If memory serves, the reason is that electrons are fermions, and the Pauli exclusion principle states that no two fermions may occupy the same quantum state (i.e. have the same quantum numbers). The electronic structure of a hydrogen-like atom is given by the solution to the Schrödinger equation, the full derivation of which I will defer to someone better versed in physical chemistry. The quantum numbers arise from the radial and angular (polar and azimuth) terms in the solution. Additional operators are not necessary.
Thank you very much for your response.
Yes, of course, the electrons are Fermion and they must obey the Pauli's exclusion principle.
The question that I would emphasize was like this.
Can the better solution that satisfies the exclusion principle be derived automatically by only solving the raw original Schroedinger equation? The Hamiltonian is given by equation 6 in the attached file "Notation." And can the electronic state of all the spiecies of atom be approximated by the hydrogen-like model? As Mr. Ayarza has indicated recently in the other blog page, what is the obstacle for approaching the exact solution is the simultaneous inter-electronic Coulomb repulsion. The equation 7 gives the representation. The usual textbooks say that in the one-body approximation scheme, an electron, as it moves, may be penetrating the total electron cloud which is assumed to be the sum charge density which is thought to be contributed from all the other electrons. But it is difficult to believe or accept for me that one particular electron can move through the electron cloud without causing the other electron to run away from that electron. On the contrary, when one particular electron moves through the electron cloud, the other electrons should run away quickly from the incoming particular electron due to the Coulomb repulsion force. So that the initial assumption of the static hydrogen-like spatial orbit must be disturbed or distorted or modified strongly. To me such a wave function as composed from the Slator determinant (eq 8) of spin-orbital seems but a severe simplification as a solution. Should we believe in the hydrogen-like approximation still?
Actually, I have ever tried to compute the ground state wave function of Hellium atom using Kimball-Shortley method 5 years ago. The lattice was 6-dimensional and the number of the lattice points was set to be 17^6, a very huge number indeed. This may be the limit of the current personal computer. In order to avoid the zero division, I made an assumption as shown below.
Vne(R1)=Qe*Qn/ABS(R1) when ABS(R1) is not equal to 0
=Qe*Qn/0.01*A when ABS(R1) is eqaul to 0
Vne(R2)= similarly defined
Vee(R1,R2)=Qe*Qe/ABS(R1-R2) when R1 is not equal to R2
=Qe*Qe/0.01*A when R1 is equal to R2
where Qe is the electronic charge, Qn the nuclear charge, R1 the 3-dimensional vector of the electron1, i.e., R1=(x1,y1,z1), the R2 the one of the second electron2,i.e, R2=(x2,y2,z2) and A the lattice spacing. The result is shown in the attached file "Photo of Hellium." The position which the one electron can take is 1st, 2nd,3rd,...,17th lattice point along one axis. So the 9th position is the center. Using the index representation, the origin corresponds to (9,9,9). I had let the position of the first electron be (6,9,9), i.e., this electron is located 6th position on the x axis. And I had visualized the electron density of the second electron on x2-y2 plane for various values of z2. In the photo file, from the upper left to upper right and then from the lower left to lower right, the index of z2 increases from 2 to 9, and the density of the second electron is shown as brightness display. You can notice the increasing blackness at the position where the first electron is located, that is at (6,9,9) (vivid in the lower far right figure as a left black spot square.) This repulsive phenomenon was able to be observed anywhere the first electron may be. Therefore, I would argue that the hypothesis of the electron cloud and moving electron that penetrates the electron cloud is dubious. Therefore, further, the hydrogen-like orbital assumption seems dubuous too.
First of all, think a case for Hellium atom. If the wave function is assumed to be given by the Slator determinant function like
then if R1=R and R2=R and f(R)<>0, then f(R)f(R)<>0. This can't be, since this situation describes the occasion when the two electrons overlap or are colliding as already negateded in the above explanation of the result of computation like such a probability does decrease as R2 approches R1. Absolutely the functional form of the wave function must be a function of both two variable R1 and R2 simultaneously.
where the orbital function F(R1,R2) approaches zero when R1 approaches R2, in accordance with the Coulomb repulsion law.
The theory of "electron penetrating electron cloud" is impossible. If it is possible, the there the electrons are violently colliding with each other. The hydrogen-like model does not hold for any atom. Even if the concept of spatial orbital is viable, the whole wave function needs to be rewritten with drastic changes.
I need advice from someone who is versed in this matter if possible.
Please someone who is acquainted with this matter, join in this discussion and tell me the proof that the hydrogen-lile, one-body approximation can be adequately justified for all of the element in the periodic table and tell me how good the wave function's degree of accuracy is in general.
Dr. Hsieh, thank you for your cooperation.
For systems containing more than one electron, one must account for the exchange operations between electrons in addition to their interactions with the nucleus. The Hartree-Fock method is one attempt to solve the problem, and Wikipedia has a serviceable article introducing the concepts:
The field of computational chemistry has advanced significantly (particularly given the improvements in both computer hardware and software in the last fifty years), and while there are still some areas where in silico predictions fail to match experimental results, computational models are certainly useful enough. If they didn't work, no one would use them.
Thank you for your further response.
Give me some time while I read the Wikipedia's article you introduced.
But, your logic that "If they didn't work, no one would use them, Actually almost everyone are using without doubting it. So that the HF method is accurate," I cannot accept such a logic soon.
Thank you again