As you know, the Schrodinger equation only explains hydrogen atom by solving its radial wave equation. But then the same method is assumed to work also for atoms other than hydrogen. The actual spherical solution of Schrodinger equtation does not agree with any experiment. If you wish, you can find my paper discussing this issue. Best wishes

This is standard quantum mechanics. The Schrödinger equation is the equation for the wavefunction of a quantum system. It can be written for *any* quantum system, it can be solved, explicitly, under certain assumptions, that can be tested by numerical methods to any degree of precision.

These facts can be found in any textbook and many lecture notes.

https://icols.berkeley.edu/sites/default/files/pdf/Udem.pdf

Hi victor and Stam,

I have been teaching quantum mechanics for 20 years and I am very familiar with most of quantum mechanics textbooks, none of these textbooks discussed the above question.

Do you think that the obtained quantum numbers for H-atom are going to be the same for other atoms. I do not think so, therefore, why we use them to find the electron configuration?

What is a well known fact is that, in a confining potential the energy levels are discrete, labeled by a non-negative integer, n and that the angular dependence can be expanded in spherical harmonics, that are labeled by two integers, l and m. So for *any* given electron, in an atom this is true.

Now for an atom more complicated than the hydrogen atom, one needs to work a bit to set up the calculation, because the wave function doesn't factorize over the electrons. But the two issues are distinct. How the factorization can be realized in practice is a separate issue from that of what are the labels of the states.

Dear Imad,

atomic physics beyond the hydrogen atom is indeed complex. The books I'd recommend are Bransden (& Joachain) for a not so introductory but still not very advanced level. The most advanced (and extremely well written) I've come across is the book by Cowan (1981, I think). The codes that were written at the time are still employed by those dealing with atomic (such as e.g. core level) spectroscopy.

The idea behind sticking with the same kind of solutions across the periodic system is the idea (or hope :-) that at least on average the many electrons live in a potential (or effective potential) of spherical symmetry, and that this average is relevant. When this is true, then -- however complex the system may be --

(1) the solution shall factorize in radial and angular parts, respectively.

(2) stationary states shall be eigenstates of total angular momentum.

(3) the principal quantum number still enumerates the solution of the radial Schrödinger equation along the increase (one by one) of the number of nodes (n-l)

However,

(4) due to the electron electron interactions, the l (or j in the presence of spin orbit coupling) degeneracy is lost.

The approximation is not so bad since closed (filled) shells indeed contribute a spherically symmetric charge (and thus electrostatic potential energy) and zero angular momentum.

(5) In not so few cases (mostly those with a single open shell), Hund's rules account quite well ****for the ground state**** of atoms and ions. More complicated configurations are hard to predict.

Edit: I should add that the single electron picture is utterly inadequate for almost everything of real contemporary interest in atomic physics!

Yes, the potential (effective potential) is usually radial and spherically symmetric, therefore the angular solution of Schrodinger equation is given in terms of spherical Harmonica and the quantum numbers l and m. But l depends on n and consequently on the radial part solution. The principle quantum number is n and this we have no idea about it for complicated atoms, we may hope that it is non-negative, integer, ....and this is if we are sure that the potential is completely radial, but it is depends on many factors, see the correction terms for H-atom (relativistic, spin-orbit, ....).

Once more: the important point here is that the potential is confining, which means that , at any given energy, the particle cannot escape to infinity. For this case the energy levels take discrete values, which means they are labeled by an integer. The fact that the ground state is of finite energy means that this integer canot take arbitrarily negative values. And this does not depend on the angular part, since this *is* compact.

Take as an example the anisotropic, three dimensional, harmonic oscillator.

Imad, it is not correct to say that l depends on n, as a quantum number it is independent. What depends on n is, which quantum numbers l and m are possible at a given n, and this indeed follows the same systematics as in the hydrogen atom.

