## Friday, March 29, 2019

### Proving the Periodic Table

The year 2019 is the International Year of the Periodic Table celebrating the 150th anniversary of Mendeleev's discovery. This prompts me to report on something that I learned in recent years when co-teaching "Mathematical Quantum Mechanics" with mathematicians in particular with Heinz Siedentop: We know less about the mathematics of the periodic table) than I thought.

In high school chemistry you learned that the periodic table comes about because of the orbitals in atoms. There is Hundt's rule that tells you the order in which you have to fill the shells in and in them the orbitals (s, p, d, f, ...). Then, in your second semester in university, you learn to derive those using Sehr\"odinger's equation: You diagonalise the Hamiltonian of the hyrdrogen atom and find the shells in terms of the main quantum number $n$ and the orbitals in terms of the angular momentum quantum number $L$ as $L=0$ corresponds to s, $L=1$ to p and so on. And you fill the orbitals thanks to the Pauli excursion principle. So, this proves the story of the chemists.

Except that it doesn't: This is only true for the hydrogen atom. But the Hamiltonian for an atom nuclear charge $Z$ and $N$ electrons (so we allow for ions) is (in convenient units)
$$a^2+b^2=c^2$$

$$H = -\sum_{i=1}^N \Delta_i -\sum_{i=1}^N \frac{Z}{|x_i|} + \sum_{i\lt j}^N\frac{1}{|x_i-x_j|}.$$

The story of the previous paragraph would be true if the last term, the Coulomb interaction between the electrons would not be there. In that case, there is no interaction between the electrons and we could solve a hydrogen type problem for each electron separately and then anti-symmetrise wave functions in the end in a Slater determinant to take into account their Fermionic nature. But of course, in the real world, the Coulomb interaction is there and it contributes like $N^2$ to the energy, so it is of the same order (for almost neutral atoms) like the $ZN$ of the electron-nucleon potential.

The approximation of dropping the electron-electron Coulomb interaction is well known in condensed matter systems where there resulting theory is known as a "Fermi gas". There it gives you band structure (which is then used to explain how a transistor works)

 Band structure in a NPN-transistor
Also in that case, you pretend there is only one electron in the world that feels the periodic electric potential created by the nuclei and all the other electrons which don't show up anymore in the wave function but only as charge density.

For atoms you could try to make a similar story by taking the inner electrons into account by saying that the most important effect of the ee-Coulomb interaction is to shield the potential of the nucleus thereby making the effective $Z$ for the outer electrons smaller. This picture would of course be true if there were no correlations between the electrons and all the inner electrons are spherically symmetric in their distribution around the nucleus and much closer to the nucleus than the outer ones.  But this sounds more like a day dream than a controlled approximation.

In the condensed matter situation, the standing for the Fermi gas is much better as there you could invoke renormalisation group arguments as the conductivities you are interested in are long wave length compared to the lattice structure, so we are in the infra red limit and the Coulomb interaction is indeed an irrelevant term in more than one euclidean dimension (and yes, in 1D, the Fermi gas is not the whole story, there is the Luttinger liquid as well).

But for atoms, I don't see how you would invoke such RG arguments.

So what can you do (with regards to actually proving the periodic table)? In our class, we teach how Lieb and Simons showed that in the $N=Z\to \infty$ limit (which in some sense can also be viewed as the semi-classical limit when you bring in $\hbar$ again) that the ground state energy $E^Q$ of the Hamiltonian above is in fact approximated by the ground state energy $E^{TF}$ of the Thomas-Fermi model (the simplest of all density functional theories, where instead of the multi-particle wave function you only use the one-particle electronic density $\rho(x)$ and approximate the kinetic energy by a term like $\int \rho^{5/3}$ which is exact for the three fermi gas in empty space):

$$E^Q(Z) = E^{TF}(Z) + O(Z^2)$$

where by a simple scaling argument $E^{TF}(Z) \sim Z^{7/3}$. More recently, people have computed more terms in these asymptotic which goes in terms of $Z^{-1/3}$, the second term ($O(Z^{6/3})= O(Z^2)$ is known and people have put a lot of effort into $O(Z^{5/3})$ but it should be clear that this technology is still very very far from proving anything "periodic" which would be $O(Z^0)$. So don't hold your breath hoping to find the periodic table from this approach.

On the other hand, chemistry of the periodic table (where the column is supposed to predict chemical properties of the atom expressed in terms of the orbitals of the "valence electrons") works best for small atoms. So, another sensible limit appears to be to keep $N$ small and fixed and only send $Z\to\infty$. Of course this is not really describing atoms but rather highly charged ions.

The advantage of this approach is that in the above Hamiltonian, you can absorb the $Z$ of the electron-nucleon interaction into a rescaling of $x$ which then let's $Z$ reappear in front of the electron-electron term as $1/Z$. Then in this limit, one can try to treat the ugly unwanted ee-term perturbatively.

Friesecke (from TUM) and collaborators have made impressive progress in this direction and in this limit they could confirm that for $N < 10$ the chemists' picture is actually correct (with some small corrections). There are very nice slides of a seminar talk by Friesecke on these results.

Of course, as a practitioner, this will not surprise you (after all, chemistry works) but it is nice to know that mathematicians can actually prove things in this direction. But it there is still some way to go even 150 years after Mendeleev.

## Saturday, March 16, 2019

### Nebelkerze CDU-Vorschlag zu "keine Uploadfilter"

Sorry, this one of the occasional posts about German politics and thus in German. This is my posting to a German speaking mailing lists discussing the upcoming EU copyright directive (must be stopped in current from!!! March 23rd international protest day) and now the CDU party has proposed how to implement it in German law, although so unspecific that all the problematic details are left out. Here is the post.

