# Atomic Shell Theory: Bohr-Sommerfeld model

1. Aug 8, 2011

### hbaromega

This is a very naive question. But I think, it's an important point that has been unattempted in textbooks. The question is:

How far should one trust the Bohr-Sommerfeld model or the atomic shell theory for all elements in the periodic table?

This question generally comes in mind, since we know that the Bohr's model was a kind of hypothesis or ansatz to explain the hydrogen atom spectra. And Sommerfeld modified the quantization condition in a spirit of generalization, using the analogy to the planetary motion under the central force due the sun. These all can be put together as the initial development of quantum mechanics and often regarded as theories of Old Quantum Mechanics.

Now we know that the H-atom can be exactly solved from the Schroedinger's equation. But what about atoms with higher atomic numbers (may be we can only expect hydrogen-like wavefunctions for alkali atoms)? Should Bohr-Sommerfeld model (apart from the relativistic correction) still be working ?

Then why do people say that in cuprates copper has d9 electronic state? Doesn't that sound imprecise?

Thanks.

2. Aug 8, 2011

### DiracRules

It depends on the precision you want.

First of all, you must know that Schroedinger's equation cannot be solved exactly for atoms different from hydrogen atom.

Actually, one of the best representations of the atoms of higher atomic number is given by the Hartree theory, which tries to approximate Schroedinger's equation by making hypothesis about the potential seen by the electrons.

Next, what do you mean when you say "How far should one trust the Bohr-Sommerfeld model or the atomic shell theory for all elements in the periodic table?" ?

You know, the Bohr model is an oversimplified model of the atom, and in the new quantum mechanic it is replaced by the model deduced by Schroedinger's equation, in which the electrons are replaced by waveforms.

Moreover, Bohr's model cannot account for effects related to the filling of the subshells, that is important to build the periodic table of elements or for a finer structure of the spectrum of atoms than the one studied by Bohr, and to the spin-orbit interaction: an atom immersed in a magnetic field has different levels of energy than a free atom because of the spin-orbit interaction.

Bohr's model works for rough calculations but, as I said, for finer calculations and for to have a (better) explanation of a large variety of phenomena, not explained by Bohr's model because, as you said, it is just an hypothesis (that works) built upon experimental data.

3. Aug 8, 2011

### hbaromega

I think, if you can answer my last question, it'll be clear.

How can one say Cu has d9 electronic state in cuprates? Or does it need to have d-state at all?

Let me put a few more questions?

1) Are there experiments that can see real electronic orbitals of H-atom? Note that the orbital photographs seen in most textbooks were engineeringly plotted by H. E. White (no real hydrogen atom experiments done).

Ref. http://prola.aps.org/abstract/PR/v37/i11/p1416_1

2) How can we define s, p, d orbitals for non-hydrogen atoms? We may have s',p', d' orbitals that may hold totally different kind of geometries.

4. Aug 8, 2011

### DiracRules

Actually it's the first time I hear about cuprate.
However, the fact is that the atomic configuration of an atom is not fixed, but it can change if a new configuration with lower energy is available.
I don't know (if you have some links please post me; I will search for myself) actually the structure of cuprate but, since the energies of the outer subshells narrows as you go away from hydrogen atom, it is possible that copper configuration in cuprate is more stable and less energetic if it is in 3d9 state.

As regards the photos or images of electron orbitals, last month a group of researchers at Politecnico di Milano (Italy) managed to "take a photo" of the orbitals in a molecule. I can't find the article in English, but http://www.galileonet.it/articles/4e2fdf4d72b7ab4b16000090" it is in Italian (maybe you can get it translated by google), with some photos.
I must remind you that after Schroedinger, it's almost wrong to think about orbitals as "paths": it is more correct if you think about where it is more probable to find an electron.

We can define the subshells from their energy. Dirac derived a formula that corrects Bohr's taking in account spin orbital interaction and other interactions in the second order. In this formula are present the quantum numbers n,l,j (where j is the total angular moment)

Actually, I've just read on Wikipedia that in cuprate copper is Cu++ and O--, so Cu++ is [Ar]3d9 because 2e- have been removed by the more electronegative oxygen.

Last edited by a moderator: Apr 26, 2017
5. Aug 8, 2011

### hbaromega

Actually it's the first time I hear about cuprate.
However, the fact is that the atomic configuration of an atom is not fixed, but it can change if a new configuration with lower energy is available.

It's not only about cuprates.We always find the oxidation state and then decide the electronic configuration, purely from the chemistry point of view. My question is:

1) Is there any physics explanation behind the atomic configuration of non-hydrogen atoms?

