momentum property in quantum mechanics describes how something transforms under rotations (see Noether's theorem). Although this is quite an abstract property, in the case of electron orbitals it relates to something extremely concrete and amenable to visualisation: namely, the shape of the angular probability distribution.

Any interpretation based on classical concepts is doomed to fail at some point. In this case the confusion arises from imagining that the electron is in a certain "place", or that it "moves". The only truly useful description of the electron that has been found is the quantum wave function, and its corresponding probability distribution.

$\endgroup$
  • 1
    $\begingroup$ How does zero angular momentum mean spherically symmetric? I suspect that the answer can be proven mathematically, but is their a physical explanation for it. $\endgroup$ – user29568 May 31 '14 at 9:04
  • 2
    $\begingroup$ @user29568 The deepest understanding available of this point is via Noether's theorem, linked in my post, which nevertheless requires some mathematics. If you want a physical analogy, note that even classically an object with non-zero angular momentum has an axis of rotation, which defines a preferred direction in space. Saying that an object has no angular momentum is equivalent to saying that no such axis exists, therefore all directions are equivalent and we have spherical symmetry. In general, zero angular momentum is equivalent to the property of perfect rotational invariance. $\endgroup$ – Mark Mitchison May 31 '14 at 9:27
  • $\begingroup$ I think I got what your saying, so since the electron has no angular momentum then it has no favored direction, which means that it can basically go in all directions which is defined by a spherical area. But, what makes the electrons defined in the s-orbital so special that they have no angular momentum. Is it their by default? $\endgroup$ – user29568 May 31 '14 at 13:23
  • 1
    $\begingroup$ The special s-orbital state is simply one of the infinitely many solutions that exist for the Schroedinger equation of an atom. The reason it can have precisely zero angular momentum is related to the quantum discreteness of other quantities like energy for bound states. The best way of understanding this is by analogy with standing waves in classical physics, which can only possess certain quantised values of frequency and wavelength. Ultimately this is related to the properties of wave-like equations when the boundary conditions require the solution to be confined to a finite region. $\endgroup$ – Mark Mitchison May 31 '14 at 17:18
  • 2
    $\begingroup$ @user29568, if anything, mathematics improves certainty, as it is hard to make testable predictions about mere intuitions. Once you can manipulate, measure and make predictions about those measurements, you can perform science. The challenge is that as we delve ever deeper, the things we are modeling are no longer the things we evolved to intuitively comprehend. We viscerally know what it's like to throw a rock and predict where it will fall, but understanding probability distributions takes a bit more work. $\endgroup$ – Dan Bryant May 31 '14 at 23:16
6
$\begingroup$

Electrons are elementary particles. Elementary particles are quantum mechanical entities, and their interactions are described very well by he solutions of quantum mechanical differential equations. Orbitals are the mathematical description of these solutions.

Elementary particles are not particles in the sense of a billiard ball, with a specific (x,y,z) trajectory. They do not move in solutions of classical mechanics problems. That is why they are called orbitals and not orbits. They have a locus in space described by a probability function which is equal to the square of the wave function of the solution.

hydrogenorbital

They have measured hydrogen orbitals with an ingenious microscope , so the electrons are not in orbit, but in a distribution in space

Atomic distances, as in the image, are of the order of 10^-10meters, the proton in the picture is localized at diameters of 10^-15 meters . As this involves volumes, the S state probability distribution does go through the locus of where the proton is , with a very small probability of interaction. At that point one should solve the problem with second quantization, no longer the simple Schroedinger one . There exist studies of this.

The case of the positronium, which has similar solutions to the hydrogen ones, and lives for a while and then annihilates is an experimental proof of the S state passing through the center of the potential.

Also electron capture from the S state is observed in proton rich nuclei.

$\endgroup$
  • $\begingroup$ I know that an electron has no specific trajectory, but aren't the orbitals, a function that allow us to calculate the probability of finding an electron in a certain region of space. My question is how does the electron have multiple possible places to be, if it can't move. Or is it just that since its impossible to predict exactly where an electron would be, then we can't predict how it evolves with time. $\endgroup$ – user29568 May 31 '14 at 13:08
  • 1
    $\begingroup$ It is quantum mechanics. If there is a time dependence in the probability distribution, a time dependence will appear. Otherwise it will just be the draw of the specific measurement ( as the ones in the picture above). Similar to a standing wave on a string, it is in the whole string. $\endgroup$ – anna v May 31 '14 at 13:10
  • $\begingroup$ I don't think I follow. As a particle, does the electron have a velocity? $\endgroup$ – user29568 May 31 '14 at 13:13
  • $\begingroup$ If yes, doesn't that mean it is in motion. So, are the electrons described in the s-orbital in motion or are they still? $\endgroup$ – user29568 May 31 '14 at 13:20
  • 1
    $\begingroup$ It depends on the experiment. Electrons in an accelerator have a velocity. Electrons in a cathode ray tube have a velocity. This is because they are not bound and are only probabilistically limited by the Heisenberg uncertainty principle. Bound electrons are completely probabilistically limited by the boundary conditions of the solution of the shrodinger equation. The wavefunction is not in an eigenstate of the velocity operator . It is in an eigenstate of energy. estimating velocity by using the 1/2mv**2 but it does not mean much : an average number to compare with the unphysical Bohr model $\endgroup$ – anna v May 31 '14 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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