5

The different probability cloud shapes of the Hydrogen atom (consisting of one proton and one electron) can be solved using the spherical wave function. The answer is often summarized as follows:

hydrogen wave functions

When plotted with 3D graphics, they look something like:

hydrogen probability cloud

My question is about the direct verification of these "strange" (so to speak) shapes. I have searched and found almost no experimental results that directly verify all these shapes, for example by taking many separate measurements and tabulating the results into a 3D cloud. There is an experiment that purportedly show the (1,0,0) and (2,0,0) and (3,0,0) shape of the probability cloud, but what about the exotic shapes like (4,1,1) and (4,2,0)?

If we show these pictures to a 6 year-old, the first thing they will probably say is: "I don't believe it. Proof it and show me!" Great claims require great experimental proofs. The greater the claim, the greater the burden of proof, right? :)

EDIT: i did read the related post Is there experimental verification of the s, p, d, f orbital shapes? and follow the links given in the answers. It is one experiment where it looks like the top-most (1,0,0) and (2,0,0) and (3,0,0) shapes are reproduced. Unfortunately, only the spherical ones are reproduced, and arguably it would easier to approximately reproduce a general "round" spherical shape compared to a very strange characteristic one like, for example (4,2,0), which if it can be reproduced very accurately is a good omen that the Schrodinger function produces this "shape", as it is not easy to produce this exact specific probability cloud with some other functions...

  • 3
    Wavefunctions are not physical observables – KP99 Oct 01 '21 at 13:06
  • 2
    Many properties of these orbital shapes are inseparable from highly verified properties like energy level splittings. – jwimberley Oct 01 '21 at 14:13
  • 2
    If you are looking to prove it to a 6 year old, your options are limited. Chemical bonds also follow the Schrodinger equation, but with multiple nuclei. They too have properties predicted by the Schrodinger equation. For example the triple bond between carbon atoms is radially symmetric around the axis between the atoms. From this, you can predict that groups bonded to the pair should be able to swivel. The triple bond does permit swiveling. – mmesser314 Oct 01 '21 at 14:31
  • i realize the indirect consequences do exist, yes... although i mean, when a lot of steps are involved in proving something indirectly; say suppose this is true, then that is also true, and then some other things related to this is also true, and finally the macroscopic properties are true... it is easier to slip up somewhere in the logic steps. The direct proof, for example by tabulating the results in 3D after many separate measurements of positions, if available, would have been the strongest, undeniable proof, no? –  Oct 01 '21 at 14:52
  • 1
    related https://physics.stackexchange.com/q/200143/ – Wihtedeka Oct 01 '21 at 14:54
  • 3
    Does this answer your question? Is there experimental verification of the s, p, d, f orbital shapes? – Ruslan Oct 01 '21 at 16:42
  • @Ruslan, thank you, i did read that post follow the links given in the answers. It is one experiment where it looks like the top-most (1,0,0) and (2,0,0) and (3,0,0) shapes are reproduced. Unfortunately, only the spherical ones are reproduced, and arguably it would easier to approximately reproduce a general "round" spherical shape compared to a very characteristic one like, for example (4,2,0), which if reproduced very accurately is a good sign that the Schrodinger function produces this "shape", as it is not easy to produce the exact specific shape with some other functions... –  Oct 01 '21 at 17:01
  • 1
    See the PhD thesis of A.S. Stodolna — the third link in the accepted answer — particularly, section 4.3. There're quite a lot of different states showed. – Ruslan Oct 01 '21 at 17:05
  • @Ruslan thank you, i have downloaded and read the paper now. Honestly I don't quite understand the experiment technique used, I see that they analyzed a lot of (6,,) orbital shapes. perhaps if any experts reading this can weigh in and help analyze it? Thank you. –  Oct 01 '21 at 17:16
  • 1
    Actually, the hydrogen atoms themselves are discussed in chapter 5. The states imaged were with $n\sim30$, not $6$. There you might be able to also see the limitations of the technique. – Ruslan Oct 01 '21 at 22:59

1 Answers1

0

I would tell the six year old that these angular shapes are not weird at all. They are the functional forms of irreducible representations of SO(3)...at least in the angular department. The fact that:

$$ \sum_{m=-l}^l|\psi_{nlm}(\rho,\theta, \phi)|^2 \propto \frac {f(\rho)} {4\pi} $$

i.e., full shells are spherically symmetric is to be expected. Moreover, should he complain the $z$-axis is arbitrary, I'd remind him these a irreducible representations. They are closed under rotations:

$$\psi_{nl'm'}(\rho',\theta', \phi') = \sum_{m=-l}^lc^{lm}_{l'm'}\psi_{nlm}(\rho,\theta, \phi)$$

Regarding the radial wave function, I would hope he expects more nodes as the energy increases.

My only concern would be if this precocious little sixth grader noticed that the the energies aren't just degenerate in $m$...which is mandatory based on the conservation of angular momentum, but also in $l$, (and similarly, planetary orbital energies are independent of eccentricity). This means there is a hidden symmetry, the Newton/Coulomb potential is separable in parabolic coordinates, and atom don't care about our coordinates. They have no obligation to be in the standard $\psi_{nlm}$ orbitals, and could be orbitals labeled by different quantum numbers (sometimes, $n_1,n_2,m$)...now those are "strange".

JEB
  • 33,420
  • yes, the mathematical properties follows from the spherical wave function after the standard derivation steps taken to get the equations in FIgure 1. But, does this actually correspond to reality out there? –  Oct 01 '21 at 17:53
  • 1
    @James Approximately. Orbitals under Schrödinger's Eq. are stationary states, they never change nor transition, which is contradicted by experiment. Including interaction with the EM field makes them approximations that are not strictly stationary. Meanwhile the $Z^0$ introduces parity violating states/transitions. Relativity/Dirac equation, finite nuclear size, all add (hyper)fine structure. But these states allow use to compute high precision energy transitions that are verified in the lab. – JEB Oct 01 '21 at 18:01