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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.

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    You might want to try posting this question on http://theoreticalphysics.stackexchange.com/ – Bill Cook Nov 05 '11 at 20:13
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    I would suggest rewriting your question in purely mathematical terms. As it stands, it is best asked in the theoretical physics stackexchange (as Bill Cook mentions). But there is indeed a mathematical question here -- but you did not ask it. – Jacques Carette Nov 05 '11 at 20:31
  • Jacques, what do you have in mind? – Sergei Akbarov Nov 05 '11 at 20:50
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    @Bill Cook - that's a terrible advice. theoreticalphysics is specifically for research-level questions, while this one seems to be more suitable for http://physics.stackexchange.com/ (difference between those two sites is the same as between MO and Math.SE) – Marcin Kotowski Nov 05 '11 at 21:14
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    "closed as off topic by Greg Kuperberg, Mark Sapir, Todd Trimble, Ryan Budney, Alain Valette 30 mins ago"

    @ Greg Kuperberg, Mark Sapir, Todd Trimble, Ryan Budney, Alain Valette - gentelemen, may I ask you, why did you decide that this is off topic?

    – Sergei Akbarov Nov 05 '11 at 22:21
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    It's off topic because the periodic table is chemistry before it is anything else, and the Schr"odinger equation is physics before it is anything else. It's a great chemistry-and-physics question, but as a strict math question it's hopelessly arcane. There could be some related math question that's reasonable, but a revision of your question is anyone's guess. – Greg Kuperberg Nov 05 '11 at 22:59
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    @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 http://theoreticalphysics.stackexchange.com/questions/473/is-the-mendeleev-table-explained-in-quantum-mechanics ). 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 06 '11 at 00:14
  • @Greg: 2) will the following revision of my question be suitable: "should we treat the postulates of quantum mechanics as axioms of an axiomatic system?" – Sergei Akbarov Nov 06 '11 at 00:19
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    @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 06 '11 at 01:03
  • @Mark: 2) concerning this: "There is no option "question does not make any sense" in the list of closing options." -- It is not me who pretend that quantum mechanics is based on "postulates", so this is pulery mathematical (if you want, logical) question, if this pretension can be considered as a presentation of an axiomatic theory. Doesn't this make sense? – Sergei Akbarov Nov 06 '11 at 01:11
  • @Mark: from this discussion it follows that there is a qualitative difference between mechanics (as a part of differential geometry) and quantum mechanics (as a part of what you say): in the first case this is an axiomatic system, while in the second one this is not. And this is not obvious for non-specialists like me. So, when appealing to colleagues, I thought I could count on their understanding, since they at least are able to understand my question. But what I see here from moderators is not called understanding, this much more resembles rudeness ("hamstvo" in Russian, is it? :). – Sergei Akbarov Nov 06 '11 at 02:40
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    @Mark: I guess, this means that you do not think I am a mathematician. You can check this by the link I gave at my page here: http://www.mathnet.ru/php/person.phtml?&personid=8763&option_lang=eng Is "Doctor of Physical and Mathematical Sciences" enough high level for you? – Sergei Akbarov Nov 06 '11 at 09:43
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    @Sergei: There is certainly a germ of a legitimate question in what you asked, but as Jacques Carette already pointed out, it should be written in purely mathematical terms, i.e., it should probably ask whether certain facts about eigenvalues and eigenfunctions of a certain Schrödinger operator have been proved (see, for example, the slides Terry Tao refers to). After you ask a rigorous question, you can mention the periodic table of the chemical elements as a background/motivation (see the faq to learn more about this). – Dmitri Pavlov Nov 06 '11 at 10:43
  • @Dmitri: 1) I asked Jacques Carette what he has in mind - he did not reply. If his suggestion was more clear, I would consider it more seriously. 2) Your reformulation of my question is not correct: I am asking about axiomatization of quantum mechanics. Formally this is not equivalent to finding eigenvalues or eigenfunctions of Schrödinger operator, since one could expect that the Mandeleev table could be proved without considering those eigenvalues. My question is absolutely clear and correct for mathematicians who have an impression of what axiomatic system is, why I should reformulate it? – Sergei Akbarov Nov 06 '11 at 13:14
  • @Mark: there was no necessity to remove your coments here. However, I appreciated your will to talk to me, but you see, the discussion could be much more interesting, if you didn't close this topic so quickly. – Sergei Akbarov Nov 06 '11 at 19:11
  • @Sergei: 1) I offered one possible explanation for Jacques' comment. As for 2), no, it is neither clear nor correct. For example, what do you mean by a “proof” of the periodic table? Also, a mathematical question should refer to a specific mathematical model of quantum mechanics (in our case, a specific mathematical model of an atom), e.g., you can ask a question about the Berezin-Shubin model, framed in specific terms (e.g., something about eigenvalues/eigenvectors, not about a “proof of the periodic table”). – Dmitri Pavlov Nov 06 '11 at 19:19
  • @Dmitri: 1) There is only one preson, who can explain what Jacques Carette had in mind - this is Jacques Carette himself. Instead of disputing what he wanted to say, it is much more clever to ask him this question. I did exactly this. Before he gives an answer it is senseless to refer to him. By the way, do you think it was nice of him that he dropped a hint, and disappeared immediately after that? – Sergei Akbarov Nov 07 '11 at 02:51
  • @Dmitri: 2) I already explained what I mean by "proof". If you ask what exactly I would like to be proved, then the answer is the following. The Mendeleev table claims, for example, that the first elecctronic orbit 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. If quantum mechanics was an axiomatic system, all those regularities would be proved. I asked, whether this is indeed so. And it is strange for me that there are mathematicians who say that this question is unclear or incorrect. – Sergei Akbarov Nov 07 '11 at 03:00
  • @Dmitri: 3) You compel me all the way to specify not only WHAT I would like to be proved, but also HOW I expect this to be proved. Initially you told that I should mention the Schrödinger operator, now you speak about the Berezin-Shubin model. Why is it so important? If you know what should be proved (example: the statement on the structure of orbits), why the question remains unclear until I explain how I think this must be proved (solving Schrödinger equation, or using the Berezin-Shubin model)? I used to think that the question HOW is extra in the formulation of problem. – Sergei Akbarov Nov 07 '11 at 07:59
