# Extracting electron wave functions from experiments

In nuclear and nucleon physics it’s quite standard to extract electromagnetic form factors – which are the Fourier transforms of charge and current distributions – from elastic electron-nucleon or photon-nucleon scattering.

First question: Is it possible to do something similar for the electromagnetic form factors of the electron shell in atoms?

That means to calculate something like "electron-hydrogen scattering minus electron-proton scattering".

Second question: Is it possible to do that for specific excited states, e.g. to "measure" the 2pz wave function in the hydrogen atom?

That includes the preparation of atoms in this excited state. It may include a similar "subtraction", i.e. "contribution excited state minus ground state".

I like this question, as former SLAC DISer (and TJNAF, MIT-Bates, DESY), where elastic-scattering was used to calibrate the instrument, I've thought about it (long ago).

Also, if you want to tackle quark structure functions, it really helps to compare and contrast with the QM view of the atom. There are similarities (I think of the Lamb Shift as "Sea $$\gamma, e+e^-$$", for example), and differences (quarks don't have wave functions, yet).

Now I thought the answer was "No", because atoms are neutral. We measure the proton elastic form factor with liquid hydrogen, and then to get neutron data, we use liquid deuterium and subtract. For magnetization, you can use polarized $$^3$$He, which to a good approximation, is singlet protons and a polarized neutron. When I looked this up, I noticed I was an author on the paper--it was that long ago that I forgot.

So how are you going to deal with a neutral atom?

Your suggesting of subtracting proton scattering is misguided: at atomic energies (say $$13.8\,$$eV), the proton form factors match the static charge and anomalous magnetic moments--there is no $$Q^2$$ dependence over the relevant range for atomic physics.

Also, the proton is not stationary in the middle--that is just one of our QM assumptions because we solve the Schrödinger Equation in the reduced mass coordinates (see: my 2nd paragraph--we forget all the things we do to make the Coloumb potential analytically solvable). In the lab, the protons are in there little tiny $$Y_l^m$$ orbitals with Laguerre radial dependence, too--and the size of the orbital is:

$$a_0 \times \frac{m_e}{m_p} \approx 25\,{\rm fm}\gg r_p \approx 0.8\,{\rm fm}$$

So you're just probing a neutral atom with enough $$Q^2$$ to resolve a Bohr radius, so:

$$|Q^2|^{\frac 1 2} \approx \frac{\hbar c}{a_0} = \frac{197\,{\rm MeV \cdot fm}}{50000\,{\rm fm}} \approx 4\,{\rm keV}$$

and that is much greater than the ionization energy for hydrogen (it is not a coincidence that it is roughly $$2R_{\infty}/\alpha$$).

Even before that, there is the experimental difficulty of getting a low enough energy electron beam that can still traverse aluminum windows and vacuum windows into your detector.

But I was wrong. People do it, and it is described here: https://en.wikipedia.org/wiki/Atomic_form_factor .

Here's a nice X-ray scattering result from Chlorine, described in the link:

Of course X-rays don't lose energy through multiple scattering, so that helps experimentally.

• Thanks. I have some background in DIS, so I know how to deal with proton and deuteron. And I agree, of course one needs to probe the atom at low energies. Anyway, I do not insist on elastic scattering, all I am interested in is to get some kind of form factor, vertex function or something like that.
– TomS
Commented Jul 6 at 19:51
• @TomS I thought it was elastic. The thing with DIS is, QCD factoring and asymptotic freedom apply: the quarks appear as free point particles in the proton--I don't think that applies to EM, esp. at high $Z$. Walter Greiner did QED in high $Z$ stuff, though I don't recall a reference.
– JEB
Commented Jul 7 at 5:16
• You mentioned DIS, and my response was just that I am quite familiar with it. For my question I see no relation to DIS. I just wanted to mention that I don’t insist on elastic scattering(off the electron shell) but that inelastic scattering would be fine; the intention is to extract information re the electron wave functions, only. Inelastic scattering off the electron shell does not automatically mean DIS in the QCD sense. And probing the electron shell structure means that we stay in a kinematic regime well below any inelastic channel for the scattering off the nucleus.
– TomS
Commented Jul 7 at 6:43

Is it possible to do something similar for the electromagnetic form factors of the electron shell in atoms?

That means to calculate something like "electron-hydrogen scattering minus electron-proton scattering".

Yes. It's also called a "form factor."

