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Added 22, November:

I've succeeded in making the question entirely unintelligible with all my additions. So I thought I would summarize it in the simplest form I could manage and add it to the title. The question is thereby somewhat narrower in scope than my original query, but I would be happy to have this focused version answered.

The main point is that the loose answer

(A) because the outer shell is filled

frequently heard is definitely wrong. If we take the usual definition of a shell, all the noble gases but Helium have just the s and p subshells (corresponding to the representations $V_0\otimes S$ and $V_1\otimes S$) filled in the outermost shell, and this is not a full shell once the atom is bigger than Argon. The wikipedia article on noble gases offers an amusing formulation whereby, for a noble gas, 'the outer shell of valence electrons is considered to be "full"' (my emphasis).

All this led to my initial confusion: I thought that the term 'shell' must mean something else for multi-electron systems in a manner adapted to the answer (A). Such an alternative definition does not seem to exist, and it is not at all obvious how to come up with one in a non-tautological way. So I believe the essence of what I am asking is whether or not there is

a natural mathematical explanation for the stability of noble gases

that is more or less independent of experiments confirmed by difficult computations using approximation schemes.

Anyone interested in the background and details is invited to read the incoherent paragraphs below, or to look up a proper reference like Atkin's book on physical chemistry.


Added, 21 November:

Having read some more here and there, I should make some terminological corrections, in case I mislead anyone with my ignorance. I will do so here, and leave the text below as written, in case the temporary confusion is helpful to other interested non-experts.

Firstly, as far as I can tell, the word 'orbital' does seem to refer to a wavefunction, not a representation. So the eigenfunctions in the second occurrence of the representation $V_1\otimes S$ will be called the $3p$-orbitals. That is, the individual wavefunction is an orbital, while the representation itself might be referred to as the so-and-so orbitals, in the plural. The definitive term for any given occurrence of an irreducible representation in $L^2\otimes S$ seems to be subshell.

Secondly, I finally read the wikipedia article on shells. It doesn't help much with my questions, but it does describe the convention regarding the term 'shell'. As far as I can tell, the shells are simply the direct sums of the following form:

K-shell: $1s$

L-shell: $2s\oplus 2p$

M-shell: $3s \oplus 3p \oplus 3d$

N-shell: $4s\oplus 4p \oplus 4d\oplus 4f $

O-shell: $5s\oplus 5p \oplus 5d\oplus 5f\oplus 5g $

and so on. (In case you're wondering, the representations after $V_3\otimes S=f$ are labeled consecutively in the alphabet.) The key point is that, in this form, a shell does not consist of a grouping of subshells of similar energies when more than a few electrons are present. Rather, the conventional description of the phenomena says that more than one shell can be incomplete in an atom. For example, in the fourth row of the periodic table, starting with potassium (K) and up to copper (Cu), they say both the $M$-shell and the $N$-shell are incomplete. If this is confusing to you, I suggest you don't worry about it. I just wanted to point out that my question `what is a shell?' has a clear-cut answer in this usage. If we don't want to go against this convention, we need to stick to the version

`What determines a period?'

I'm very sorry for all the confusion.


Added, 19 November:

After receiving the nice answers from Jeff and Antoine, I realized that I should sharpen the question somewhat more. I like Neil's formulation, but I think I am asking something much more naive. Allow me to start by providing more background for mathematicians whose knowledge is as hazy as mine. As mentioned by Jeff and Antoine, the Hamiltonian for a system of $N$ electrons moving around a nucleus of charge $Z$ looks like

$$H=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N \frac{Ze^2}{r_i}+ \sum \frac{e^2}{r_{ij}}.$$

In principle, one would like to understand the structure of discrete spectrum solutions to the eigenvalue equation $$H\psi=E\psi.$$ One characterizes solutions using labels called 'quantum numbers'. In general, there is an objective label often denoted $l$, the angular momentum quantum number. This comes from the fact that there is an $SO(3)\times SU(2)$ symmetry, breaking up solutions into the irreducible representations $V_l\otimes S$, mentioned earlier. All the vectors in a given irreducible representation must have the same energy level $E$, because a basis for the space can be obtained from a highest weight vector by applying elements of $LieSO(3)$. Any given $V_l\otimes S$ is called a subshell or an orbital (I'm a bit unclear about this because the term 'orbital' is also used for the individual eigenfunctions) and is described using the (historical) labels

$$V_0\otimes S \leftrightarrow s$$ $$V_1\otimes S \leftrightarrow p$$ $$V_2\otimes S \leftrightarrow d$$ $$V_3\otimes S \leftrightarrow f$$