Actually one may find that the hydrogen derived nomenclature of quantum numbers is not the most fortunate one. Historically, I would think, it derives from the degeneracy of all E(n,l,ml) for a given principal quantum number, which leads to the relatively simple hydrogen spectrum. [The same applies for E(n,j,mj), if we do not enter the level of precision requiring Lamb shift and hyperfine coupling]. This degeneracy is unique to the Coulomb potential.

In nuclear physics, nomenclature evolved in a different manner. there you will find quantum numbers like 1s, 1p, 1d, etc. the principal quantum number then takes a different meaning: 1X is just the lowest energy state of symmetry (angular momentum) X, 2X the second lowest etc. The above conflict does never arise.

Kai, I agree with the first paragraph, you are right. but still I need to be convinced and to convince my students. Why we use n= 1,2,3,.... and l= 0,1,2, ....(n-1), and m= -1,..,0,...+l for Ca, Na, and other atoms....These numbers are due to H-atom only and for radial potential.

One way to see this is by realizing that the one-electron wavefunctions, of the hydrogen atom, labeled by n, l and m are a complete set of functions in the one-particle Hilbert space. Therefore it's possible to express any *other* one-particle wave function, namely that of an electron in any other atom, as a linear combination of the wavefunctions of the electron in the hydrogen atom. The fundamental feature that allows this is the possibility of identifying the electrons in these atoms as one-particle states. That's all.

I am not entirely happy with Stam's answer, but in many introductory courses you'll not get far beyond that. Whenever there is more than one electron (or hole) in open shells, then the one electron picture essentially breaks down. The "real" states then are the properly constructed, antisymmetrized states involving all open shells. This construction shall employ the single electron states, though, which Stam rightfully mentions. In the simplest case you get Slater determinants but many states (take a singlet state made from 2 electrons in different orbitals) are not Slater determinants.

Fundamentally there is no such thing as "an electron in (single/independent particle) state XY". There are the many N-electron states. Nevertheless there are quite a few cases, where single particle language is nevertheless appropriate.

But as a fairly simple counter example, consider photoionization from neutral noble metal atoms (ground configuration d10 s1). If the process removes "the s electron", then you're left with an ion in d10 configuration, which is a singlet. The photoelectron spectrum therefore has only a single peak of this character. Let's have "one of the d electrons" removed, resulting in the config d9s1). In a single electron picture without spin orbit interaction (soc) you'll expect a single peak. Admitting soc, you'll expect a spin orbit split doublet. None of the two is observed, there are actually four peaks, representing four energetically different (groups of) states, they can readily be understood in terms of the exchange interaction between d9 and s1, the relative importance of soc and exchange may nicely be evaluated across the series Cu-Ag-Au. Hence, the electrons cannot be considered independent!

I do try to include this information in lectures on both, atomic physics and magnetism.

For a deeper understanding of your primary question, I really recommend studying the first couple of chapters of Cowan's book. It's tough but really worth it.

Edit: going through the physics of He and its excited states actually is another very useful chapter in an atomic physics course, also nicely highlighting the necessity of thinking beyond the single particle approximation. At the same time, it is also fairly tractable since "only" 2 electrons are involved.

Indeed, what I wrote assumed that we are working below the ionization threshold, so

we are considering only the discrete part of the spectrum. I agree that in a multi-electron system, talking about single electrons is delicate. But I don't think it's that hard, either.

As long as one is in the framework of non-relativistic quantum mechanics, the wave function of the multi electron system is some complicated function of the position of the electrons. It's hard to compute it explicitly-but it's possible to draw certain conclusions, that are dependent only on the fact that any one-particle state feels a confining potential and that the angular dependence can be expanded in spherical harmonics. That the particular ``one-particle state'' will be, generically, some superposition doesn't alter the conclusions qualitatively.

I guess that the answer is different for other atoms. But the similarity is that they could all be solved in the spherical coordinates for single atom. And we label n l m for different radial and angle solutions, which is expanded in the same basis function, like Bessel function. While for the atom in crystal structure, the n l m label could be used to have the actual orbital expanded in the orbitals that an atom has when it is free of interaction with other atom. N l m is just something they use to label the basis function to expand the actual orbital.

