Is the Mendeleev table explained in quantum mechanics? - MathOverflow

Is the Mendeleev table explained in quantum mechanics?

2011-11-05T19:36:52Z

Does anybody know if there exists a mathematical explanation of Mendeleev table in quantum mechanics? In some textbooks (for example in "F.A.Berezin, M.A.Shubin. The Schrödinger Equation") the authors present quantum mechanics as an axiomatic system, so one could expect that there is a deduction from the axioms to the main results of the discipline. I wonder if there is a mathematical proof of the Mendeleev table?

P.S. I hope the following will not be offensive for physicists: by a mathematical proof I mean a chain of logical implications from axioms of the theory to its theorem. :) This is the standard approach everywhere in mathematics. For instance, in Griffiths' book I do not see axioms at all, therefore I can't treat the reasonings at pages 186-193 as a proof of Mendeleev table. By the way, that is why I did not want to ask this question at a physical forum: I do not think that people there will even understand my question. However, after Bill Cook's suggestion I made an experiment - and you can look at the results here: http://theoreticalphysics.stackexchange.com/questions/473/is-the-mendeleev-table-explained-in-quantum-mechanics

So I ask my colleagues-mathematicians to be tolerant.

P.P.S. After closing this topic and reopening it again I received a lot of suggestions to reformulate my question, since in its original form it might seem too vague for mathematicians. So I suppose it will be useful to add here, that by the Mendeleev table I mean (not just a picture, as one can think, but) a system of propositions about the structure of atoms. For example, as I wrote here in comments, the Mendeleev table claims that the first electronic orbit (shell) can have only 2 electrons, the second - 8, the third - again 8, the fourth - 18, and so on. Another regularity is the structure of subshells, etc. So my question is whether it is proved by now that these regularities (perhaps not all but some of them) are corollaries of the system of axioms like those from Berezin-Shubin book. Of course, this assumes that the notions like atoms, shells, etc. must be properly defined, otherwise the corresponding statements could not be formulated. I consider this as a part of my question -- if experts will explain that the reasonable definitions are not found by now, this automatically will mean that the answer is 'no'.

The following reformulation of my question was suggested by Scott Carnahan at http://meta.mathoverflow.net/discussion/1202/should-a-mathematician-be-a-robot/#Item_0 : "Do we have the mathematical means to give a sufficiently precise description of the chemical properties of elements from quantum-mechanical first principles, such that the Mendeleev table becomes a natural organizational scheme?"

I hope, this makes the question more clear.

Answer by Physicist for Is the Mendeleev table explained in quantum mechanics?

2011-11-05T20:36:35Z

It depends on what you mean by proof. Even the helium atom wavefunction cannot be obtained in closed form (the way the hydrogen atom wavefunction is), so any results about the periodic table will have some level of approximation or phenomenological assumptions in them. That said, there do exists references that explain the qualitative (and quantitative) features of the periodic table based on quantum mechanics principles. Griffiths' Quantum Mechanics for instance has a very quick discussion of the periodic table around pages 186-193. It's not very complete, and also mostly not quantitative, but it nicely illustrates how quantum mechanics gives rise to the structure of the periodic table.

Answer by Terry Tao for Is the Mendeleev table explained in quantum mechanics?

2011-11-05T20:50:37Z

There is some rigorous work by Goddard and Friesecke on this, see

http://www.ma.hw.ac.uk/~chris/icms/GeomAnal/friesecke.pdf

My understanding is that even getting accurate numerics for the Schrodinger equation becomes very difficult once one has more than 10 or so electrons in play. The one regime where we do seem to have good asymptotics is when the atomic number is large but the number of electrons are small (i.e. extremely highly ionized heavy atoms).

At any rate, the foundations of the periodic table are pretty much uncontested (i.e. N-body fermionic Schrodinger equation with semi-classical Coulomb interactions as the only significant force). The main difficulty is being able to solve the resulting equations mathematically (or even numerically).

Answer by Denis Serre for Is the Mendeleev table explained in quantum mechanics?

2011-11-05T21:18:03Z

I doubt any answer will be satisfactory. My opinion is that we are still very far from a mathematical justification. If we accept the mathematical foundations of quantum mechanics, and if we make the approximation that the nucleus of the atom is just one heavy thing with $N$ positive charges, then the motion of the $N$ electrons is governed by a linear equation (Schrödinger) in ${\mathbb R}^{3N}$. The unknown is a function $\psi(r^1,\ldots,r^N,t)$ with the property (Pauli exclusion) that it has full skew-symmetry. For instance, $$\psi(r^2,r^1,\ldots,r^N,t)=-\psi(r^1,r^2,\ldots,r^N,t).$$ In practice, we look for steady states $e^{i\omega t}\phi(r^1,r^2,\ldots,r^N)$. Then $\omega$ is the energy level.

Because of the very large space dimension, one cannot perform reliable calculations on computer, when $N$ is larger than a few units. One attempt to simplify the problem has been to postulate that $\phi$ is a Slatter determinant, which means that $$\phi(r^1,r^2,\ldots,r^N)=\|a_i(r^j)\|_{1\le i,j\le N}.$$ The unknown is then an $N$-tuple of functions $a_i$ over ${\mathbb R}^3$. Of course, we do not expect that steady states be really Slater determinants; after all, the Schrödinger equation does not preserve the class of Slater determinants. Thus there is a price to pay, which is to replace the Schrödinger equation by an other one, obtained by an averaging process (Hartree--Fock model). The drawback is that the new equation is non-linear. Such approximate states have been studied by P.-L. Lions & I. Catto in the 90's.