As regards the photos or images of electron orbitals, last month a group of researchers at Politecnico di Milano (Italy) managed to "take a photo" of the orbitals in a molecule. I can't find the article in English, but here it is in Italian (maybe you can get it translated by google), with some photos.
I must remind you that after Schroedinger, it's almost wrong to think about orbitals as "paths": it is more correct if you think about where it is more probable to find an electron.

This nature paper has the experimental detail. By orbitals I meant the probability density distribution, not paths (see the H. E. White's paper). And as I said, I wish to see the proof of electron's probability density distribution for the H-atom (apart from the indirect way from the spectroscopy results) that we solve through Schroedinger's equation in the textbooks.

Can you see the gap between chemistry and physics? I can believe atomic number, but how can I believe the atomic configuration and hence the s,p,d,f nomenclature?

We can define the subshells from their energy. Dirac derived a formula that corrects Bohr's taking in account spin orbital interaction and other interactions in the second order. In this formula are present the quantum numbers n,l,j (where j is the total angular moment)

Did you mean subshells for the Bohr-Sommerfeld theory, i.e. the old quantum mechanics?
I know Dirac's relativistic correction and that is for the H-atom.

Actually, I've just read on Wikipedia that in cuprate copper is Cu++ and O--, so Cu++ is [Ar]3d9 because 2e- have been removed by the more electronegative oxygen.

Again, this is ridiculous. Electrons are not removed. We can say, there's no significant probability density. But in what region? I can think of chemical bonding as overlap or distribution of electronic wavefunction.

Now the questions:

1) How can I calculate the atomic wavefunction or the energy (i. e. eigenvalue) of a non-hydrogen atom?

2) Even if I succeed to do for atoms, how should I calculate the redistribution of wavefunctions when they form a molecule?

6. Aug 8, 2011

### cgk

Both of those are done using techniques of Quantum Chemistry. This is basically hardcore numerical quantum many body physics, and there are (large and very complex) software packages which can do such calculations (e.g., Molpro, CFOUR and MRCC). Using these techniques, relative energies and properties can be calculated to around 0.1 kJ/mol ... 4 kJ/mol (depending on your patience. 1 kJ/mol is about 0.010 eV) for sufficiently friendly atoms and small molecules (e.g., energy differences between states, between products, transition states, and educts of a chemical reaction etc.). Unfortunatelly, understanding how these many-body method work require an intimate knowledge of quantum mechanics, many-body theory, and numerics, and this is not easily explained. Some terms to get you started are "Coupled cluster theory", "Multirefernce configuration interaction", "correlation consistent basis set" and "basis set extrapolation".

7. Aug 9, 2011

### hbaromega

As far as I know, the quantum chemistry methods use Hartree-Fock or post-Hartree-Fock methods that use linear combinations of atomic orbitals and again those orbitals are treated like s, p, d kind of orbitals, which are hard to accept.

Hope you got my point. Again my question is very simple:

How can we believe that s, p, d, ... orbitals exist beyond hydrogen atom? Any experimental or theoretical evidence?

8. Aug 9, 2011

### abhi2005singh

Well, one can also ask: "How do we know that these orbitals exist for H atom?". The only answer which comes to mind is that they are predicted using theory and that no experimental evidence is available against it. The theory explains well the experimental facts. This is true not only in the present case but for any theory in general.

For atoms with Z > 1, as mentioned previously, HF theory is used (other approaches are also available) which assumes existence of such orbitals. The results of these calculations predict some results. If those results are proved experimentally, then the theory stands.

In short, if the experimental results can be explained using the assumption of existence of such orbitals, then those experimental results can be treated as the proof for the existence of these orbitals.

9. Aug 9, 2011

### DiracRules

If subshells exists for multielectron atoms, it will result in atomic spectra: there are transition rules that allow only certain decay. You can look very carefully at the spectrum and calculate the energy of the gap.

I'm quite sure it happens, and they already found evidence, but I cannot be sure 100% having not my book under hand :D

Moreover, probably this is a stronger evidence: they found two types of helium, ortohelium and parahelium, with different physical properties. This is due to the fact that parahelium has both the electron in state 1s2 (a state of singulet), whilst ortohelium is in the state 1s1 2s2 (a state of triplet). The transition rules shows that is impossible for optical transition to transform a singulet state to a triplet state and vice versa.
I think this is a proof of the existence of subshells, or not?