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    @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 07 '11 at 09:42
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    @Sergei: I think you got it right in 3), modulo terminology. “HOW” in your description refers to the fact that you must always specify the mathematical model. It is important because otherwise it is not a mathematical question. And of course I strongly disagree with your “HOW”. Even if you prove the same statement from a physical point view in two different models of quantum mechanics, form a mathematical point of view you proved two completely different statements. – Dmitri Pavlov Nov 07 '11 at 09:49
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    @Sergei: To sum up, a mathematical formulation of your question should include a specific mathematical model of quantum mechanics, mathematical definitions of all terms like “electron”, “subshell”, “orbit”, and (rigorous) mathematical statements or conjectures that you want to be proved. – Dmitri Pavlov Nov 07 '11 at 09:55
  • @Dmitri: 1) I already asked Jacques, now it's his turn. 2) That's the main point: "you must always specify the mathematical model... otherwise it is not a mathematical question". I don't agree. I can ask WHETHER THERE EXISTS A MATHEMATICAL MODEL, describing this or that phenomena, and this also will be a mathematical question for the mathematicians working in close fields (especially in fields with the same name, like "quantum mechanics"). I can also ask a mathematician WHICH THEORIES IN HIS FIELD ARE AXIOMATIC SYSTEMS, and this also will be a mathematical question. Should I explain, why? – Sergei Akbarov Nov 07 '11 at 11:19
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    @Dmitri: In addition to what I said: when I see a text with words like "postulate" or "axiom", I can ask a specialist, whether this is an axiomatic theory or not, and whether it describes the phenomena I am interested in. Dima, I think, I understand your idea: you think that a specialist must be a robot. Moreover a badly adjusted robot, who can't understand normal language which people use for communication. Such a robot that for asking him something you should write a special program, like in FORTRAN, otherwise he will be answering only: "this is not my field, rewrite your question!". – Sergei Akbarov Nov 07 '11 at 12:14
  • @Dmitri: An explanation to non-robots, why this is bad: this makes impossible the communication between specialists and non-specialists (because to ask a robot anything, you must be enough sophisticated in his specific language, so you must be a specialist), this leads to the lack of understanding between different groups of people in society, and, being a kind of rudeness ("hamstvo" in Russian), this contradicts to the ethic norms. Is that clear, or I should reformulate this in some special langauge? – Sergei Akbarov Nov 07 '11 at 12:25
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    @Sergei: 1) The point is that once you post a reply to somebody's comment, there is no way for this person to find out about this unless he looks at this page again or you inform him personally. This problem is fixed in SE 2.0 software, but MO still uses SE 1.0. – Dmitri Pavlov Nov 07 '11 at 12:31
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    @Sergei: The question “WHETHER THERE EXISTS A MATHEMATICAL MODEL describing this or that phenomenon” clearly belongs to the domain of physics, not mathematics. You other comments about robots also apply to physicists, not mathematicians. After all, we have two different sites (MathOverflow and Physics / Theoretical Physics SE) and two different sciences (mathematics and physics) for a reason, otherwise there would be just one site and one science. – Dmitri Pavlov Nov 07 '11 at 12:44
  • @Sergei: I also find your claims of rudeness completely inappropriate. Rudeness is by definition a violation of community's widely accepted norms. It is you who is violating these norms, not the MO community. Anyway, you are always welcome to start a thread on Meta about this. This comment thread is turning into a discussion, which is by itself a violation of MO rules. – Dmitri Pavlov Nov 07 '11 at 12:47
  • @Dmitri: You didn't comment the other two questions, which I posted: "WHICH THEORIES IN HIS FIELD ARE AXIOMATIC", and "whether this is an axiomatic theory or not...". Does this mean that you agree with me that these are mathematical questions? As to the first one - “WHETHER THERE EXISTS A MATHEMATICAL MODEL describing this or that phenomenon” - I don't agree with you, because a mathematician who work in this field should know the applications of his activity. On the contrary, a physisist very likely will say that this is not physics, since the question is about mathematical models. – Sergei Akbarov Nov 07 '11 at 15:38
  • @Dmitri: About ethics. That's interesting: "It is you who is violating these norms". What do you mean? And I did not understand this: "you are always welcome to start a thread on Meta about this". I am not a programmer, where can I start a thread? – Sergei Akbarov Nov 07 '11 at 15:45
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    @Sergei: I suggest that you first read the FAQ: http://mathoverflow.net/faq. It explains what Meta is and why discussions like this are discouraged here. It also explains what questions are appropriate here and why your question in its current form is inappropriate for MO, which should address your questions about rudeness/ethics. Also, you can see a link to Meta at the very top of any MO page. Finally, your other two questions are just as bad for MO for similar reasons, but I reiterate my request to transfer this discussion to Meta if you want to continue it. – Dmitri Pavlov Nov 07 '11 at 16:57
  • "I reiterate my request to transfer this discussion to Meta if you want to continue it" - does this mean that our discussion here will be deleted? – Sergei Akbarov Nov 07 '11 at 22:09
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    Dima, and this statement needs proof: "It is you who is violating these norms". As well as those: "your question in its current form is inappropriate", "your other two questions are just as bad for MO for similar reasons". – Sergei Akbarov Nov 08 '11 at 06:52
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    @Sergei: You are trying to continue the discussion here despite being explicitly told not to do so. I give up. – Dmitri Pavlov Nov 08 '11 at 12:56
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    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
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    This is a brilliant question, if I could only upvote it more than once. @Dimitri, the whole point is this is partially reverse mathematics. I.e. what axioms must we postulate in order to mathematically derive the structure of the period table which is already known experimentally to be correct and if the existing postulates of QM (appropriately formulated) are sufficient to do this. The negative reaction here is ridiculous and I also agree that you will get nonsense from physics se on this and that it will be mathematically very challenging. – Benjamin Sep 14 '15 at 18:10

