I have been reading a journal article on ab-initio pseudo-potentials, and I need some help understanding it. The article is C. F. Melius and W. A. Goddard, III. Phys. Rev. A 1974, 10, 1528.

A summary of the article for those that do not have access follows. The authors present a way to construct a pseudo-potential. Basically, their argument goes like this (according to my understanding):

Suppose that we are using the Hartree Fock method. We can represent the ground-state wavefunction of some atom as a function of the valence orbitals, core orbitals, and spin.

$$\Psi= f (\phi_{core}, \varphi_{valence},X)$$

Note: in the article they use a weird letter instead of $f$. After we obtain the ground-state occupied orbitals, we solve for the valence orbitals. We obtain the valence orbital by solving the variational equations $$H^{HF}\varphi_{i}=\epsilon_i\varphi_{i}$$ Where $H^{HF}$ is the Hartree Fock Hamiltonian and is given by $$H^{HF}=-\frac{\nabla^2}{2}-\frac{Z}{r}+2J_{\alpha}-K_{\alpha}$$ For the valence orbital, $J_{\alpha}$ is equal to the classical Coloumb potential due to a charge density corresponding to the $\phi_{core}$. The exchange operator $K_{alpha}$ is an integral operator resulting from the antisymmetric form of the wavefunction.

The authors then state that the valence orbital of the ground-state wavefunction is NOT determined uniquely by the variational principle. This is because the solutions of the variational equations may not yield a wavefunction that is purely a valence orbital. In other words, the mixing of core orbitals with a valence orbitals does not change its energy. Hence the solution $\varphi_i$ may be of the form $$\varphi_i=\varphi_{valence}+\sum_i c_i\phi_{core}$$ The authors then impose the constraint that $\varphi_{i}$ be orthogonal to other occupied orbitals.

What I do not understand is this. How does orthogonality allow for a unique solution? Also later on, the authors state "Although this orthogonality restriction has the desired consequence of leading to a prescription for a unique valence orbital, there is no reason to consider $\varphi_i$ to have a special significance over any other combination of $\varphi_i$ with the various core orbitals." What do they mean by this statement?


1 Answer 1


First off, I would like to note that you wrongly interpreted the very first equation. Below I quote the relevant part of the paper:

In the Hartree-Fock (HF) approximation for, say, the $\ce{Na}$ atom, the ground-state wave function has the form

$$ \Psi = \mathcal{A} (\Phi_{\mathrm{core}} \phi_\nu X) \, , \tag{1} $$

where $\Phi_{\mathrm{core}}$ is a product of (ten) spatial orbitals very similar to the orbitals of $\ce{Na+}$ and $\phi_\nu$ is the valence orbital (the one removed in ionizing to $\ce{Na+}$). ($X$ is an appropriate product of spin functions.)

So, $\Psi$ is not just some function of spin orbitals designated by a weird letter as you said, it is, as usual, their antisymmetrized product, where I bet the calligraphic A symbol ($\mathcal{A}$) as usual stands for the antisymmetrizer. The only small difference is that small calligraphic a is used in the text rather then much more common capital one.

Now to you actual question. Both core and valence orbitals are thought in the derivation as the solutions of the Hartree-Fock equations, which have a very well-known feature: they are defined to within a unitary transformation (see, for instance, here). So, orbitals which are solutions of the Hartree-Fock equations are not unique: there are infinite different sets of orbitals that minimize the Hartree-Fock energy and you are free to choose any of these set for any further work. This is, essentially, what is said in the paper:

It will be important in our later analysis to note that the valence orbital in (1) is not determined uniquely by the variational principle. One can modify $\phi_\nu$ by mixing in an arbitrary amount of any core orbital (doubly occupied. in $\Phi_{\mathrm{core}}$) without changing the energy.

Just note that this is true not only for the valence orbital: all orbitals (i.e. core as well) are not uniquely determined by the variational principle, as I mentioned above.

Now, if the goal is to have a unique valence orbital, then it can be achieved by restricting it to be orthogonal to core orbitals. In that case when transforming orbitals you can not mix any amount of any core orbital into the valence one since it will immediately make it non-orthogonal with this particular core orbital. And since mixing in core orbitals is the only way you can change the valence one, orthogonality of the valence orbital implies its uniqueness.

In order to obtain unique solutions for $\phi_\nu$, one generally requires that $\phi_\nu$ be orthogonal to the other occupied orbitals.

And it is important to note (as it is done in the next sentence of the paper) that :

Although this orthogonality restriction has the desired consequence of leading to a prescription for a unique valence orbital, there is no reason to consider the orthogonal valence orbital $\phi_\nu$, to have a special significance over any other combination of $\phi_\nu$, with the various core.

Note finally, that due to the orthogonality restriction you can not also mix any amount of the valence orbital into the core ones, but you can mix core orbitals with themselves anyway you like since it won't do any harm for their orthogonality with the valence orbital. Thus, while the valence orbital is uniquely defined by the orthogonality requirement, the core orbitals are still defined to within a unitary transformation.

  • $\begingroup$ Very nice answer, +1. Just want to ask for some clarification: When the authors of the article state One can modify $\phi_\nu$ by mixing in an arbitrary amount of any core orbital (doubly occupied in $\Phi_{\mathrm{core}}$) without changing the energy. they are talking about the total energy of the whole molecule/system not the energy eigenvalue of the valence orbital, right? $\endgroup$
    – Philipp
    Mar 12, 2016 at 12:02
  • $\begingroup$ @Philipp, they are surely talking about the total energy (or just its electronic part). When orbitals are transformed so does their energies: in matrix notation when spin orbitals are transformed as follows $\mathbf{\Psi}' = \mathbf{U} \mathbf{\Psi}$, their energies become $\mathbf{E}' = \mathbf{U} \mathbf{E} \mathbf{U}^{\dagger}$. It is the Slater determinant, and thus, its energy, which is invariant to a transformation, not individual spin orbitals and their energies. $\endgroup$
    – Wildcat
    Mar 12, 2016 at 12:15
  • $\begingroup$ Thanks, for clearing that up. Just wanted to make sure that my basic knowledge still works. I'm getting a bit rusty :) $\endgroup$
    – Philipp
    Mar 12, 2016 at 12:19

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