Another proof could that the energy levels within each subshell are very narrow, while between shells are broader, and you can see this when you ionize the atom.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html" [Broken]
http://ibchem.com/IB/ibnotes/full/ato_htm/12.2.htm" [Broken]

I'm sorry for my poor speech: "because 2e- have been removed by the more electronegative oxygen." Obviously electrons aren't removed in the meaning we give to this word :D

You are right, sorry :D

Actually I can't, probably because I don't like chemistry too much :D

When I treated Hartree-Fock method, we studied that it was based on Schroedinger's equation. That is, Hartree hypothesized the effective potential Z(r) and put it into Schroedinger's equation. Through iterations, he arrived to a waveform that was in accord with experimental evidence. In this waveform, we have subshells.

Last edited by a moderator: May 5, 2017
10. Aug 9, 2011

### hbaromega

@DiracRules

Sorry. A severe objection. We can think of subshells. But they don't need to have the same geometry of the hydrogen atom s,p,d orbitals.

Moreover, transition doesn't show the nature of the orbitals. It just reflects the difference between two energy levels.

So it seems that we are still using the Bohr's hypothesis (a bit modification due to Sommerfeld) to decide the electronic configuration of elements. So it's an empirical fitting of the spectroscopy results to the Bohr-Sommerfeld energy spectra formula, which is true only for the hydrogen atom.

It's strange to see that shell theory (old quantum mechanics) still works for many elements. And there exists no true quantum theory to support or deny the theory.

So I can say, we still trust the old quantum theory (since in modern quantum mechanics, non-hydrogen atom problem is non-trivial). In that sense it's as good/bad as we have the Boyle's law to describe gases.

Hope you'll agree with my point.

11. Aug 9, 2011

### hbaromega

See, the Bohr's model as an ansatz gives correct values for the hydrogen atom spectra (apart from the fine-structure correction). And solving Schroedinger equation for the H-atom reproduces Bohr's formula. In that sense, we have a microscopic theory that supports Bohr's formula for the H-atom. But there's no such quantum theory that support Bohr-Sommerfeld formula for other atoms. However, Bohr-Sommerfeld formula seems to work well (I hope so though I never have tested). It's similar to the shell theory that works in nuclear physics.

So my point is that we still use an ansatz (which is not true quantum mechanics) to infer
the atomic orbitals and hence the elctronic configurations.

12. Aug 9, 2011

### abhi2005singh

No, it doesn't work well. There are many things which are not explained: http://en.wikipedia.org/wiki/Bohr_model#Shortcomings"

This is also not correct. We are not using the atomic orbitals because of the BS model. These orbitals follow from quantum mechanics. Please read what @DiracRules have written in this regard.

Transition do show the nature of the orbitals. They tell you about the symmetry of the orbitals by utilizing the selection rules, which in most cases is good enough to infer the initial and final state orbitals.

No ways. Spectroscopy cannot be described using the BS model and no one actually use it in the field of spectroscopy, apart from obtaining some initial guesses. This is specially true for elements with Z > 1.

Last edited by a moderator: Apr 26, 2017
13. Aug 9, 2011

### hbaromega

I think, s,p,d orbitals were defined only for the hydrogen or hydrogen-like (alkali) atoms.
How do you determine probability densities for other atoms?

I guess, the wiki link tells about shortcoming of the Bohr's model, not about the BS model.

How can you be sure there are no other orbitals 'except' s,p,d,f,g, etc? Note that all these orbital geometries are defined from hydrogen atom wavefunctions. Also remember that in the hydrogen atom problem in quantum mechanics, we can actually separate the differential equation into radial and spherical harmonics part. It may not be possible to do the same for other atoms.

In atomic spectroscopy, I guess, we look at emission lines. Or may be absorption in some cases, I'm not sure. Now probably I can calculate the wavelengths of those lines and hence differences between energy levels. Now could you tell me, what formula I'm going to use next and how I connect to symmetry of orbitals from these?

Last edited by a moderator: Apr 26, 2017
14. Aug 9, 2011

### DiracRules

I have to correct what I said previously and say that you are right saying that we cannot do the same for multielectron atoms if we want to solve it exactly. The problem is the presence of the potential interaction between electrons.
However, this problem is, in a certain way, overcome by making hypothesis about the potential that affects the electrons ( Z(r) ). By making this hypothesis, we can split the equation in an angular and radial part.
I realize this cannot satisfy you, but it's the way things are going since 1930! If it was wrong, it would have been corrected :D

If you do not think that approximation is a sort of trickery that hides true reality, think of this: physicists approximated the potential interaction between electrons in the way I told you also to treat identical particles, and to take into account the spin. And it works very well too :D

But I think that the one of the best way to "believe" into the existence of subshells comes from spectroscopy.

