Why is a LCAO necessary within Hartree Fock?Functional variation in derivation of the Hartree-Fock equationsWhy are excited Slater determinants used to describe electron correlation?What is the point of introducing virtual orbitals in Hartree-Fock calculations?Slater determinant as an unperturbed atomic wave functionIs HF the only method for trying to solve the many-electron Schrödinger equation?Determining Kohn-Sham and Hartree Fock virtual orbitals: The underlying fieldWhat exactly is meant by 'multi-configurational' and 'multireference'?Hartree product and the Slater determinantLinear Combination of Atomic Orbitals
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Why is a LCAO necessary within Hartree Fock?
Functional variation in derivation of the Hartree-Fock equationsWhy are excited Slater determinants used to describe electron correlation?What is the point of introducing virtual orbitals in Hartree-Fock calculations?Slater determinant as an unperturbed atomic wave functionIs HF the only method for trying to solve the many-electron Schrödinger equation?Determining Kohn-Sham and Hartree Fock virtual orbitals: The underlying fieldWhat exactly is meant by 'multi-configurational' and 'multireference'?Hartree product and the Slater determinantLinear Combination of Atomic Orbitals
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As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.
What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?
physical-chemistry quantum-chemistry computational-chemistry molecular-orbital-theory
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
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As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.
What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?
physical-chemistry quantum-chemistry computational-chemistry molecular-orbital-theory
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
$endgroup$
$begingroup$
(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
3
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02
add a comment
|
$begingroup$
As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.
What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?
physical-chemistry quantum-chemistry computational-chemistry molecular-orbital-theory
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
$endgroup$
As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.
What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?
physical-chemistry quantum-chemistry computational-chemistry molecular-orbital-theory
physical-chemistry quantum-chemistry computational-chemistry molecular-orbital-theory
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
asked May 6 at 12:46
JacobJacob
4814 silver badges18 bronze badges
4814 silver badges18 bronze badges
$begingroup$
(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
3
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02
add a comment
|
$begingroup$
(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
3
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02
$begingroup$
(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
$begingroup$
(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
3
3
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02
add a comment
|
1 Answer
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Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set – is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$hat1 = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat1|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text where c_i = langle i | f rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
$endgroup$
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The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
add a comment
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Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set – is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$hat1 = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat1|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text where c_i = langle i | f rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
$endgroup$
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
add a comment
|
$begingroup$
Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set – is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$hat1 = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat1|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text where c_i = langle i | f rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
$endgroup$
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
add a comment
|
$begingroup$
Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set – is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$hat1 = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat1|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text where c_i = langle i | f rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
$endgroup$
Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set – is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$hat1 = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat1|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text where c_i = langle i | f rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
answered May 6 at 13:41
orthocresol♦orthocresol
43.6k7 gold badges136 silver badges267 bronze badges
43.6k7 gold badges136 silver badges267 bronze badges
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
add a comment
|
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
$begingroup$
The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?
$endgroup$
– Jacob
May 6 at 13:47
3
3
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
$begingroup$
I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.
$endgroup$
– orthocresol♦
May 6 at 13:48
add a comment
|
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(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).
$endgroup$
– Jon Custer
May 6 at 13:02
3
$begingroup$
LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.
$endgroup$
– Feodoran
May 7 at 8:02