6 Answers6

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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.

Jeff Harvey
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    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
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    @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
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    @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
  • http://www.pnas.org/content/97/1/28.full – Steve Huntsman Mar 30 '16 at 14:16
  • Jeff, it's about responsibility for the spoken word. If a scientist pronounces the word "postulate" or the word "axiom", putting into it the meaning that mathematicians usually put into it, he is obliged to give an axiomatic system in which what he calls axioms will be axioms. This is not a matter of whim, this is his duty. On the other hand, if he puts into these words some unusual meaning, not the one that mathematicians put in, then he is obliged to explain what he means. – Sergei Akbarov Jul 01 '22 at 07:11
  • There would be no problem if Berezin and Shubin wrote "we call these statements postulates, but note that these are not postulates in the mathematical sense, we understand this and that by this word" - and then an explanation of what they mean. But they do not say this, making the reader think that they mean postulates in the usual mathematical sense, and this is a breach of the reader's trust, because this is not true. – Sergei Akbarov Jul 01 '22 at 07:11
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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 Tao
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  • 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 05 '11 at 21:59
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    References to the work given in the above slides can be found at http://www-m7.ma.tum.de/bin/view/Analysis/ElectronicStructure – Terry Tao Nov 06 '11 at 04:26
  • 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
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    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
  • @TerryTao, I think it's worth remarking also that Goddard-Friesecke's work, being an attempt to introduce an order, still does not give an axiomatization of quantum mechanics. Since they do not define atoms, electrons, shells, etc. And don't introduce them axiomatically. – Sergei Akbarov Sep 16 '17 at 09:06
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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 $2$.

