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I was reading that besides measuring the angle of ricocheted electrons bouncing off the proton to pin down its size, it is also possible to excite the electron and then measure the frequency of the light emitted by the excited electron. Why would the gap between ground state and excited state tell us the size of a proton? Is there something I have missed?

Qmechanic
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user6760
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  • On a related note, measurements of the proton size in muonic hydrogen (i.e., a hydrogen atom where the electron is replaced by a muon) yield a slightly smaller size than in normal hydrogen. See https://journals.aps.org/prd/abstract/10.1103/PhysRevD.98.013002 – PM 2Ring Dec 21 '19 at 11:52
  • @PM2Ring: I read the muon orbits much closer than electron and they can measure the proton charge radius better but muon is too short lived despite being heavy a bad choice no? – user6760 Dec 21 '19 at 12:29
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    True, it is a problem that the muon has a short mean lifetime (around 2.2 microseconds), but we don't have much choice about that. Most subatomic particles have short lifetimes, so particle physicists are used to doing experiments that operate on short timescales. A couple of microseconds is actually quite long when you're used to working with processes that happen in nanoseconds and shorter. ;) – PM 2Ring Dec 21 '19 at 12:43
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    @PM 2Ring, I think that discrepancy was resolved just this past year. See https://science.sciencemag.org/content/365/6457/1007 – KF Gauss Dec 21 '19 at 16:26
  • Do you really mean size, or do you mean mass? You can get the mass because the energy levels depend on the reduced mass. –  Dec 22 '19 at 03:14
  • @BenCrowell: I'm interested in knowing the radius of the proton accurately so I'm still trying to understand the relationship between proton charge radius and energy gap. – user6760 Dec 22 '19 at 03:21
  • @user6760 It’s a typical problem in perturbation theory. Most QM textbooks will have something on this, under the name of “volume effect of nucleus” or something like this. – ZeroTheHero Dec 22 '19 at 04:22

2 Answers2

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This is an interesting and non-trivial problem. Basically the Coulomb potential assumes a point particle but, if the proton is modelled as a solid sphere of finite radius, part of the electron wave function would be "inside" the proton, where the assumption of point charge no longer holds.

To account for this one must modify the Coulomb potential from $1/r$ outside the proton to (basically) $C r^2$ inside, where $C$ is some constant. The simplest model is to think of the proton as a uniformly charged sphere (constant volume charge density) so the $Cr^2$ term comes from Gauss's law for the potential inside this type of sphere.

This small perturbation in the potential will affect slightly the energy level. Since for small distances the radial probability density generally goes like $r^{2(\ell+1)}$, the smaller values of $\ell$ will produce wave functions with larger probabilities of having the electron "inside" the proton, so experiments were done measuring the energy difference between $2S_{1/2}$ and $2P_{1/2}$ which have $\ell=0$ and $\ell=1$ respectively. These states would normally have the same energy under the pure Coulomb potential since both are $n=2$ states, but they are affected differently under the assumption that the proton has a non-zero volume.

The story of the "proton problem" goes back 10 years or so, when a group in Geneva made extremely accurate measurements of the size of the nucleus. Basically, they deduced what value of the radius of the proton (assumed as a uniform spherical charge distribution) was needed to reproduce their experimental measurements of energy levels, and it didn't agree with the accepted value. There's a good synopsis of this

The proton -- smaller than thought: Scientists measure charge radius of hydrogen nucleus and stumble across physics mysteries https://www.sciencedaily.com/releases/2010/07/100712103339.htm

(They used muonic hydrogen since the Bohr radius of this system is smaller than the usual electron-proton system, thus enhancing the portion of the wavefunction inside the nucleus.)

The unexpected result was only confirmed this year. A summary of new results can be found here and the actual paper of the experiment

Bezginov, N., Valdez, T., Horbatsch, M., Marsman, A., Vutha, A.C. and Hessels, E.A., 2019. A measurement of the atomic hydrogen Lamb shift and the proton charge radius. Science, 365(6457), pp.1007-1012

appears to be available online from this link provided courtesy of GoogleScholar.

Note there are other perturbations in hydrogen - the fine and hyperfine structure - which have to be accounted for as well, making this volume effect non-trivial to isolate.

I love this stuff. It shows that the hydrogen atom is not completely archeological but there's still some interesting surprises to be found in this canonical example of undergraduate level quantum mechanics

ZeroTheHero
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  • But why is charged sphere a good model of proton charge distribution? Why not uniformly-charged ball, for example? – Ruslan Dec 21 '19 at 16:40
  • @Ruslan not sure I follow. The uniformly charged sphere is the simplest assumption. – ZeroTheHero Dec 21 '19 at 16:47
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    It might be simplest to analyze, but it seems somewhat unphysical (at least to me, who doesn't know much about proton structure) that all (or most of) the charge would reside on some particular distance from the center. – Ruslan Dec 21 '19 at 16:53
  • @Ruslan uniformly charged sphere means constant charge density. Sorry if there’s confusion on this. not the surface of the sphere but through volume. – ZeroTheHero Dec 21 '19 at 16:54
  • So you meant uniformly charged ball, not sphere, right? I suppose this should be edited into the answer. – Ruslan Dec 21 '19 at 16:54
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    @Ruslan yes Zero means a spherical volume with uniform 3D charge density, and not a spherical shell or a solid conducting ball where the charge would migrate to the surface. – uhoh Dec 21 '19 at 17:01
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    @ZeroTheHero +1 fantastic summary of the problem, thanks! btw is my comment above correct? – uhoh Dec 21 '19 at 17:05
  • @Ruslan constant volume charge density. The charge is uniformly distributed in the volume. – ZeroTheHero Dec 21 '19 at 17:17
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    @uhoh yes you are correct. – ZeroTheHero Dec 21 '19 at 17:18
  • Ruslan's point is the sphere is the boundary of the ball. – Eric Towers Dec 22 '19 at 23:59
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    @EricTowers right. That would be a spherical shell, not a sphere. – ZeroTheHero Dec 23 '19 at 00:42
  • I think we're seeing a terminological discrepancy between physics and mathematics here. Mathematicians use "ball" to mean the solid volume and "sphere" to mean the boundary of a ball. – Andreas Blass Dec 23 '19 at 01:26
  • @AndreasBlass you could well be right. I did edit to “solid sphere” etc after the comments and hopefully this is enough. Certainly in my corner of the physics world “ball” is not used. We speak of shells (v.g. the shell model) for a hollow sphere, and sphere otherwise. In Canada we refer to egg shells, not egg spheres ; – ZeroTheHero Dec 23 '19 at 02:11
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Usually when determining the energies of the hydrogen atom we assume the the proton is a point charge. By changing that to a finite charge distribution the potential is altered for small electron-proton distances. The energies are sensitive to the extent of the distribution. By comparing simulations to very accurate measurements of excitation energies information is found on the proton's charge distribution.

my2cts
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