The principal quantum number is determined as follows. We order all the orbitals by energy levels, and the $n$-th time $V_0 \otimes S$ occurs, we call that subspace $ns$. However, the $n$-th time that $V_1\otimes S$ occurs, we call it $(n+1)p$. In general, the $n$-th time $V_l\otimes S$ occurs, the label is

$(n+l+1)$(whatever letter $l$ corresponds to).

For example, the orbital $3d$ is the first occurrence of the representation $V_2\otimes S$. When there is just one electron in a $1/r$ potential, these $n$ really label the order of the energy levels, and $d$ really occurs the first time in the third energy level. This is the reason for the shift in labeling, even though this correspondence with energy levels breaks down for multi-electron systems. In fact, in neutral atoms with $N=Z$, the energy levels are ordered like $$1s<2s<2p<3s<3p<4s<3d <4p<5s<4d<5p<6s< \ldots$$ (It's not supposed to be obvious how to fill in the $\ldots$. However, I've been told that knowing $4s<3d$ is essential to passing A-level chemistry in England.) I hope it is clear from this that the orbitals (or the subshells) are completely well-defined and have a labeling scheme that is a bit odd, but makes sense when the obvious translation is combined with history. So the real question is

What determines a shell?

Some energy-non-decreasing consecutive sequence of subshells is a shell. The shells then determine the rows of the periodic table (if we ignore the added complication that $Z$ is also increasing as we move right and down).

Now, it is tempting to conclude that the standard convention is a bit arbitrary. After all, the chemical similarity of the columns could possibly be explained simply by the fact that

they have the same representation of $SO(3)\times SU(2)$ at the uppermost energy level, and the same number of electrons in this representation

without any reference to an outermost shell at all. This pattern can be easily seen by the arrangement into blocks shown here. And then, unlike the $N=1$ case, the energy levels in one shell are not even the same, just rather close to each other.

Unfortunately, the grouping into rows represents real phenomena. This comes out very clearly, for example, in the graph of ionization energies, with the noticeable peaks at the end of the rows immediately followed by precipitous drops. So I believe a bit of thought reduces a good deal of my original question to two parts:

  1. Is there some reason for a big gap in ionization energy when one moves to an $s$-orbital?

  2. Can one show that two orbitals of the same type are necessarily separated by a huge energy gap, so that it is highly unlikely for them to occur in the same shell?

1 and 2 together make it natural that the dimensions we see in each shell will be of the form

2, 2+6, 2+6+10, 2+6+10+14, etc,

even in the general case. Of course, this doesn't say anything about how many times each combination is likely to occur.

In any case, I hope there is a mathematical answer to these two questions that doesn't involve a full-blown programme in hard analysis.

At the more speculative end, one might analyze a family of operators like $$H_{\epsilon}=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N\frac{Ze^2}{r_i}+\epsilon \sum \frac{e^2}{r_{ij}}.$$ Is there a way to see the numbers we see occurring via the spectral flow of this family as we go from $\epsilon =0 $ to $\epsilon=1$? But maybe this is just as difficult as giving a full account of the structures using analysis.

By the way, I certainly wouldn't like to complain, but it is a bit puzzling to me why some people regard this question as inappropriate for the site. Since I don't keep too well in touch with cultural trends in the mathematical world, maybe I am unaware of how much things have changed since I was a Ph.D. student. In those days, a programme like

Prove the stability of matter, based only on the Schroedinger equation

was regarded as an example of an important mathematical problem motivated by atomic structure, tackled by people like Fefferman and Lieb. The questions I ask here are hopefully much easier, but still research-level mathematics to my mind.


Original question:

This question prompts me to ask something more specific about the periodic table. As far as I know, the main significance of the periodic table is that

The elements in the same column have similar chemical properties.

