# Is the Mendeleev table explained in quantum mechanics?

Does anybody know if there exists a mathematical explanation of the 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 normal in mathematics. As an example, 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 the 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: https://physics.stackexchange.com/questions/16647/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 states 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 a system of axioms like those from the 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://mathoverflow.tqft.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.

• You might want to try posting this question on theoreticalphysics.stackexchange.com – Bill Cook Nov 5 '11 at 20:13
• @Greg: 1) if you ask this "great chemistry-and-physics question" to chemists or physicists, they will treat you as an idiot (and this is what happened to me at theoreticalphysics.stackexchange.com/questions/473/… ). Because they do not understand what logic is. If this were not so, there would not be contradictions between what people write here and what they write there: "Yes, quantum mechanics... – fully, quantitatively, and comprehensively explains all of chemistry..." So I still think that I should address this to mathematicians. – Sergei Akbarov Nov 6 '11 at 0:14
• @Mark: 1) if physicists did not convince me, like Luboš Motl at their physical forum, that everything is "fully, quantitatively, and comprehensively" explained, I would not ask this question here. But their arguments are always so definite, uncompromising, unequivocal (and I would say in some sense offensive, don't you find them so? :), that you begin to think that maybe you read wrong books, and if you ask mathematicians who are interested in tags like "quantum-mechanics", they will give an explanation, which could be verified (as this usually happens with mathematicians). – Sergei Akbarov Nov 6 '11 at 1:03
• @Sergei: 1) There is no way to keep track of replies to your comment. You should contact Jacques personally if you want a reply. 2) “I already explained what I mean by "proof"”: No, you didn't. You can prove a theorem, but the periodic table is not a theorem. You have to state exactly what do you want to prove. “The Mendeleev table claims…”: You have to define in mathematical terms what do you mean by an “orbit”, “electron”, “subshell”, etc. Right now there is no mathematical content in your statements. “If you know what should be proved”: No, you don't, at least in mathematical terms. – Dmitri Pavlov Nov 7 '11 at 9:42
• Too bad I missed the party. I think this is great question! And I can't understand the reaction of some people here and on Physics.SE. Every serious mathematician, physicist, or chemist is aware of the merits of Mendeleev's table and its experimental validation. And every serious mathematician, physicist, or chemist knows similarly well that Quantum Mechanics is an idealisation. Furthermore we can not even give a closed solution for the helium atom. So I think it is a good test of QM and our mathmatical or computational skills, if we can derive the properties of M's table from SE. – Uwe Franz Jan 20 '13 at 17:10

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 $2$.

• Even if no answer will be satisfactory, I have found the answers to this question to be very interesting... – Tom Church Nov 9 '11 at 16:52
• The OP asks two completely different questions: (1) "[...] if there exists a mathematical explanation of Mendeleev table in quantum mechanics? " (2) "if there is a mathematical proof of the Mendeleev table? [...] by a mathematical proof I mean a chain of logical implications from axioms of the theory [...]" This answer discusses the use of approximations, which is irrelevant to both 1 and 2. The answer to 1 is yes, and the necessity of making approximations doesn't affect the validity of the explanation. The answer to 2 is no, simply because physical theories aren't axiomatic systems. – Ben Crowell Jun 18 '13 at 14:13
• Ben, you should explain yourself, this sounds strong: "The answer to 1 is yes". And this also: "physical theories aren't axiomatic systems". What about classical mechanics? Or probability theory? – Sergei Akbarov Jun 20 '13 at 11:38
• @SergeiAkbarov: What makes you think classical mechanics is an axiomatic system? Are you claiming that Newton's laws are an axiomatic system? I think that's clearly untrue. Or are you thinking of some restricted physical model within classical mechanics, which can be described axiomatically? That doesn't mean that classical mechanics in general is an axiomatic system. (I also don't know what you mean by the question about probability theory, since that isn't a physical theory.) – Ben Crowell Mar 6 '16 at 19:03
• This answer is inaccurate and pretty naive. It gives a relatively decent description of atomic physics at the level of an introductory undergraduate textbook, but nothing more. Yes, solutions are not single Slater determinants, which is why we have post-Hartree-Fock methods for describing multi-configuration electronic states, which have been staples of atomic physics for many decades; as a whole, they refine the quantitative accuracy of HF but rarely change its qualitative features (and then, typically in molecules more than atoms). – Emilio Pisanty Sep 12 '17 at 16:10

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.

