04172017, 08:42 AM



19 posts, read 15,457 times
Reputation: 20


Quote:
Originally Posted by Iaskwhy
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.

The principle of quantum superposition makes it possible to predict the results of measuring a physical quantity in a certain quantum state (for orbitals, for example, we take the energy of oneelectron orbitals, any other physical quantity appropriate for a given state can be considered, consider this a "thought experiment").
The Hamiltonian must be "written out" for a particular problem, and it is written out considering from what depends on the total energy of the system, actually "works" physical advisability. But if the premises are not correct (for example, MO can not be obtained by a linear combination of AO), then what is generally calculated? In addition, the principle of quantum superposition is more fundamental (if we may say so), the Hamiltonian is more "manmade".

04172017, 12:14 PM



1,770 posts, read 1,287,149 times
Reputation: 1694


Quote:
Originally Posted by chemist777
The principle of quantum superposition makes it possible to predict the results of measuring a physical quantity in a certain quantum state (for orbitals, for example, we take the energy of oneelectron orbitals, any other physical quantity appropriate for a given state can be considered, consider this a "thought experiment").

You cannot get the energy of a quantum system from the wave function alone. It simply doesn't work that way. You need the Hamiltonian. You are too hung up on superposition.
Consider this, if you take the wave function for a proton in free space and for an electron in free space you can combine them into a single wave function. You can then use the Hamiltonian of an electron and proton in free space to get the energy of the system. However, this system is not a hydrogen atom.
Quote:
The Hamiltonian must be "written out" for a particular problem, and it is written out considering from what depends on the total energy of the system, actually "works" physical advisability. But if the premises are not correct (for example, MO can not be obtained by a linear combination of AO), then what is generally calculated? In addition, the principle of quantum superposition is more fundamental (if we may say so), the Hamiltonian is more "manmade".

The Hamiltonian isn't anymore manmade than the wavefunction of any other quality of the system.

04172017, 12:43 PM



19 posts, read 15,457 times
Reputation: 20


Quote:
Originally Posted by Iaskwhy
You cannot get the energy of a quantum system from the wave function alone. It simply doesn't work that way. You need the Hamiltonian. You are too hung up on superposition.
Consider this, if you take the wave function for a proton in free space and for an electron in free space you can combine them into a single wave function. You can then use the Hamiltonian of an electron and proton in free space to get the energy of the system. However, this system is not a hydrogen atom.

That's right, that's why in the VB method the chemical bond is "lost" (isolated atoms but not molecules),
see p. 6 http://vixra.org/pdf/1704.0068v1.pdf
Carefully analyze the principle of quantum superposition, where the electron is in the 1 quantum state (the first orbital) and in the 2 quantum state (the second orbital) means there is a 3 quantum state (linear combination) where we can "calculate" (measure) the orbital energy. And proceeding from the principle of superposition this energy is not the energy of the binding molecular orbitals (and should be such (the main assumption of the MO method)) (in MO, the energy would be lower in value), but simply the spectrum of the discrete values of the original atomic orbitals. Here, instead of energy, we can take any physically correct value that can be measured, even "mentally". See p. 4 http://vixra.org/pdf/1704.0068v1.pdf
Last edited by chemist777; 04172017 at 12:53 PM..

04172017, 12:53 PM



1,770 posts, read 1,287,149 times
Reputation: 1694


Quote:
Originally Posted by chemist777
That's right, that's why in the VB method the chemical bond is "lost" (isolated atoms but not molecules),
see p. 6 http://vixra.org/pdf/1704.0068v1.pdf
Carefully analyze the principle of quantum superposition, where the electron is in the 1 quantum state (the first orbital) and in the 2 quantum state (the second orbital) means there is a 3 quantum state (linear combination) where we can "calculate" the orbital energy. And proceeding from the principle of superposition this energy is not the energy of the binding molecular orbitals (and should be such (the main assumption of the MO method)) (in MO, the energy would be lower in value), but simply the spectrum of the discrete values of the original atomic orbitals. Here, instead of energy, we can take any physically correct value that can be measured, even "mentally". See p. 4 http://vixra.org/pdf/1704.0068v1.pdf

It's not lost, you are just incorrectly applying the physics. You seem to know chemistry quite well, but I think you have fundamental gaps in your understanding of quantum mechanics. How on earth can you calculate the energy of the system without the Hamiltonian?