For example, the noble gases at the far right are all pretty stable on their own. The explanation for this is that they have (in the neutral state) a full outermost shell. Now my question is

Is there an explanation, at some reasonable level of mathematical rigor, of when a new shell starts?

That is, where do the lengths of the periods

2, 8, 8, 18, 18, 32, 32

come from? I confess I've been puzzled by this ever since my university physics course.

Allow me to pinpoint my confusion a bit more. I understand that there are some numbers that are important in atomic structure, and these are 2, 6, 10, 14, and so on. This is because of the occurrence of the representations

$$V_l\otimes S$$

inside the Hilbert space for a single particle moving in a central potential. These are the orbitals one hears about in physical chemistry courses. Here, the $V_l$ are the representations of $SO(3)$ of odd-dimension $2l+l$, while $S$ is the standard two-dim representation of $SU(2)$. Since all the states in a single orbital have the same energy, it is natural that the breaks will occur after some collection of orbitals are all filled.

Thus, for a hydrogen atom, the successive shells have dimensions

$2n^2$

for $n=1, 2, 3, \ldots$ because the representations $V_l\otimes S$ for $l=1, 2,\ldots, n-1$ occur each with multiplicity one inside the $n$-th shell.

So if the periods in the table were of lengths

2, 8, 18, 32,

I would have vaguely assumed that the pattern of shells even in general looks like that of the hydrogen atom. But of course, among the known elements, each period length that occurs in the hydrogen atom is repeated twice, except for the first one. So a more precise question is

Is there a reasonable mathematical explanation of this `multiplicity two' of the periods?

The little I recall of discussions in standard textbooks were quite unclear. There are various rules by the name of Hund's rule, the Aufbau principle, and so on, but I couldn't gather from any of them

*Where the breaks should occur. *

What I do see is that periods end when the orbitals 1p, 2p, 3p, etc. get filled following the Aufbau principle. (Here, p is the chemist's label for $V_1\otimes S$.) So perhaps another version of the question is

Is there some reason that the p-orbitals mark the end of the periods?

To be honest, I've never understood the Aufbau principle either, because I don't know the rationale behind the principal quantum number for the larger atoms. That is, the number 4 in the orbital 4p refers to the 4-th energy level in the case of the hydrogen atom. But for larger nuclei, the $n $ in orbital '$np$' does not refer to the energy level. (This discrepancy is in fact implied by the Aufbau principle.) So what is the significance of the $n$ in general that enables them to play some role in a physical principle?

I realize this question is becoming incoherent already. Nevertheless, I would very much appreciate clarification on any sensible version of it at a level of mathematical rigor of your choice. (I am not asking for any axiomatics.) Pointers to an accessible reference would be equally welcome.


As with the earlier question, a word of explanation is in order on the decision to post this on Math Overflow. I will draw upon an analogy I read long ago in an article of George Mackey's that went something like this: Say my mother tongue is Korean. If I would really like to use English fluently, it is probably best eventually to learn from native speakers of English. On the other hand, if I would like a good translation into Korean of English literature, it is better to consult an educated Korean who knows a lot about English. Of course answers from real physicists will be very gratefully received, especially if they bear in mind that the query comes from someone who struggles against a serious language handicap.

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The word of explanation is nice, but I still don't think this has anything to do with the subject of this group, so I am voting to close. –  Igor Rivin Nov 18 '11 at 16:26
    
You have not taken spin into account. Each of your states appears twice because there are two values for the spin. I am also voting to close because I don't think this is the right site. –  Bruce Westbury Nov 18 '11 at 16:31
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Oh well, I understand how it can appear that way. I think I was hoping for some reason based purely upon representation theory, making it suitable for this site. Certainly there must be many mathematical physicists here that know something about this. Bruce: I did take spin into account, as you can see from the description of the orbitals. I'm asking about the double occurrence of the periods, not the even dimensions of the orbitals. –  Minhyong Kim Nov 18 '11 at 16:36
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I would summarise the problem as follows: what is the rigorous mathematical question for which the combinatorial structure of the periodic table is the right answer? I think this is an excellent question, and I would love to know the answer. I also think it is fully appropriate for mathoverflow. –  Neil Strickland Nov 18 '11 at 17:03
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Antoine: Many thanks for your input. Actually even in the precise model, the angular momentum quantum number is well-defined and has physical meaning. In my addition on 19 November, you'll see the precise general definition of the principal quantum number, which I think agrees with standard usage. I agree that this number does not label energy levels in multi-electron systems. You could say that this whole long question is running around this last thorny issue. –  Minhyong Kim Nov 22 '11 at 1:35

