Hellmann-Feynman theorem

If the Hamiltonian of a system changes, the change in energy should be related to both the change in the Hamiltonian and the response of the wave function. The Hellmann-Feynman theorem states that for an exact eigenfunction, only the derivative of the Hamiltonian is decisive. This statement can be generalized to any kind of variational wave function. And since I finally kind of understand the math behind it, I will put up some formulas. The derivation is similar to this Ref. [1].

We start with some parameters describing our wave function, (e.g. orbital coefficients and expansion coefficients ).



And we need some kind of a rule to make an n-electron wave function from λ at a geometry R [2].



We can compute the energy of this wave function in the usual way



We may formally view this as a function of λ (not any more a bilinear function) and rewrite it according to



This way all geometry dependence is in the Hamiltonian - the actual geometry dependence that comes from the motion of the nuclear charges and the geometry dependence that arises from the wave function expansion (e.g. motion of atom centered basis sets).

The above expression gives you the energy for any kind of expansion λ at any geometry R. Now in practice we will have some kind of a rule (or mapping) on how to create the wave function (e.g. MCSCF) at every geometry.



Now, the geometry dependent energy with respect to this rule is given according to:



Next we can compute the derivative of this expression with respect to a coordinate x in R (derivatives will be denoted with a superscript (x) or (λ)). This requires the use of the chain rule [3]



or with the definition from above

The first term evaluates the change in the Hamiltonian with the reference wave function. The second term is the gradient of the energy with respect to the expansion parameters multiplied with the change in expansion parameters.

If λ was determined in a completely variational way, then

The corresponding second term cancels and we arrive at the Hellmann-Feynman theorem.

The significance is that with a completely variational method, we do not have to worry about the response of the wave function to the geometrical perturbation and the formulas simplify. In real life MCSCF (and the special case SCF) is the expansion where this holds. For such a wave function the gradient comes directly from combining the density matrix (from last post) with the atomic derivative integrals. CI gradients are more complex because the orbitals are not optimal with respect to the CI expansion coefficients. And in coupled cluster not even the amplitudes are variational.

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[1] R. Shepard. The Analytic Gradient Method for Configuration Interaction Wave Functions. in Modern electronic structure theory. 1995 ed. D.R. Yarkony

[2] Note that there are three types of coordinates: wave function coordinates λ, coordinates of the Hamiltonian R (I am thinking here mostly about the nuclear geometry but it could be anything else as well) and implicit electronic coordinates that I am not writing out.
- The λ in the wikipedia article corresponds to my R actually (I hope it is still understandable).
- A slightly tricky thing in this context is that in general the same λ can give a different wave function at a different geometry. This happens because atom centered basis sets move around in space with the atoms.

[3] The notation is kind of difficult with all these different types of coordinates (maybe not written in the optimal way here) but in principle this is a straight forward application of the chain rule. You could also do it with Taylor expansions truncated to first order - but chain rule is prettier ...

Density

Density matrices are not only nice for mathematical reasons, they also offer a stringent physical definition of orbitals. The point of this post is to show how you can create what is called "natural orbitals" from any kind of wave function. These things are also interesting with respect to orbital imaging (as introduced here and discussed for example here).

In order not to make any prior assumptions, I have to use general wave functions. In real life you would usually look at these things from the perspective of an orbital basis. And then the expressions would be simpler (maybe I will show that, too).

The one particle density matrix γ₁ of an n-particle wave function Ψ is defined as:



This quantity is a "convenient partial sum". It lets you evaluate all the properties of 1-particle operators without the need of the whole wave function. This operation (in a finite orbital basis) is carried out in many quantum chemical codes.

Because the particles are interchangable you only need one such density matrix. (An n-particle density matrix which is analogously defined, lets you evaluate the matrix element of any n-particle operator.)

First we have to define what a one particle operator is. I think the most straight forward (and hopefully correct) definition ist the following:
An operator A is said to act only on r₁ if

The idea is that you can do all the integrations with respect to the other variables ahead. You need some variable renaming to make sure that the operator only acts on the second function.

This contains already the density matrix



From the density matrix we can construct, what is called "natural orbitals" without the need of any a priory definition of an orbital basis.

Maybe it would be easier to understand it in the matrix representation but in the general case we can also define an operator of the following form.



We may also symmetrize it (again easier in cartesian coordinates; edit: the 1-particle density of a wave function is already symmetric). This symmetrized operator has clearly defined orthogonal 1-particle eigenfunctions, the so called "natural orbitals" (and occupation numbers as eigenvalues). The natural orbitals represent the 1-particle density matrix. And this one particle density matrix contains all the information needed for 1-particle operators (like e.g. the dipole). So natural orbitals are something kind of physcial in my opinion.

The eigenvalues of the density matrix are the occupations. In a Hartree-Fock wave function the natural orbitals correspond to the usual HF orbitals and the eigenvalue of occupied orbitals is one (zero for virtuals). The density matrix is idempotent (or equivalently a projection). In the general case there is no clear distinction between occupied and virtual orbitals and there is strictly spoken no such thing as a HOMO or LUMO. But then we notice that HF represents most wave functions quite well in "well behaved" cases. So you may still think about HOMOs and LUMOs even in a more strict sense.

cgi

This page is my contribution to a world without spam.

The idea is that it gives you a new random email address everytime you visit it. So in principle such things could fill up databases of spam companies who are using robots to search the web for email addresses.

And I wanted to play with cgi a little bit.

"Guests"

Here is how Argonne National Laboratory, Argonne, IL (just to make sure) treats their "guests":

"The Guest hereby agrees on behalf of the Guest, his/her heirs, administrators,
executors, and assigns to release the Department, Argonne, the UChicago and its Board
of Trustees, and the officers, agents, and employees of the foregoing from all liability for
any personal injuries to the Guest which may arise out of the Guest’s access to Argonne
and the use of its facilities. The Guest further hereby covenants and agrees that neither
the Guest nor the Guest’s heirs, administrators, executors, or assigns will bring any suit
in a court of law or equity or before any administrative agency for any such injury
(including injury causing death) against any of the aforementioned parties."

Did I misunderstand something or is this just crazy and way out of line? The way this is formulated Argonne staff could actively attempt to murder me and I could not sue them. I assume that such a waiver is not in accordance with US law anyway but why do people write such a crap then?

Things like that just piss me off. But I guess it's probably not enough to make me really cancel the trip.