Why is a LCAO necessary within Hartree Fock?
$begingroup$
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
476417
$endgroup$
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
476417
$endgroup$
$begingroup$
(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
476417
$endgroup$
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
476417
asked May 6 at 12:46
JacobJacob
476417
asked May 6 at 12:46
JacobJacob
476417
asked May 6 at 12:46
JacobJacob
476417
asked May 6 at 12:46
JacobJacob
476417
476417
$begingroup$
(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
1
active
oldest
votes
$begingroup$
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":
$$hat{1} = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat{1}|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text{ where }c_i = langle i | f rangle$$
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
41.1k7124255
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f114916%2fwhy-is-a-lcao-necessary-within-hartree-fock%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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":
$$hat{1} = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat{1}|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text{ where }c_i = langle i | f rangle$$
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
41.1k7124255
$endgroup$
$begingroup$
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":
$$hat{1} = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat{1}|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text{ where }c_i = langle i | f rangle$$
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
41.1k7124255
$endgroup$
$begingroup$
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":
$$hat{1} = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat{1}|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text{ where }c_i = langle i | f rangle$$
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
41.1k7124255
$endgroup$
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":
$$hat{1} = sum_i |iranglelangle i| quad Longleftrightarrow quad |frangle = hat{1}|frangle = sum_i |iranglelangle i | f rangle = sum_i c_i|irangle text{ where }c_i = langle i | f rangle$$
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
41.1k7124255
answered May 6 at 13:41
orthocresol♦orthocresol
41.1k7124255
answered May 6 at 13:41
orthocresol♦orthocresol
41.1k7124255
answered May 6 at 13:41
orthocresol♦orthocresol
41.1k7124255
41.1k7124255
$begingroup$
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 |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f114916%2fwhy-is-a-lcao-necessary-within-hartree-fock%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
(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