Once more: The fundamental issue is that we're talking about a *finite* number of electrons. Therefore the wave function of the Z-electron state can be expanded in the basis of 1-electron wavefunctions, which are labeled by n,l,m (and s for the spin). Below the ionization threshold, the energy levels are discrete and are labeled by an integer, n.

That's all there is. All the non-trivial information is contained in the ``overlap'' coefficients. And while these considerations do not take into account, explicitly, the nucleus, its contribution will only change the overlap coefficients, not the fact that the states can be, consistently, labeled by n, l, m and s.

This is no longer true when ionization occurs, because, then the scattering states are not countable, they're a continuum.

1) hydrogen-like one-electron wave functions for an arbitrary nuclear charge Z for the discrete spectrum depend on n, l, m as in the hydrogen case Z=1. They do not (!), however, form a complete basis set for the space of allowable one-particle functions.

2) the reason why it still is useful to teach orbitals from the atomic point of view (i.e. using spherical symmetry) is two-fold:

2.1) molecular orbitals may be approximated as linear combinations of atomic orbitals (LCAO) with good accuracy even for finite basis sets (here, nowadays mostly gaussian-type basis sets are used). The physical reason for this is the divergence of the coulomb potential at the position of the atomic nuclei.

2.2) the slater-rules derived from calculations on atoms and ions essentially say that the main effect of closed inner shells is a shielding of the nuclear charge as regards the orbitals of electrons in the valence shell: Z becomes dependent on n and l. Thus, qualitatively, the valence orbitals resemble the hydrogen-like orbitals for such effective Z in a rotational symmetry.

3) of course, one has to keep in mind that orbitals are not the whole story. They provide a clue to the many-electron wave functions that may be thought of as l inear combinations of slater determinants built from suitable orbitals.

Dear Imad

Notice that the solution of Schrodinger equation for a 1-electron atom with nuclear charge +Ze is the same as the solution for the H-atom (with different energies, mean orbital radius,... of course). The integer n (principal quantum number) and l which has a maximum value of (n-1) is a direct consequence of the series solution of the radial part of the hydrogen-like atom. The values of ml are a consequence of space quantization as you know. Therefore, the electron in such an atom can, in principle, occupy any of these hydrogen-like states.

Now if we add a second electron, it is true that the potential seen by this electron (as well as that seen by the first electron) deviates from spherical symmetry. If we attempt to determine the atomic orbitals for the second electron with precision, we should consider the charge distribution of the first, and include it in the Coulomb potential arising from the nucleus. But this contribution cannot be known without a priori knowledge of the precise wave function, which is now perturbed by the presence of the second electron. Thus one is led to use self-consistent methods and the resulting orbitals are approximate ones with a degree of precision dictated by the calculation method. In this picture, the orbitals are observed as perturbed atomic ones.

Further, when we encounter atoms with more and more electrons, we can always estimate the potential as a spherical part due to the screened nuclear potential for the nucleus and filled shells, and consider the valence electron as an electron moving in a spherically symmetric potential arising from an effective positive charge. The orbitals are, to very good approximation, hydrogen-like. But the problem arises when we have more than one electron outside a closed, noble-gas configuration. Self-consistent methods can come to the rescue, at least partially.

In all the above discussion we do not take into consideration non-spherical nuclear charge distributions, which could be the case in heavy nuclei. Also we do not consider interactions beyond Coulomb's. Other interactions could be treated as perturbations, which modify the orbitals further

Note that there is a class of atoms or molecules states where the H atom wave functions can work in a fairly nice manner: the highly excited Rydberg states. Then the core/ external electron correlation is weak and one gets mostly the H atom levels up to a quantum defect term in the energy. You have also the H+ H- system where these states appears (see papers of Ubachs for instance)