• The Mendeleev table is not a mathematical theorem. It is a method of organizing atoms into groups with similar chemical properties. Quantum Mechanics is a framework which includes the nonrelativistic Schrödinger equation as well as quantum field theory. If the OP wants to know if X can be proved in a mathematically rigorous way from Y then isn't it reasonable to ask for a mathematically precise definition of both X and Y? – Jeff Harvey Nov 11 '11 at 12:56
• @Jeff Harvey: I think it’s quite reasonable to ask without making the statements precise. Finding the right formulation of an informal idea is often as hard as proving it, or even harder. Of course, many statements are too vague to make an interesting question; but what makes this one good, I think, is that while we don’t necessarily have a precise formal statement of it in mind, “we know it when we see it”… as in Terry Tao’s answer, we do start to see it. – Peter LeFanu Lumsdaine Nov 11 '11 at 23:41
• @Jeff Hervey: Asking questions about things which are not precisely defined is a tradition in mathematics. For example, mathematicians discussed the problems of probability theory long before Kolmogorov (in 1933) gave his axioms of probability (only after that probability got a precise definition). My question is just another example: in books on mathematical physics (e.g., in the Berezin-Shubin book) they speak very often about atoms, which as far as I understand, are not defined by now. If they can discuss atoms, why can't they discuss the Mendeleev table which describes properties of atoms? – Sergei Akbarov Nov 12 '11 at 21:13
• pnas.org/content/97/1/28.full – Steve Huntsman Mar 30 '16 at 14:16

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).

• Terry, thank you very much for your answer, but your link is just a presentation, it does not contain even references... I would like to thank also the other people who wrote the answers. So, dear colleagues, as far as I understand, we came to a conclusion that from the point of view of logic the Mendeleev table is not explained in quantum mechanics? :) – Sergei Akbarov Nov 5 '11 at 21:59
• References to the work given in the above slides can be found at www-m7.ma.tum.de/bin/view/Analysis/ElectronicStructure – Terry Tao Nov 6 '11 at 4:26
• It's strange to call "rigorous" a treatment where the charge of the nucleus is much larger then the number of electrons. In the cases considered both are around 10 - some sort of large number! Also, in chemical reactions the atoms are weakly ionized, almost neutral, so in fact $N\approx Z$. The words "use the model of non-interacting fermions" clearly hide much of the work and make a huge leap of faith from the initial equation to the answer. – Anton Fetisov Nov 13 '11 at 11:29
• Goddard-Freiseke's work is rigorous in the regime where N is at most 10 and Z is sufficiently large; however, the predictions of that paper agree quite well with the experimental data when N and Z are both near 10, which suggests that the limitation to sufficiently large Z is a technical one rather than a fundamental one. Also, their analysis does consider Coulomb interactions between the fermions (otherwise the problem would be very easy). As I understand it, the non-interacting model is only used as a base model from which one applies rigorous perturbation theory (see p.36 of slides). – Terry Tao Nov 13 '11 at 16:59
• It's worth remarking, perhaps, that the asymptotic regime that you mention breaks the physical assumptions that underpin the electrostatic Schrödinger equation in use: for highly ionized atoms, you expect spin-orbit coupling and other relativistic effects to take on significant roles; the asymptotics might be relevant for the mathematical proof but they do step outside of the physical envelope. This is particularly relevant given the John Ball quote at the end of the Friesecke presentation you linked to. – Emilio Pisanty Sep 13 '17 at 13:57

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.

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.