From spectroscopy, you can see that each level is split into other sublevel: for example, transition rules says that you can't have a transition 3s->2s but 3p->2s or 3d->2p.
Now, since $\Delta E_{3p\rightarrow 2s}$ is different from $\Delta E_{3d\rightarrow 2p}$, you can't explain this if you don't consider subshells into your model.

Now you can say, why do we not consider subshells as proper shells?
The mathematical answer can be "because they share the same principal quantum number n".
The experimental answer can be "because going up through the periodic table, we see that removing certain electrons requires more energy than the trend. We say thus that we passed into another shell".

15. Aug 9, 2011

### hbaromega

Now the discussion has become pretty interesting.

If an approximation works, then there should be arguments for it. If we lack reasoning, then it's just a fitting formula.

I find no reasoning to extend the same hydrogen atomic orbitals (electronic probability densities) and corresponding energy eigen values for other atoms.

It may be satisfactory for chemists, but not for physicists.

Question is : While calculating the energy of the levels (s, p, d, whatever), are we using the hydrogen-atom energy eigenvalue formula? Don't we require to modify that since many electrons are talking to each other now?

Again filling the shells by putting electrons one by one doesn't make any physics sense.
Electrons are always interacting and they are in a cloud, not in isolated islands.

I believe, this is a serious issue which has been neglected since 1930. We should try to solve two, three, four atoms problems numerically and figure out true atomic orbitals for them.

And even if a large community accepts a theory, without a justification, the theory cannot be regarded as the correct or true theory.

16. Aug 9, 2011

### abhi2005singh

There is not much difference between the two.
It has been working so far. Do not freak out if someone gives you this answer. This is true with every theory/model.
Most of the theories (if not all) are "fitting formula". If the fitting works for all/most cases, then it is a good theory/model.
This is done in all the multi-fermion cases. If this does not make sense to you in atomic theory, then this should not make sense to you in all branches of Physics involving fermions.
People have tried, they do not know enough mathematics.

Again, answer to all your questions seems to be that "it works".

17. Aug 9, 2011

### cgk

hbaromega.
you are confusing lots of things. (I admit it's simple with this topic, as there is lots of information around which is misleading or plainly wrong, even in textbooks). Let's clarify some things:

1) The spherical harmonics (s,p,d,f,...) have /nothing/ to do with the hydrogen atom. These are the solutions of the homogenous angular Laplace equation, and they form (exactly) the angular part of any Laplace problem with a spherical potential. If a potential is slightly non-spherical (think of a atom in a molecule), it is often a good idea to expand the solutions into spherical harmonics anyway, because these form a convenient complete orthonormal set of functions on the sphere (i.e., for angular functions). Note that this is /NOT/ an approximation as long as the expansion is not truncated (complete orthonormal set!). In practice that means for example that the orbitals of the O atom in H2O have not only s and p character, but also a bit of d, of f, of g etc character, with increasingly negligible weight. This is realized in practical calculations by using systematic series of basis sets for expaning the one-particle wave functions (for example, the mentioned correlation consistent basis sets). By doing that you can approach the infinite basis limit to any degree you like.

2) There are no "true orbitals" for systems with more than one electron. Orbitals are, by definition, ONE PARTICLE wave functions. They are DEFINED in terms of some kind of Hartree-Fock or Kohn-Sham mean field picture (there are also "natural orbitals", which are something different, but let's ignore them for now). Orbitals are not true wave functions either[1]. Rather, they are used to BUILD wave functions by plugging them into Slater determinants or configuration state functions.

3) Of course 3--6 electron systems can be calculated effectively exactly, and this has been done long time ago. Some people are still doing it now. The most accurate methods for that are typically of the variational monte carlo or iterative complent class. For systems with more than 6 electrons the mentioned quantum chemistry methods come into play. Note that also these methods are accurate to more than 0.01% in total energies, often much more (1 kJ/mol is a VERY VERY SMALL energy compared to the total energies of the systems calculated! Even Hartree-Fock typically gets total energies right to around 1%).

4) The aufbau principle and shell filling you mention is based on the Hund rules, which were originally empirically derived. Nevertheless, the theoretical reasoning behind them is sound (go look up Hund rules in Wikipedia), and almost always works. If you don't like the reasoning, there is no reason to believe it: You can /calculate/ the energies of the different states (using quantum chemistry) and see that it works. You don't have to believe it, you can test it.

