Quantum mechanics, Quantum-mechanical analysis of the MO method and VB method from the position of PQS.
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You misunderstood what was written in the paper about multiplication of functions. It means system consisting of two parts, and then the wave functions that describe these parts that allow us to obtain wave information that describes the system as a whole.
ψ12 (q1, q2, t) = ψ1 (q1, t) * ψ2 (q2, t)
All this is known and elaborated in quantum mechanics for probably about 100 years. And then parts can be (and will be) different. About the identity of particles is not here. Moreover, electrons are fermions and they are indistinguishable, that is, identical.
Your statement "The wave function for identical bosons looks the same as for MO and VB theories" is not quite correct (I note once again, electrons are fermions and they are identical, that is, indistinguishable). Here everything is more complicated. The point is that the Schrödinger equation does not take into account the spin of the particles. But the electrical interaction does not depend on the spin, and so when the spin is not important, everything is OK (Hamiltonian of a system of electrically interacting particles (In the absence of a magnetic field) does not contain spin operators, and therefore, when applied to the wave function, it does not affect the spin variables in any way).
But if the spin should be taken into account (as in the formation of a chemical bond), then one must understand that the Schrödinger equation Satisfy both the coordinate component (the wave function that depends on the coordinates) of the wave function and the spin component (the wave function depends only on the spin of the particles). That is, the Schrödinger equation can be written in the form:
where ψ1 is the wave function of the system depending only on the coordinates of the particles,
and ψ2 is the wave function of the system which depends only on the spin of the particles,
Ψ12 (ξ1, ξ2) is, naturally, the wave function of the system, but here it must be noted,
That ξ1, ξ2, ξ3, ξ4 conditionally designate the sets of three coordinates and the projection of the spin of each of the particles.
Now it is necessary to recall that for fermions the principle of indistinguishability of particles leads to the fact that the wave function must be antisymmetric (When the permutation changes sign). For bosons it is symmetric (it does not change sign when it is permuted). This is easy to understand if we write the wave function of the system ψ12 (ξ1, ξ2) as follows:
ψ12 (ξ1, ξ2) = e^(i * α) * ψ12 (ξ1, ξ2)
Where α — is a real constant.
e^(2*i * α) = 1,
e^(i * α) = +, - 1
ψ12 (ξ1, ξ2, ξ3, ξ4) = +, - ψ12 (ξ1, ξ2, ξ3, ξ4)
(In the permutation of indistinguishable particles the wave function of the system can only change to an unimportant phase factor; this follows from the fact that when the particles are rearrangedIdentical states are obtained, that is, the particle systems physically must be completely equivalent).
And it is natural that for fermions (the principle of indistinguishability) the total function of the system must be antisymmetric (it changes sign on a permutation), but this function is the product of the coordinate wave function of the system (ψ1 (r1, r2, r3, r4 ...)) and the spin wave function of system (ψ2 (σ1, σ2, σ3, σ4, ...)). Therefore, for a symmetric coordinate function, the spin function must be antisymmetric, and vice versa.
But in quantum mechanics it is shown that a symmetric spinor of the second rank describes a system with a total spin equal to unity (for two particles), and an antisymmetric spinor reduces to a scalar, which corresponds to zero spin.
Therefore, when the chemical bond is formed (two electrons, the spins are opposite, the resultant spin is zero), the spinor ψ2 (σ1, σ2) is antisymmetric, and hence from the equation (see above) we get that ψ1 (r1, r2) is symmetric.
I note that ψ1 (r1, r2) is symmetric, and this is for fermions.
The paper you posted didn't mention spin once. The word "spin" wasn't used a single time in the paper.
In the use of "spin" in this work there was no need. Much of the consideration was understood by default.
Really? It was? Hahaha how many English speakers have read this paper of yours. I would imagine not very many. I don't think you have a firm grasp of the language.
You say things like
"It is shown that the main assumption of the molecular orbitals method (namely, that the molecular orbital can be represented like a linear combination of overlapping atomic orbitals) enters into an insurmountable contradiction with the principle of quantum superposition."
LCAO is BASED on quantum superposition, it cannot be contradictory.
Really? It was? Hahaha how many English speakers have read this paper of yours. I would imagine not very many. I don't think you have a firm grasp of the language.
You say things like
"It is shown that the main assumption of the molecular orbitals method (namely, that the molecular orbital can be represented like a linear combination of overlapping atomic orbitals) enters into an insurmountable contradiction with the principle of quantum superposition."