Denis Serre
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    Even if no answer will be satisfactory, I have found the answers to this question to be very interesting... – Tom Church Nov 09 '11 at 16:52
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    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. –  Jun 18 '13 at 14:13
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    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
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    @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.) –  Mar 06 '16 at 19:03
  • According to V.I.Arnold, http://www.springer.com/us/book/9780387968902, Classical Mechanics can be at least considered as an axiomatic system of second order, i.e. an axiomatic system inside another axiomatic system, like General Topology, or Theory of Real numbers, or Probability Theory, etc. Hilbert's axiomatization of Euclid's geometry, https://en.wikipedia.org/wiki/Hilbert%27s_axioms, is another example. – Sergei Akbarov Mar 06 '16 at 19:14
  • As to Probability Theory, I am sorry, this is Mathematics. – Sergei Akbarov Mar 06 '16 at 19:31
  • Anyway, the words "Postulates of quantum mechanics", https://en.wikipedia.org/wiki/Mathematical_formulation_of_quantum_mechanics#Postulates_of_quantum_mechanics, must mean something. If they are not the same as postulates (or axioms) in Mathematics, then there must be an explanation of their meaning. Otherwise this becomes an abuse of the trust of a listener, isn't it? – Sergei Akbarov Mar 06 '16 at 19:41
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    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
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    Now, you might say that post-HF methods are not "proof" because they're based on numerical diagonalizations (in which case one has to wonder why you're OK with HF, which also uses numerical and approximate methods to find the ground states within the Ansatz manifold), and you mathematicians can spend as long as you want debating what makes a "proof" in this regard (and what it is you're "proving", exactly), but what doesn't fly is ignoring fifty+ years of research and pretending physics is still stuck at undergraduate level. – Emilio Pisanty Sep 12 '17 at 16:12
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    @Emilio. I didn't pretend that my answer be the only possible one. Why not write your own answer, that of a Physicist ? It would be valuable. I should be happy to vote for it. – Denis Serre Sep 12 '17 at 18:38
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    I didn't write an answer when I first saw this, and I'm not writing one now, because there isn't really a question. But this answer is misleading at the very best, and it is dangerously incomplete given its visibility. "We do not expect that steady states be really Slater determinants"... so why not talk about the huge range of methods that don't restrict themselves to that manifold? So in e.g. your last paragraph, yes, if you artificially restrict the solution space, you get restricted solutions, but you act like that's surprising in some way? – Emilio Pisanty Sep 12 '17 at 22:39
  • I accepted this answer because it is the most honest: "My opinion is that we are still very far from a mathematical justification." :) The others are a bit different from this point of view. – Sergei Akbarov Sep 16 '17 at 09:13
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    @SergeiAkbarov This answer consists of (i) a personal opinion, followed by (ii) a statement of the problem (that you've elsewhere described as insufficient), and (iii) a misrepresentation of the state of atomic physics, without offering any justification for the personal opinion. The only reason I can see for accepting it is that it agrees with your own existing preconceptions. – Emilio Pisanty Sep 18 '17 at 12:54
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    I believe this answer contains serious misconceptions both about the power of modern computers and about quantum mechanics. If the dimension of the state space was only $3N$, we could actually solve it easily on modern computers. Unfortunately, the state space has dimension exponential in $N$. – Peter Shor Nov 02 '18 at 14:11
  • @EmilioPisanty you can give a better answer here. You can also open a question here at MathOverflow about mathematicians "preconceptions" on axiomatic method. – Sergei Akbarov Jul 01 '22 at 05:58
  • @Sergei Until my technical comments above are addressed, I see no reason to conclude that this discussion is being carried out in good faith, and therefore no reason to engage with it any further. – Emilio Pisanty Jul 01 '22 at 12:39
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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.

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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.

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    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 07:38
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    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 09:19
  • 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 09:22
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    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
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    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
  • I'm not saying that things are simple, not by a long shot, just that the theory under this particular approximation is simple. The model it's derived from is just Schrodinger's equation for an atom, whose details can be found in any quantum mechanics textbook. Assuming no electron repulsion leads to decoupling, and one can then solve according to the theory of the hydrogen atom, also standard. Any textbook will do. – Antoine Levitt Nov 18 '11 at 19:10
  • "Any textbook will do" - this sends us back to the begining of my post. Antoine, you can just try to find a book on quantum mechanics, where there is 1) a system of axioms (or postulates), 2) a phrase like "by atom we mean..." (i.e., a definition of atom), and 3) a Proposition or a Theorem about atoms. Warning: in the Berezin-Shubin book they mention atoms in Proposition 2.2 (chapter 4), but these are atoms in the sense of measure theory, they have nothing in common with physical atoms. – Sergei Akbarov Nov 18 '11 at 20:04
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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:

    http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations#Exceptions

    http://en.wikipedia.org/wiki/Relativistic_quantum_chemistry#Periodic_table_deviations

    – 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
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    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
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    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