Quantum/Computational Chemistry

IPnano
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Ok guys so I know you don't like us quantum chemists and the approximate methods I'm about to talk about isn't "really physics" but I am looking for some good answers and if you guys have anything to offer BEYOND HYDROGEN then perhaps you can assist me.

I am wondering if someone can explain exactly how we (or the computer more specifically) move from the basis set to the molecular orbitals.

For example,

If we use a 3-21G basis set this means the following:

1. We are approximating three slater type orbitals (STOs) using contracted Gaussian functions.

2. The 3 in the basis set means we are using three contracted Gaussians to approximate the STO of the inner electrons, the 2 in the basis set means two contracted Gaussians for the inner-valence electrons and the 1 in the basis set is for using 1 contract Gaussian for the outer-valence electrons.

3. These three STOs, which we have approximated using contracted Gaussians, are brought together using a Linear Combination of Atomic Orbitals (LCAO) in order to approximate the molecular orbitals. In other words, the STOs are our basis functions, which we can assume are our atomic orbitals.

Now, if we use LCAO to create the molecular obritals (MOs), we should have 3 MOs since we are combining 3 atomic orbitals (the three STOs). Number of MOs equals the number of AOs.

Questions:

1. How does the computer generate three MOs? I can see how the LCAO would give us a SINGLE new MO for a SINGLE linear combination of basis functions. Where do the other two come from? If it has to do with manipulating the exponents, how are those exponents decided? Even if they are optimized, where do they come from and how are they limited to 3 MOs?

2. If we have a molecule with, say, 7 carbon atoms and we want to make an MO for this molecule, the inner electrons (1s2) of the carbon atom will be approximated using the 3 from the basis set. Is this saying that we will take an LCAO approach with SEVEN STOs all approximated using the 3 from our basis set? And the inner-valence electrons would be the 2s2 electrons approximated taking a linear combination of SEVEN STOs using the 2 from our basis set? And finally our outer-valence electrons using a linear combination of SEVEN STOs using the 1 from our basis set?

In other words, if we have 7 carbons atoms, are we using the basis set SEVEN times for EACH linear combination of STOs?

3. If the above is true, what does it mean to use the 3-21G on hydrogen since hydrogen only has 1 electron? By default is it only using the 3 from the basis set?

Thanks,
 
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IPnano said:
Ok guys so I know you don't like us quantum chemists and the approximate methods I'm about to talk about isn't "really physics" but I am looking for some good answers and if you guys have anything to offer BEYOND HYDROGEN then perhaps you can assist me.
Who is saying that? Quantum chemistry is real and important physics, and until your post I've never heard anyone claim otherwise.

If we use a 3-21G basis set this means the following:

1. We are approximating three slater type orbitals (STOs) using contracted Gaussian functions.
Actually, no. This would be the case for the STO-3G "basis set". For the 3-21G set, however, the exponents are "optimized" to represent actual AO radial functions (which are not STOs except for H and He).

2. The 3 in the basis set means we are using three contracted Gaussians to approximate the STO of the inner electrons, the 2 in the basis set means two contracted Gaussians for the inner-valence electrons and the 1 in the basis set is for using 1 contract Gaussian for the outer-valence electrons.
I don't know this Pople notation of basis sets well, but I was of the impression that the -21 had something to do with being split-exponent sets, and the 1 something with p functions.

Anyway: Knowing this kind of notation for this kind of basis set is not terribly important. Not in the least because, apart from the Gaussian program, most other software based on GTOs use Dunning-type (cc-pVnZ) oder Karlsruhe-type (def2-nZVPP) basis sets nowadays. For those no obscure letter/number/symbol (*+ anyone?) combinations are required in the notation.

3. These three STOs, which we have approximated using contracted Gaussians, are brought together using a Linear Combination of Atomic Orbitals (LCAO) in order to approximate the molecular orbitals. In other words, the STOs are our basis functions, which we can assume are our atomic orbitals.
Yes. Apart from them not being STOs for most basis sets (including the one you quote), but actual AO radial functions.