04172017, 01:07 PM



19 posts, read 15,457 times
Reputation: 20


Quote:
Originally Posted by Iaskwhy
It's not lost, you are just incorrectly applying the physics. You seem to know chemistry quite well, but I think you have fundamental gaps in your understanding of quantum mechanics. How on earth can you calculate the energy of the system without the Hamiltonian?

It's you don't understand the principle of quantum superposition.
You need to read carefully the principle of superposition (this definition from quantum mechanics):
"For example, consider two quantum states (actually existing) are described by wave functions ψ1 and ψ2. From the principle of superposition [1, p. 21] it should be clearly, that their linear combination (ψ3 = C1ψ1 + C2ψ2) will be the third quantum state (as actually existing), which will be described by a wave function ψ3. What does it mean? The fact that the measurement of a certain physical value d in the state ψ1> will result d1, and for measure a value for of d in the state ψ2> will result d2. When the third quantum state ψ3> is realized, then when measuring a physical quantity, the quantum system will take the values d1 and d2 with probabilities, respectively, C1^2 and C2^2. That is, in a quantum state ψ3> when we will have many dimensions sometimes d1 value and sometimes d2 (with certain known frequency)."
p. 12 http://vixra.org/pdf/1702.0333v2.pdf
And now instead of "d" substitute what you want (the correct value), length, energy, etc. And the principle of superposition will give an answer.
I already told you to figure it out in the PQS.
P.S. The definition of quantum superposition is given in the book on quantum mechanics L.D. Landau (see link). Lev Davidovich Landau apprentice of Niels Bohr (he thought so, Landau), except that it is constantly communicated and worked with such prominent physicists like Heisenberg, Dirac, Pauli and others (they were friendly and working relations). Believe me, these scientists understood the fundamental foundations of quantum mechanics becouse they created these foundations.
Last edited by chemist777; 04172017 at 02:31 PM..

04172017, 03:09 PM



19 posts, read 15,457 times
Reputation: 20


L. D. Landau (09.01.1908  01.04.1968, a Soviet Physicist, Nobel Prize in Physics (1962)) worked in Copenhagen with Bohr. He also worked in Moscow, Leningrad (Russia) and Kharkov (Ukraine).

04172017, 04:44 PM



1,770 posts, read 1,287,149 times
Reputation: 1694


Quote:
Originally Posted by chemist777
It's you don't understand the principle of quantum superposition.
You need to read carefully the principle of superposition (this definition from quantum mechanics):
"For example, consider two quantum states (actually existing) are described by wave functions ψ1 and ψ2. From the principle of superposition [1, p. 21] it should be clearly, that their linear combination (ψ3 = C1ψ1 + C2ψ2) will be the third quantum state (as actually existing), which will be described by a wave function ψ3. What does it mean? The fact that the measurement of a certain physical value d in the state ψ1> will result d1, and for measure a value for of d in the state ψ2> will result d2. When the third quantum state ψ3> is realized, then when measuring a physical quantity, the quantum system will take the values d1 and d2 with probabilities, respectively, C1^2 and C2^2. That is, in a quantum state ψ3> when we will have many dimensions sometimes d1 value and sometimes d2 (with certain known frequency)."
p. 12 http://vixra.org/pdf/1702.0333v2.pdf
And now instead of "d" substitute what you want (the correct value), length, energy, etc. And the principle of superposition will give an answer.
I already told you to figure it out in the PQS.
P.S. The definition of quantum superposition is given in the book on quantum mechanics L.D. Landau (see link). Lev Davidovich Landau apprentice of Niels Bohr (he thought so, Landau), except that it is constantly communicated and worked with such prominent physicists like Heisenberg, Dirac, Pauli and others (they were friendly and working relations). Believe me, these scientists understood the fundamental foundations of quantum mechanics becouse they created these foundations.