• Antoine, I am not against approximations. And I know that it's not necessary to solve equation for gathering information about its solutions (people in qualitative theory of differential equations demonstrate this all the way). But in any case there must be a system of axioms, definitions and propositions, otherwise it would be impossible to understand this theory. For example, Berezin and Shubin in their book speak very often about atoms, but they do not define atoms and do not make propositions about them (if not counting atoms in math. sense). Do you know a text, where this is not so? – Sergei Akbarov Nov 18 '11 at 7:38
• Well, as I said, you can postulate the following: An atom (or molecule, for that matter), is a core of Z protons and M neutrons described by their position, plus N electrons, each described by a wave function $\phi_i$. The electrons arrange themselve as the N lowest eigenfunctions of the eigenproblem on $H^1(R^3,C^2)$ : $-\Delta \phi_i + V(x) \phi_i = \lambda_i \phi_i$, where $V(x) = Z/|x|$ for an atom at $(0,0,0)$. Then, proposition : the eigenfunctions are organised in shells of same $\lambda_i$. The first eigenvalue has multiplicity 2, the second, multiplicity 8, etc. – Antoine Levitt Nov 18 '11 at 9:19
• This is a toy model with little relevance for a quantitative (or even qualitative, for larger atoms) understanding of the true nature of elements. I don't think you have the correct approach, though. Physics is not a subset of mathematics, and cannot be understood as such. – Antoine Levitt Nov 18 '11 at 9:22
• I doubt that mathematical physicists will agree with this: "Physics is not a subset of mathematics, and cannot be understood as such". But anyway if you say that everything is so simple, then there must be a reference. What you say (your definition, proposition) - where is this written? – Sergei Akbarov Nov 18 '11 at 12:40
• I'm actually pretty sure most mathematical physicists will agree with the sentence. Physics aims to understand the physical world, which no model alone can do without (physical, ie approximate and not rigorous) analysis. For instance, the full Schrodinger equation for an atom is mathematically perfect, and physically useless, because of its sheer complexity. So is a description of a gas by its individual molecules, which is why physicists have invented thermodynamics. But that's philosophy, and I'm not about to get dragged into that debate. – Antoine Levitt Nov 18 '11 at 19:09

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.

• Anton, I did not understand this: "As usually stated, it is a vague and strictly speaking false". Vague and even false? – Sergei Akbarov Nov 13 '11 at 17:44
• Trivially, it's not a mathematically precise statement, so by the standards of total rigour it's vague. It's more precise (very precise) in predicting the ground state electron configurations, but less precise in predictions of chemical properties. Exceptions are rare, especially for low atomic weights, but they exits and well-known to chemists. I'm not a chemist myself, so I'll just give a few simple links: chemwiki.ucdavis.edu/Inorganic_Chemistry/… en.wikipedia.org/wiki/… – Anton Fetisov Nov 15 '11 at 23:30
• As it turned out (and this was not obvious for me at the beginning), what the Mendeleev table states cannot be mathematically statements at all, since the notions like atoms, electrons, shells, etc., are not precisely defined. I think those participants of this discussion, who protest against the "vague formulation of the question", should write a collective letter to a journal like "Notices of the AMS" with protest against using the words like "atom", "electron", etc., in mathematical physics, since these notions have no precise definitions. – Sergei Akbarov Nov 16 '11 at 12:24
• No, I don't get your sarcasm. As said before, if you're fine with the common amount of rigour in Mendeleev's table, then you should be just as fine with the common explanations in QM textbooks, so what's the point? – Anton Fetisov Nov 16 '11 at 18:06
• Anton, in my opinion there cannot be some amount of rigor, enough amount, common amount, etc. Either there is rigor, or there isn't, that's what I used to think about it. My aim was to understand, if there were successful attempts to interpret in the mathematical language what physicists say about what I asked. Physicists themselves can't explain this, that's why I asked this here. From what people told me I deduce that the attempts were not successful. That's enough for me. – Sergei Akbarov Nov 16 '11 at 21:44