Vielleicht bin ich zu doof, aber ich verstehe nicht, wo der genaue Fortschritt zu dem, was auf EU-Ebene diskutiert wird, sein soll. Ausser dass der CDU-Vorschlag so unkonkret ist, dass alle internen Widersprüche im Nebel verschwinden. Auch auf EU-Ebene sagen doch die Befuerworter, dass man viel lieber Lizenzen erwerben soll, als filtern. Das an sich ist nicht neu.

Neu, zumindest in diesem Handelsblatt-Artikel, aber sonst habe ich das nirgends gefunden, ist die Erwähnung von Hashsummen („digitaler Fingerabdruck“) oder soll das eher sowas wie ein digitales Wasserzeichen sein? Das wäre eine echte Neuerung, würde das ganze Verfahren aber sofort im Keim ersticken, da damit nur die Originaldatei geschützt wäre (das waere ja auch trivial festzustellen), aber jede Form des abgeleiteten Werkes komplett durch die Maschen fallen würde und man durch eine Trivialänderung Werke „befreien“ könnte. Ansonsten sind wir wieder bei den zweifelhaften, auf heute noch nicht existierender KI-Technologie beruhenden Filtern.

Das andere ist die Pauschallizenz. Ich müsste also nicht mehr mit allen Urhebern Verträge abschliessen, sondern nur noch mit der VG Internet. Da ist aber wieder die grosse Preisfrage, für wen die gelten soll. Intendiert sind natürlich wieder Youtube, Google und FB. Aber wie formuliert man das? Das ist ja auch der zentrale Stein des Anstoßes der EU-Direktive: Eine Pauschallizenz brauchen all, ausser sie sind nichtkommerziell (wer ist das schon), oder (jünger als drei Jahre und mit wenigen Benutzern und kleinem Umsatz) oder man ist Wikipedia oder man ist GitHub? Das waere wieder die „Internet ist wie Fernsehen - mit wenigen grossen Sendern und so - nur eben anders“-Sichtweise, wie sie von Leuten, die das Internet aus der Ferne betrachten so gerne propagiert wird. Weil sie eben alles andere praktisch platt macht. Was ist denn eben mit den Foren oder Fotohostern? Müssten die alle eine Pauschallizenz erwerben (die eben so hoch sein müsste, dass sie alle Film- und Musikrechte der ganzen Welt pauschal abdeckt)? Was verhindert, dass das am Ende ein „wer einen Dienst im Internet betreibt, der muss eben eine kostenpflichtige Internetlizenz erwerben, bevor er online gehen kann“-Gesetz wird, das bei jeder nichttrivialen Höhe der Lizenzgebühr das Ende jeder gras roots Innovation waere?

Interessant waere natuerlich auch, wie die Einnahmen der VG Internet verteilt werden. Ein Schelm waere, wenn das nicht in großen Teilen zB bei Presseverlegern landen würde. Das waere doch dann endlich das „nehmt denjenigen, die im Internet Geld verdienen dieses weg und gebt es und, die nicht mehr so viel Geld verdienen“-Gesetz. Dann müsste die Lizenzgebühr am besten ein Prozentsatz des Umsatz sein, am besten also eine Internet-Steuer.

Und ich fange nicht damit an, wozu das führt, wenn alle europäischen Länder so krass ihre eigene Umsetzungssuppe kochen.

Alles in allem ein ziemlich gelungener Coup der CDU, der es schaffen kann, den Kritikern von Artikel 13 in der öffentlichen Meinung den Wind aus den Segeln zu nehmen, indem man es alles in eine inkonkrete Nebelwolke packt, wobei die ganzen problematischen Regelungen in den Details liegen dürften.

## Wednesday, March 06, 2019

### Challenge: How to talk to a flat earther?

Further down the rabbit hole, over lunch I finished watching "Behind the Curve", a Netflix documentary on people believing the earth is a flat disk. According to them, the north pole is in the center, while Antarctica is an ice wall at the boundary. Sun and moon are much closer and flying above this disk while the stars are on some huge dome like in a planetarium. NASA is a fake agency promoting the doctrine and airlines must be part of the conspiracy as they know that you cannot directly fly between continents on the southern hemisphere (really?).

These people are happily using GPS for navigation but have a general mistrust in the science (and their teachers) of at least two centuries.

Besides the obvious "I don't see curvature of the horizon" they are even conducting experiments to prove their point (fighting with laser beams not being as parallel over miles of distance as they had hoped for). So at least some of them might be open to empirical disprove.

So here is my challenge: Which experiment would you conduct with them to convince them? Warning: Everything involving stuff disappearing at the horizon (ships sailing away, being able to see further from a tower) are complicated by non-trivial diffraction in the atmosphere which would very likely turn this observation inconclusive. The sun being at different declination (height) at different places might also be explained by being much closer and a Foucault pendulum might be too indirect to really convince them (plus it requires some non-elementary math to analyse).

My personal solution is to point to the observation that the declination of Polaris (around which I hope they can agree the night sky rotates) is given my the geographical latitude: At the north pole it is right above you but is has to go down the more south you get. I cannot see how this could be reconciled with a dome projection.

How would you approach this? The rules are that it must only involve observations available to everyone, no spaceflight, no extra high altitude planes. You are allowed to make use of the phone, cameras, you can travel (say by car or commercial flight but you cannot influence the flight route). It does not involve lots of money or higher math.