3 Answers 3

The question is probably not appropriate for MO, but it's interesting nevertheless. I think (but have no proper reference nor physical knowledge for this) the reason for these different shells is that, loosely speaking, the inner electrons screen the potential, such that the outer electrons see an effective potential of charge Z - Ninner. This shifts the eigenvalues, and accordingly the order in which the shells are filled. This is probably discussed in quantum chemistry books, although I haven't found anything satisfactory yet.

The appropriate theory to discuss this is Hartree-Fock.

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You are completely correct in your analysis of the structure one obtains by considering the Schrodinger equation for Z electrons in a central potential due to a nucleus of charge Ze when the Coulomb interaction between electrons as well as relativistic effects such as spin-orbit coupling are ignored. In such a model the periods would indeed be 2, 12, 18 etc. for exactly the reasons you have described. In physics jargon the energy in this model depends only on the principal quantum number $n \in {\mathbb Z}, n>0$ and the allowed $\ell$ values are $\ell\le n-1$. Orbitals with $\ell=0,1,2,3,4, \cdots$ are labelled by $s,p,d,f \cdots$ for historical reasons. Thus at $n=1$ one can have one or two states in the $(1s)$ orbital (accounting for spin), at $n=2$ one has the $(2s)$ and $(2p)$ orbitals with $2(1+3)=8$ states. And at $n=3$ one should have $18$ states by filling the $(3s),(3p),(3d)$ orbitals. But in the real world this is not what happens. This simple model does not give a correct description of atoms in the real world once you get past Argon. I believe the main effect leading to the breakdown is the Coulomb interaction between the electrons.

So there is no simple mathematical model based say just on the representation theory of $SU(2)$ and simple solutions to the Schrodinger equation which will account for the structure of the periodic table past Argon. However one could ask whether including Coulomb interactions between electrons does gives a model which correctly reproduces the next few rows of the periodic table past Argon. I am not an expert on this, but since I doubt there are physical chemists on MO I'll just give my rough sense of things.

To approach this problem with some level of rigor probably requires difficult numerical work and my impression is that this is beyond the current state of the art. However there are rough models which try to approximate what is going on by assuming that the interactions between electrons can be replaced by a spherically symmetric potential which is no longer of the $1/r$ form. This leaves the shell structure as is, but can change the ordering of which shells are filled first. In such a model instead of filling the $(3d)$ shell after Argon one starts to fill the $(4s)$ and $(3d)$ shells in a somewhat complicated order. Eventually one fills the $(4s),(3d),(4p)$ shells and this leads to the line of the periodic table starting at K and ending at Krypton.

Added note: There is one nice piece of mathematics associated with this problem that I should have mentioned, even if it doesn't by itself explain the detailed structure of the periodic table. When Coulomb interactions between electrons and relativistic effects are ignored the energy levels of the Schrodinger equation with a central $1/r$ potential depend only the quantum number $n$, but not on the quantum number $\ell$ which determines the representation $V_\ell$ of $SO(3)$ referred to above. When screening is included this is no longer the case and the energies depend on both $n$ and $\ell$. Why is this? With a $1/r$ central potential there is an additional vector $\vec D$ which commutes with the Hamiltonian. Classically this vector is the Runge-Lenz vector and its conservation explains why the perihelion of elliptical orbits in a $1/r$ potential do not precess. Quantum mechanically the commutation relations of the operators $\vec D$ along with the angular momentum operators $\vec L$ are those of the Lie algebra of $SO(4)$ (for bound states with negative energy). There are two Casimir invariants, one vanishes and the other is proportional to the energy. As a result the energy spectrum depends only on $n$ and can be computed using group theory without ever solving the Schrodinger equation explicitly. Perturbations due to screening, that is from some averaged effect of the Coulomb interactions between electrons, change the $1/r$ potential to some more general function of $r$ and break the symmetry generated by the Runge-Lenz vector $\vec D$. As a result the energy levels depend on both $n$ and $\ell$.