I know what quantum superposition is, I don't need your poor translated definition. Quantum superposition alone doesn't give you the energy of the system. Please show me how you work out the energy of the system alone without the Hamiltonian. Do it step by step with all the math included like a true scientist would do it.
Last edited by Iaskwhy; 04172017 at 05:00 PM..

04172017, 04:58 PM



19 posts, read 15,457 times
Reputation: 20


Quote:
Originally Posted by Iaskwhy
Quantum superposition alone doesn't give you the energy of the system. Please show me how you work out the energy of the system alone without the Hamiltonian.

Sorry, but the work does not solve the Schrodinger equation, but the MO method and the VB method are analyzed from the position of the principle of quantum superposition (for this it is not necessary to write out the Hamiltonian).

04172017, 05:03 PM



1,770 posts, read 1,287,149 times
Reputation: 1694


Quote:
Originally Posted by chemist777
Sorry, but the work does not solve the Schrodinger equation, but the MO method and the VB method are analyzed from the position of the principle of quantum superposition (for this it is not necessary to write out the Hamiltonian).

And that's where you are having this problem with the "bond disappearing".

04172017, 05:16 PM



1,770 posts, read 1,287,149 times
Reputation: 1694


Quote:
The energy of an electron in one of the atomic orbitals is α, the Coulomb integral.
α = ∫φ1Hφ1dτ = ∫φ2Hφ2dτ
where H is the Hamiltonian operator. Essentially, α represents the ionization energy of an electron in atomic orbital φ1 or φ2.
The energy difference between an electron in the AO’s and the MO’s is determined by the exchange integral β,
β = ∫φ1Hφ2dτ
β is an important quantity, because it tells us about the bonding energy of the molecule, and also the difference in energy between bonding and antibonding orbitals. Calculating β is not straightforward for multielectron molecules because we cannot solve the Schrödinger equation analytically for the wavefunctions. We can however make some approximations to calculate the energies and wavefunctions numerically. In the Hückel approximation, which can be used to obtain approximate solutions for π molecular orbitals in organic molecules, we simplify the math by taking S=0 and setting H=0 for any porbitals that are not adjacent to each other. The extended Hückel method,[2] developed by Roald Hoffmann, and other semiempirical methods can be used to rapidly obtain relative orbital energies, approximate wavefunctions, and degeneracies of molecular orbitals for a wide variety of molecules and extended solids. More sophisticated ab initio methods are now readily available in software packages and can be used to compute accurate orbital energies for molecules and solids.
We can get the coefficients c1 and c2 for the hydrogen molecule by applying the normalization criterion:
ψ1 = (φ1+φ2)/(√2(1+S)) (bonding orbital)
and
ψ2 = (φ1φ2)/(√2(1S)) (antibonding orbital)
In the case where S≈0, we can eliminate the 1S terms and both coefficients become 1/√2
Note that the bonding orbital in the MO diagram of H2 is stabilized by an energy β/1+S and the antibonding orbital is destabilized by β/1S. That is, the antibonding orbital goes up in energy more than the bonding orbital goes down. This means that H2 (ψ12ψ20) is energetically more stable than two H atoms, but He2 with four electrons (ψ12ψ22) is unstable relative to two He atoms.

https://en.wikibooks.org/wiki/Introd...Orbital_Theory
Notice, you need the Hamiltonian.

Please register to post and access all features of our very popular forum. It is free and quick. Over $68,000 in prizes has already been given out to active posters on our forum. Additional giveaways are planned.
Detailed information about all U.S. cities, counties, and zip codes on our site: Citydata.com.