LCAO is BASED on quantum superposition, it cannot be contradictory.
Quantum superposition "does not create a" new MO quality, it leads to a spectrum of AO (BY DEFINITION). And in the MO method from AO get MO, here is the contradiction. Everything is shown in the paper.
MO and AO "from different Universes", this is a different ''quality'', and one can not be deduced from one another. There are no strict theoretical prerequisites to bring the MO from the AO. Therefore, MO is better to just postulate and not try to deduce from something.
P.S. Everyone likes english style of work except of you.
Last edited by chemist777; 04-15-2017 at 12:33 PM..
Quantum superposition "does not create a" new MO quality, it leads to a spectrum of AO (BY DEFINITION). And in the MO method from AO get MO, here is the contradiction. Everything is shown in the paper.
MO and AO "from different Universes", this is a different ''quality'', and one can not be deduced from one another. There are no strict theoretical prerequisites to bring the MO from the AO. Therefore, MO is better to just postulate and not try to deduce from something.
P.S. Everyone likes english style of work except of you.
The bold just proved my point. I have no idea what you are trying to say. It looks like it's copied and pasted from Google Translate. Citing your own papers does nothing to prove the point you are trying to make.
The bold just proved my point. I have no idea what you are trying to say. It looks like it's copied and pasted from Google Translate. Citing your own papers does nothing to prove the point you are trying to make.
I quite simply explained my idea, it is a pity that you do not understand. And let's stop this useless discussion (as the last posts show).
P.S. You greatly overestimate the capabilities of Google translator.
By the way, I just had two more people with physics degrees (three including myself), all from top tier schools, read your paper and none of us could understand what you are trying to say. The discussion isn't useless when it directly impacts who can understand your paper.
Proceeding from this equation ψ3 = С1ψАО А(1) + С2ψАО В(1) and according to the principle of quantum superposition, when the third quantum state is realized |ψ3> (which is described by the wave function ψ3), then when measuring a physical quantity, for example, the orbital energy, the quantum system will take the values of E1 (the energy of the atomic orbital of atom A) and E2 (the energy of the atomic orbital of atom B) with frequency respectively |С1|^2 and |С2|^2, that is, it will have a discrete description. When measuring the energy of a given orbital, we sometimes register the value of E1, and sometimes the value of E2. But this directly contradicts the idea of the MO method, since one-electron MO should be formed with an energy lower (if it is a bonding MO) than the energies of individual AO (according to the idea of linear combination of atomic orbitals, two MOs are formed from two AO, one MO with reduced energy and the other MO increased energy). But the principle of quantum superposition prohibits this. With a linear combination of one-electron atomic orbitals, we will not be able to obtain a "new quality", that is, a one- electron molecular orbital, but we will have the spectrum of AO.
Who says it can't make a lower energy level? The energy levels come from <ψ|H|ψ>=E. They are dependent on the wave function as well as the Hamiltonian. Obviously when there is a bond, the Hamiltonian is going to change. I'd assume that is where "the chemical bond is lost", you didn't account for it in the Hamiltonian.
The Hamiltonian is where this "new quality" comes about. For example, an electron and a proton in free space have Hamiltonians that are different than the Hamiltonian for a hydrogen atom.
Principle of quantum superposition. And that's enough.
The idea is very simple. In the MO method from the AO, receive MO by certain operations. But the principle of quantum superposition (by its definition) clearly indicates that we will receive only AO from AO, and nothing else will happen. That's all. If we go to a comparison, whatever manipulations we make with the "meters" we do not get "kilograms", since they are different "qualities" (length and mass, as well as AO and MO). This is indicated by the quantum superposition principle.
Last edited by chemist777; 04-17-2017 at 02:54 AM..
Principle of quantum superposition. And that's enough.
The idea is very simple. In the MO method from the AO, receive MO by certain operations. But the principle of quantum superposition (by its definition) clearly indicates that we will receive only AO from AO, and nothing else will happen. That's all. If we go to a comparison, whatever manipulations we make with the "meters" we do not get "kilograms", since they are different "qualities" (length and mass, as well as AO and MO). This is indicated by the quantum superposition principle.
The principle of quantum superposition doesn't automatically give you the energies. You need the Hamiltonian as well. When you go from AO to MO, the Hamiltonian changes.
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