The differential electron scattering cross-section can be written as: $$\frac{d\sigma}{d\omega d\Omega} = S(\vec q,\omega)\frac{4\pi}{q^2}\;,$$ where $$\frac{4\pi}{q^2}$$ part is basically from the Coulomb potential of the ion core and the $$S(\vec q, \omega)$$ accounts for everything else. It can be measured in transmission electron microscope scattering experiments.

Second question: Is it possible to do that for specific excited states, e.g. to "measure" the 2pz wave function in the hydrogen atom?

In general, the electronic scattering form factor measures the Fourier transform of the average of the density-density operator. The average is typically (see below for further discussion) performed with respect to the full ground state of the $$N$$-electron system: $$S_0(t, q) = \langle \Psi_0|\hat n_{\vec q}(t)\hat n_{\vec q}(0)|\Psi_0\rangle\;,\tag{1}$$ where $$\hat n_{\vec q} = \sum_{i=1}^N e^{-i\vec q\cdot \hat{\vec r_i}}\;.$$

With respect to single-partial orbitals in Hydrogen: It is not possible to directly measure specific states like the "2pz" orbital. For example, it is only the absolute square of the orbital that determines the probability density, so you can't measure phase information. Further, the single-particle "orbitals" are not actually eigenstates of the Hamiltonian for any interacting system, so such a measurement would only be applicable to atomic hydrogen. That being said, if you do happen to have some way to prepare the initial state of your sample of atomic Hydrogen in the 2pz state, then that is the state that would appear in the average of Eq. (1) rather than the ground state.

It is generally difficult to consistently prepare the initial state in any state other than the ground state. Or, more precisely, in any other ensemble than the canonical ensemble at temperature $$T$$. If you work in the canonical ensemble, you have: $$S_T(t,q) = \frac{1}{Z}Tr(e^{-\beta \hat H}\hat n_{\vec q}(t)\hat n_{\vec q}(0))\;,$$ which will have contributions from higher energy states. But the contributions from higher energy states are small when the temperature is small compared to electronic excitation energy, the scale of which is typically electron-volts (1 eV is about twelve thousand degrees Kelvin).

Alternatively, if you do happen to have some consistent way to create an initial state $$|\Psi_m\rangle$$, then, yes, you will measure $$S_m(t,q)= \langle \Psi_m|\hat n_{\vec q}(t)\hat n_{\vec q}(0)|\Psi_m\rangle\;,$$ rather than the $$S_0$$ of Eq. (1).

Also, in theory, you can project out parts of the form factor in an angular momentum basis and thereby obtain an angular-momentum resolved density of states. These components of the density of states are often called the "s-DOS," the "p-DOS," the "d-DOS," and so on, using the same letter-to-angular-momentum relationship as in atomic orbitals (s=0, p=1, d=2, etc.).

This is not the same thing as measuring specific orbitals, although some people might still vaguely and confusingly refer to the angular-momentum components of the density of states as somehow related to specific orbitals...

• I added one question that focuses on specific excited states, not just the ground state.
– TomS
Commented Jul 7 at 8:26
• @TomS It is contrary to the rules of this website to add additional questions to an original question. This website is not meant to be a discussion forum. The editing of questions and answers are not meant to be a proxy for discussion. Nevertheless, I have added an update regarding your additional question.
– hft
Commented Jul 7 at 19:40
• Thanks for the hint, and for your answer. But I cannot agree to "the single-particle orbitals are only fiction for any interacting system". Suppose you have identically prepared atoms in the |n,0,0> state in a trap; using polarized photons you can create excited states in exactly one |n,l=1,m> state; this is not fiction. You are saying that it's not possible for atoms; but they do it for nucleons, they measure the form factors of the proton and of the Delta+ resonance. What's the problem for atoms? (I agree, that you can only measure charge and current density, that's clear).
– TomS
Commented Jul 8 at 21:01
• Photons are neutral particles, so they only interact very weakly with each other (this is why classical electromagnetism is linear). Electrons, on the other hand, are charged and can interact strongly with each other. For any neutral atom with atomic number greater Hydrogen there are clearly multiple electrons on the atom, which interact with each other. Thus, only neutral hydrogen atoms (or single-electron hydrogenic ions) can have their electron adequately described by single-particle orbitals.
– hft
Commented Jul 8 at 21:12
• For all other neutral atoms (helium, lithium, etc) using Slater determinants of single-particle orbitals is just a (sometimes very bad) approximation. I'm sorry to be the bearer of this bad news, but it is not possible to exactly solve for the electronic wavefunction of even very "simple" atoms like lithium.
– hft
Commented Jul 8 at 21:13