Now, if we use LCAO to create the molecular obritals (MOs), we should have 3 MOs since we are combining 3 atomic orbitals (the three STOs). Number of MOs equals the number of AOs.

Questions:

1. How does the computer generate three MOs? I can see how the LCAO would give us a SINGLE new MO for a SINGLE linear combination of basis functions. Where do the other two come from?
They come from the other linear combinations. As you just quoted, there are as many MOs as AOs. Both produce the same linear span in 3D-function space.
For example, if you have two basis functions b1 and b2, you can form m1 = b1 + b2 and m2 = b1 - b2. Two basis functions, two orbitals. In the actual Hartree-Fock process what is happening is that the MO coefficients of the basis functions are variationally optimized (subject to an MO orthogonality constraint). That is, there is an orbital matrix C^\mu_r which contains exactly one expansion coefficient for each MO r and each AO \mu. The basis functions themselves (exponents, contraction coefficients) are left untouched.


2. If we have a molecule with, say, 7 carbon atoms and we want to make an MO for this molecule, the inner electrons (1s2) of the carbon atom will be approximated using the 3 from the basis set. Is this saying that we will take an LCAO approach with SEVEN STOs all approximated using the 3 from our basis set? And the inner-valence electrons would be the 2s2 electrons approximated taking a linear combination of SEVEN STOs using the 2 from our basis set? And finally our outer-valence electrons using a linear combination of SEVEN STOs using the 1 from our basis set?
Of course you have one set of basis functions on each single atom; so you have 7 * (3-21g). The MOs are still expanded globally; that is, a single MO can have AO components on all of the atoms.
 
Glad to see you know the importance of QCHEM...apologies for the sarcasm. I always assumed the purists would never jump on board the QCHEM train. Also thanks for the informative reply.

Continued convo...

"Actually, no. This would be the case for the STO-3G "basis set". For the 3-21G set, however, the exponents are "optimized" to represent actual AO radial functions (which are not STOs except for H and He)."

I understand that we are not using ACTUAL STOs but, as I understand, we use approximations to STOs. The STO gives us the best representation of our electron orbital (cusp etc) but are very time-consuming to calculate. So we do a linear combination of Gaussians (contracted Gaussians) to approximate the shape of the STO. The integrals are much easier to calculate with Gaussians. This is why I was saying "We are approximating three slater type orbitals (STOs) using contracted Gaussian functions." Therefore the 3-21G approximates the STOs for the core electrons with 3 contacted Gaussians whereas the STO-3G, for example, would be using ACTUAL STOs for the core electrons.

So if I understand, the Hartree-Fock method allows for the optimization of the coefficients of our linear expansion of basis functions chosen from the basis set. This is done by iteratively calculating, and therefore optimizing, the guess wave function constructed from these basis functions (technically using a Slater determinant). The Roothan equations are used along with the variational theorem to give us an improved wave function constructed from our linear combination of basis functions. BUT...the exponents of our radial functions are NOT optimized in this process because these were PRE-optimized. This is why basis-sets are pre-made for certain chemical situations.

So the PRE-OPTIMIZED basis functions are used and what is optimized are the coefficients of the linear combination. The coefficients tell use the contribution (how much one function adds to the total) of each function.

The computer uses orbital matrices since a matrix equals a linear combination of its elements (or perhaps more specifically a determinant does?) And of course a matrix makes a lot more sense for numerical computation.

Now you said BOTH the AOs and the MOs get coefficients in the expansion. So are you saying that the AO matrix (which represents a linear combination of our basis functions) has a coefficient which is optimized for the constructed MO matrix which ALSO has its own coefficient?

Q1: Why would the MO matrix need a coefficient? Is this because we are technically taking a linear combination of MOs when the Slater determinant is calculated?

Finally, you said "Of course you have one set of basis functions on each single atom; so you have 7 * (3-21g). The MOs are still expanded globally; that is, a single MO can have AO components on all of the atoms."

Q2: So we are technically taking a linear combination of basis sets as well?

Cheers and sorry for the lengthly discussion.
 