Update. Suppose $N=2$ only. If we think to $\phi$ as a finite-dimensional object instead of an $L^2$-function, then it is nothing but a skew-symmetric matrix $A$. Approximation à la Slater consists in writing $A\sim XY^T-YX^T$, where $X$ and $Y$ are vectors. In other words, one approximate $A$ by a rank-two skew-symmetric matrix. The approximation must be in terms of the Hilbert-Schmidt norm (also named Frobenius, Schur): this norm is natural because of the requirement $\|\phi\|_{L^2}=N$. If $\pm a_1,\ldots,\pm a_m$ are the pairs of eigenvalues of $A$, with $0\le a_1\le\ldots\le a_m$, then the best Slater approximation $B$ satisfies $\|B\|^2=2a_m^2$, $\|A-B\|^2=2(a_1^2+\cdots+a_{m-1}^2)$. Not that good. Imagine how much worse it can be if $N$ is larger than $2$.

Answer by Jeff Harvey for Is the Mendeleev table explained in quantum mechanics?

2011-11-09T22:52:35Z

I am not offended by the suggestion that physicists should follow the standards of mathematical proof, but I think this suggestion and the phrasing of the question demonstrate a lack of understanding of how physicists think about such things and more importantly why they put such little emphasis on axioms.

In my view it is rarely useful to think of physics as an axiomatic system, and I think this question reflects the difficulty with thinking of it as such. A different question, which is much more in tune with a physicist's point of view, would be to ask what physical description is required to explain various features of the structure of atoms as reflected in the periodic table at a prescribed level of accuracy. Until you specify what features you want to understand, and at what level of accuracy, you don't even know what the correct starting point should be. If you want just the crudest structure of the periodic table, then indeed non-relativistic quantum mechanics along with the Pauli exclusion principle will give you the rough structure as described in any standard QM textbook. If you want to understand the detailed quantum numbers of large atoms then you have to start including relativistic effects. Spin-orbit coupling is one of the most important and its effects are often summarized by a set of Hund's rules which are described in many QM textbooks or physical chemistry textbooks. If you want very accurate numerical values for ionization energies or the detailed structure of wave functions then one must do hard numerical work which probably becomes impossibly difficult for large atoms. As you ask for greater and greater precision you should eventually use a fully relativistic description. This is even harder. The Dirac equation is not sufficient, one cannot restrict to a Hilbert space with a finite number of particles in a relativistic quantum theory, and bound state problems in Quantum Field Theory are notoriously difficult. So as one asks more detailed and more precise questions, one has to keep changing the mathematical framework used to formulate the theory. Of course this process could end and there could be an axiomatic formulation of some ultimate theory of physics, but even if this were the case this would undoubtedly not be the most useful formulation for most problems of practical interest.

Answer by Anton Fetisov for Is the Mendeleev table explained in quantum mechanics?

2011-11-13T11:13:18Z

It clearly depends upon what do you mean by the Periodic Table of elements? As usually stated, it is a vague and strictly speaking false, yet usually sufficient statement about the similarities of chemical properties of different atoms. In any case they don't repeat exactly, only with a given degree of accuracy and if you forget about some of the much more exotic behaviour, not common in reactions. If you really try to specify all of these, you'll be much better off with the common perturbation theory approach found in QM textbooks. Sure, in a sense it also defines what is being calculated, but there's also no other way to define these properties (at least I don't know any). Analysing the second-or-somewhat order of perturbation theory is a mathematically trivial, yet tedious task, but there can barely be a way to justify the order of PT rigorously, it just works. Or doesn't.

Answer by Antoine Levitt for Is the Mendeleev table explained in quantum mechanics?

2011-11-17T23:16:52Z

I'm arriving after the war, but this is an interesting question, so I'm going to write up what I understand about it.

First of all, for a comprehensive mathematical understanding of the periodic table, you have to settle on a model. The relevant one here is quantum mechanics (for large atoms, relativistic effects start to become important, and that's a whole mess). It's entirely axiomatic, and requires no further tweaking. Then you basically have to solve an eigenvalue on a space of functions of $6N$ coordinates (ignoring spin). That gives you a "mathematical explanation" of the table, in the sense that knowledge of the solution $\psi(x_1,x_2,\dots,x_N)$ is all there is to know about the static structure of an atom. Notice that in this formulation, all electrons are tied together inside one big wavefunctions, so an "electronic state" has no meaning. Mendeleev table is not even compatible with this formulation.

Of course, solving the full eigenproblem is not possible, so all you can do is mess around with approximations. A simplistic but illuminating approximation is to completely neglect electron repulsion. Great simplification occurs, and it turns out one can speak of "electronic states". Non-trivial behaviour occurs because of the Pauli exclusion principle. This is known as the "Aufbau" principle: one builds atoms by successively adding electrons. The first electron gets itself into the lowest energy shell, then the second one gets into the same state, but with opposite spin. The third begins to fill the second shell (which has three spaces, times two because of spin), and so on. This is the basic idea behind the table, and provides a clue as to why it is organised the way it is. So this might be the theory you're looking for. It's explicitely solvable, and only requires the theory of the hydrogenoid atoms.

Of course, because of the approximations, the quantitative results are all wrong, but the organisation is still there. Except for larger elements, where the Mendeleev table is, from what I understand, an ad-hoc hack. You can improve the approximation using ideas like "screening", and this leads to the Hartree-Fock method, which still preserves the notion of shells.

Hope that helps. Then again, if you're looking for a completely logical approach to physics that'll readily explain real life, you're bound to be disappointed. Even simple theories such as the quantum mechanics of atoms are too hard to be solved exactly, which is why we have to compromise and make approximations.