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Thanks very much for this explanation. It is reassuring to know one is not missing something completely simple. I have added something to the question. I don't want to take up your time, but any comments on it would be very much appreciated. –  Minhyong Kim Nov 19 '11 at 7:00
    
I added a note in my answer. –  Jeff Harvey Nov 19 '11 at 14:34
    
Thanks a lot. That's also very helpful. –  Minhyong Kim Nov 19 '11 at 15:03

I will rearrange the questions you asked into a language I prefer. As Jeff Harvey mentioned, there doesn't seem to be a mathematically precise solution, but I think there are "chemically rigorous" arguments in terms of screening.

  • Why the do the high angular momentum orbitals (in a fixed principal energy level) have higher energy than low angular momentum orbitals?

(revised 21 Nov.) One explanation is that in the no-Coulomb-repulsion limit, an electron in a high angular momentum orbital comes close to the nucleus much less often than an electron in a low angular momentum orbital. There is a nice set of pictures of orbital densities in the Wikipedia article on atomic orbitals where you can get an intuitive idea about the radial distribution.

When Coulomb repulsion is turned off, each electron sees a potential function $Z/r$, and the energy of an orbital depends only on the principal quantum number, not the angular momentum. Suppose you turn on Coulomb repulsion in an approximate way, by assuming you have an ion with $k$ electrons with some a priori charge density, and then add an electron. (In particular, we ignore any effect the extra electron has on the electrons already there, together with the fact that electrons don't really have well-defined individuality.) The screening effect then causes the potential energy function to interpolate between $(Z-k)/r$ far away and $Z/r$ near the nucleus. If we assume the wavefunctions of orbitals don't change too much under this perturbation of potential, we may attempt to compare the wavefunctions of two orbitals with the same principal quantum number but different angular momenta. Unless the interpolation is quite exotic, we should expect the orbital with more density near the center to have lower potential energy.

  • Why is it that the 3d level is higher-energy than the 4s level, and more generally, why do the energy levels have a progression of (energy, angular momentum) values like $(2m,0), (m+1, m), (m+2, m-1), \ldots (2m,1)$ and $(2m+1,0), (m+2, m), \ldots, (2m+1,1)$?

This is a quantitative version of the previous question, which I don't know how to answer in a satisfactory way. Experimentally, it seems that the d and f orbitals are energetically close to the corresponding s orbital, since one often sees s electrons stolen by seemingly higher-energy orbitals in cases like copper, silver, gold, palladium, etc. This suggests that numerical approximations need high precision.

  • Why do rows end with full p orbitals?

The choice of "shell being full" at angular momentum eigenvalue $\ell=1$ is a convention derived from the experimental fact that the corresponding elements are relatively chemically inert. To treat that mathematically, you have to approximate ionization energies and electronegativity (which requires some modeling of bonding behavior). It seems to happen that p-orbital electrons are hard to steal, and s-orbital electrons are not so well-attached.

A rough a posteriori explanation is that for fixed $n$, the energy gap between $(n,1)$ and $(n+1,0)$ is comparatively large and robust under perturbations like adding Coulomb repulsion terms of electrons of level less than $n$. This gap then becomes robust under change of row in the table, once you assume the energies of all other angular momentum eigenvalues fall into the gap between $(n,0)$ and $(n,1)$, as given in the previous question.

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Thank you for your nice answer. There is something I don't understand about the explanation towards the beginning however. It makes intuitive sense, but recall that in the $1/r$ case with one electron, the actual energy does not change as you move to higher angular momentum within a shell. –  Minhyong Kim Nov 19 '11 at 15:10
    
Indeed. The statement "high angular momentum orbitals (in a fixed principal energy level) have higher energy than low angular momentum orbitals" is not correct in the no-Coulomb repulsion limit as I explained in my added note. –  Jeff Harvey Nov 19 '11 at 15:30
    
Scott: Thanks for the revision. That makes a lot of sense. –  Minhyong Kim Nov 22 '11 at 1:30

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