Sorry, but you're sounding very confused. The most productive way for you to proceed would probably be to get a textbook on quantum chemistry to get you started on the basics.
 
IPnano said:
I always assumed the purists would never jump on board the QCHEM train.

Why would that be? Quantum chemistry is not very different from, say, solid-state physics. (Physically it's the same basic interactions with different boundary conditions) It says 'chemical physics' on my degree, which is in quantum chemistry. My PhD advisor's degree was in QC and his said 'theoretical physics', and I'm not sure you're much more 'purist' than that. All of physics uses approximate methods and some methods, like perturbation theory, are universal. (In any case, I've never run into the attitude either. I suspect that implying people have disdain for you before they've even said anything, like that, is liable to put people off a lot more; it's kind of 'guilt-tripping'.)
I understand that we are not using ACTUAL STOs but, as I understand, we use approximations to STOs. The STO gives us the best representation of our electron orbital (cusp etc) but are very time-consuming to calculate. So we do a linear combination of Gaussians (contracted Gaussians) to approximate the shape of the STO.

I know what you're saying and I believe cgk knew what you were saying. But what you're saying is wrong. Only the STO-nG basis set is explicitly fitted to STOs. Contracted/split-valence basis aren't. It's not an 'actual STO', since it's composed of Gaussian-type functions.

It sounds like you're confusing orbitals and basis functions. Despite the names (which are more due to history) the word 'orbital' in STO, GTO and LCAO does not really imply an orbital in the sense of a single-electron state. A more accurate description would be to say Slater and Gaussian-type basis functions. Because that's what they are. They don't need to be fitted to a particular orbital, or anything at all, actually. Solid-state people often use plane waves as basis sets.

Now, the Pople-type basis sets are constructed in such a way that certain functions represent certain (actual) orbitals in the unbound atom, and represents them as a set of gaussians, with different functions representing the 'core' and 'valence' electrons. That's what the whole 6-31G etc is all about.

But that doesn't mean those particular functions alone are representing the actual orbitals in the actual molecule. The actual single-electron states (orbitals) are the MOs. An MO is a linear combination of all the basis functions (including ones located on atoms far, far away). An electron can be as spread out over the entire molecule. Mostly they're not, of course, so most coefficients in the expansion for a given MO are near zero.

E.g. for a HF/6-31G calculation on water, the lowest-energy MO has the coefficients:
Code:
O (s-type): 0.9957719
O (s-type) 0.02200829
O (p-type) 0.0000000  0.0000000 0.0020469
O (s-type) -0.0080623
O (p-type) 0.0000000 0.0000000 -0.00183750
H1 (s-type) 0.0000655
H1 (s-type) 0.0020079
H2 (s-type) 0.0000655
H2 (s-type) 0.0020079

So this, the lowest, MO is almost entirely represented by an s-type function located on the oxygen atom. Obviously, it belongs to the oxygen's 1s core electron. The fact that it's almost entirely in the basis function intended to represent it is evidence that the basis set is well-optimized. But it's not the case that we're only using that one function for that electron. Every electron is represented by a whole LCAO expansion.

The computer uses orbital matrices since a matrix equals a linear combination of its elements (or perhaps more specifically a determinant does?) And of course a matrix makes a lot more sense for numerical computation.

It's more the other way around. You're solving a set of coupled non-linear differential equations. I'm assuming you're acquainted with how you deal with the linear case; you construct the secular matrix and determine the eigenvalues and eigenvectors. In this case, we construct a linear set of nonlinear functions (exponential equations) and then treat the problem as if it were linear. If you have a good starting guess and good basis functions, this will converge to the correct solution via SCF.
Now you said BOTH the AOs and the MOs get coefficients in the expansion.

No, the MOs don't get coefficients (in Hartree-Fock). They're either occupied or unoccupied. The basis functions (AOs) that form the MOs have coefficients.
Those are (basically) the eigenvectors of the matrix. The eigenvalues correspond to the orbital energy. (once you add the core Hamiltonian)

Clearer?
 
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If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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