5) Ab initio quantum chemistry is not an empirical science. It calculates properties of molecules using /nothing/ as input execept for the Schroedinger equation and fundamental constants (hbar, electron mass, electron charge etc.). It can calculate true wave functions to any desired degree of accuracy, the limits are only given by the computational power which you are willing to put into it.

[1] (and never probability densities. There abs square gives a density, but not the orbitals themselves),

18. Aug 10, 2011

### hbaromega

I think, hydrogen atom and one electron atom are equivalent. Now once you put one extra electron, trouble enters. Now the Schroedinger equation has two coordinates: $r_1$ and $r_2$, and due to Coulomb repulsion they are coupled, i.e. we have a term like $$e^2/|r_{1}-r_{2}|$$. So unless we get rid of electronic repulsion, we cannot do separation of variables and hence cannot have the spherical harmonics for the angular part.

Before we go to H2O, I am curious about knowing orbitals of O (oxygen).

I agree. But again the single-body wavefunction in the Hartree-Fock or the Slater determinants are ambiguous (even the shape).

I must underline the word effectively. I'm just curious to know, did the results show (effective) orbitals? How do they look like? Could you provide some references where I can see that?

I agree that it was originally empirically derived. But probably Schroedinger or Hartree-Fock equation cannot take care of that.

How does the ab initio method distinguish a hydrogen and a oxygen atom? Note that I'm only interested about orbitals in an atom, not in a molecule.

I didn't get what you wanted to mean here.

19. Aug 10, 2011

### cgk

hbaromega,
before we go on with this discussion, you should look up the following points:
1) What is a Slater determinant?
2) What is the difference between a N-electron wave function (e.g., a Slater determinant) and a one-electron wave function (e.g., a Hartree-Fock orbital)
3) What is Hartree-Fock and how does it allow you to get an approximate N-electron wave function (a Slater determinant) using one-particle wave functions (occupied HF orbitals)?

I'm not saying this to tease you. Understanding these points is absolutely essential to understand the answers to the questions you are asking, or to understand why other questions you are asking don't make sense.

I just told you that the spherical harmonics have nothing to do with hydrogen or the number of electrons. They are related to the spherical symmetry of the problem, nothing more. In particular, they form the angular part of solutions of any one-particle Laplace equation (e.g., a one-particle Schrödinger equation) with a spherical potential. The one-particle Schrödinger equations in this case are the Hartree-Fock equations for the orbitals, which for spherical systems (e.g. atoms like N) feature an spherically symmetric potential: The sum of the nuclear attraction potential (a spherically symmetric one-particle potential) and the mean field repulsion of the electrons (also a spherically symmetric one-particle potential)[1]. Both these potentials and the orbitals themselves are determined self-consistently in the Hartree-Fock procedure. Please, forget everything about hydrogen or "hydrogen-like atoms". These solutions do not help in any way in understanding the many-electron problem.

[1] In reality it's a bit more complicated due to the presence of the exchange potential, but this is not essential for understanding this point.

No, they are not. They are determined uniquely (up to unitary transformations amongst themselves) by the Hartree-Fock solution. In particular, for any ground state spherical atom, there is only one set of canonical Hartree-Fock orbitals. The orbitals are determined by Hartree-Fock, they are the OUTPUT of the Hartree-Fock calculation.

Being able to calculate something to any desired accuracy is equivalent to being able to calculate it exactly. And no, these most accurate methods for 2--6 electron systems do not use orbital expansions but rather direct real space wave function ansaetze. Asking about orbitals for such wave functions is not the right question.

Of course they can. I just told you that. If you don't believe the empirical occupation rules, you can calculate the state energies for the other, non-Hund occupations, and then see whether or not they are lower in energy than the Hund-occupation solutions. In almost all cases they are not.

H and O have different nuclear potentials (one 1/r, the other 8/r) and a different number of electrons. That are (formally) the only two inputs to HF.

20. Aug 11, 2011

### hbaromega

1) I know the Slater determinant. It is a many-body wave function constructed upon single-particle wave function so that the Pauli exclusion principle (or the exchange interaction) is taken care (using the property of exchanging row/column of a determinant).

2) and 3) I think, we get Hartree-Fock (HF) equation by minimizing the Slater determinant and HF wave functions satisfy the HF equations.

Tell me anything further I need to know.
Did you mean self-consistent Hartree-Fock? And do you start with trial single-body wave functions, e,g, Gaussians?

Sorry, how do you check the accuracy? I thought, once you know the wave function, you can construct the orbitals (sorry again H-atom analogy).

Once we have Hartree-Fock wavefunctions, are not their energy-levels defined? But Hunds rule is not included inside the Hartree-Fock!