Why do we choose specific spin orbitals in Hartree Fock theory?

[Derivator]

hi,

why does one choose in Hartree Fock theory the following type of spin orbitals?

spinorbital(r,s) = spacialOrbital(r)*alpha(s)
or
spinorbital(r,s) = spacialOrbital(r)*beta(s)

where alpha(s) is the spin up function and beta(s) the spin down function.

Why does one not choose the spin part of the spin orbitals as a linear combination of spin up and down, that is:

spinorbital(r,s) = spacialOrbital(r)*(a*alpha(s) + b*beta(s) )

--derivator

 

[cgk]

The short answer is: Because is most cases both approaches would give identical or same-quality results (like open-shell RHF vs UHF), and in the cases there they don't, most likely both are wrong. Additionally, this form of the wave function is a serious hassle for post-HF methods.

Long answer:
I think there are some programs which can do that ansaty. I think they call themselves generalized Hartree-Fock/DFT or something to that degree. Note that you need to allow complex/imaginary alpha/beta linear combinations to allow for new physics (in this case: spin polarization in arbitrary directions).

That approach is only very rarely used in Quantum Chemistry because we are dealing with a spin-free Hamiltonian (in the absence of magnetic fields etc). That implies that proper wave functions are supposed to be eigenfunctions of the Sz and S^2 operators. Such eigenfunctions do not need this generality, and, in fact, don't even need spatial wave functions which differ for alpha and beta spin (as you'd get in UHF).

Allowing these arbirary linear combinations allows you to create wave functions which break these spin symmetries even more than normal UHF wave functions do. While there may be a few cases where this allows you to get energies which are somewhat less wrong than RHF or UHF (e.g., think of frustrated systems like a cyclopropane triradical), in general the situation is similar to that in the other cases: If RHF and UHF give qualitatively different results, then most likely both are wrong. The same would apply for this generalized HF: if it gives qualitatively different numbers than RHF or UHF, then it does so by breaking the physical spin symmetry, and thus cannot be trusted.

 

[cgk]

sorry, I might have got your question wrong: my answer applies to the more general spin orbitals phi(r,s) = phi1(r)*A(s) + phi2(r)*B(s), with phi1 and phi2 being different functions (or the a/b in your formulas being space-dependent functions). If such mixing is not allowed (a and b are global, not space-dependent), your ansatz is equivalent to the standard UHF ansatz, as lineraly-dependent components of the ansatz orbitals get canceled when put into a Slater determinant. I.e., whatever you express by such a linear combinatin can also be achieved by another linear combination of orbitals with pure A or B spin part and maybe different spatial parts.

 

[SpectraCat]

To state cgk's answer another way: they use the first version you gave because it makes the mathematics simpler. The second version is completely equivalent to the first, it just is messier to deal with, and offers no obvious advantage.

Actually, now that I think about it .. your second formulation might represent a nice basis for understanding spin contamination in UHF calculations. Instead of having the spatial parts of the orbitals be different for the alpha and beta electrons, you could instead have a single set of spatial orbitals, but have them be populated by mixtures of alpha and beta electrons. Of course that shouldn't affect the results of the calculation, so it could probably be done as an orbital transform after the calculation was complete. I wonder if anyone has done this? These sound kind of like natural orbitals, and kind of like natural spin orbitals, but I think they are distinct from either of those.

 

[Derivator]

... your ansatz is equivalent to the standard UHF ansatz, as lineraly-dependent components of the ansatz orbitals get canceled when put into a Slater determinant...

I don't see this. Adding one column to another would change nothing. But my second ansatz is not equivalent to adding one column to another.

For example in a closed shell case, adding a multiple of one column (eg. the 2nd) to another (eg, the 1st) would mean:

stuff in first column looks like: spacialOrbital(r)*(alpha(s)+ b*beta(s))
where b is a complex number.

but the stuff in the second column still looks like: spacialOrbital(r)*beta(s), so here is no linear combination of spin wavefunctions, as my second ansatz would imply.

 

[alxm]

It seems to me you're saying you'd have a Slater determinant that looked something like this:

[tex]\begin{vmatrix} a\alpha(1) + b\beta(1) & \beta(1)\\a\alpha(2) + b\beta(2) & \beta(2) \end{vmatrix} = a(\alpha(1)\beta(2) - \alpha(2)\beta(1)) + b(\beta(1)\beta(2) - \beta(2)\beta(1))[/tex]

You don't see the problem here?

 

[Derivator]

It seems to me you're saying you'd have a Slater determinant that looked something like this:

[tex]\begin{vmatrix} a\alpha(1) + b\beta(1) & \beta(1)\\a\alpha(2) + b\beta(2) & \beta(2) \end{vmatrix} = a(\alpha(1)\beta(2) - \alpha(2)\beta(1)) + b(\beta(1)\beta(2) - \beta(2)\beta(1))[/tex]

You don't see the problem here?

no, my proposal would look like (in a closed shell case):

[tex]\begin{vmatrix} \Psi(\vec{r})(a\alpha(1) + b\beta(1)) & \Psi(\vec{r})(c\alpha(1) + d\beta(1))\\\Psi(\vec{r})((a\alpha(2) + b\beta(2)) & \Psi(\vec{r})(c\alpha(2) + d\beta(2)) \end{vmatrix}[/tex]

 

[Derivator]

*push*

 

[SpectraCat]

Why are you bumping the thread? What more do you want to know?

Remember that [itex]\alpha[/itex] and [itex]\beta[/itex] form a complete basis for all the possible 1-electron spin states. So the simplest choice for forming spin orbitals from a given spatial orbital [itex]\phi_i(\vec{r})[/itex] is just to form [itex]\phi_i(\vec{r})\alpha[/itex] and [itex]\phi_i(\vec{r})\beta[/itex]. As cgk already pointed out, if you allow arbitrary combinations of [itex]\alpha[/itex] and [itex]\beta[/itex], then you will end up with linear dependencies (in the spin degrees of freedom) between the columns of your Slater determinant.

 

[Derivator]

I somehow had in mind, that I didn't understand this topic completely. But thinking about it again, you are right. It was already explained completely.

sorry for that.