The simplest ab initio methods for describing excited electronic states are based on Hartree-Fock theory, namely configuration interaction singles (CIS) and time-dependent Hartree-Fock (TDHF), which is also known as the random-phase approximation (RPA). The purpose of this project is to develop simple implementations of these two methods and to understand the basic differences between them. This project assumes you have completed at least the CCSD programming project.
You can read more about these methods in J.B. Foresman, M. Head-Gordon, J.A. Pople, and M.J. Frisch, J. Phys. Chem. 96, 135-141 (1992) (CIS) and P. Jørgensen and J. Simons, "Second-Quantization Based Methods in Quantum Chemistry", Academic Press, New York, 1981. (RPA)
The fundamental idea behind CIS is the representation of the excited-state wave functions as linear combinations of singly excited determinants relative to the Hartree-Fock reference wave function, viz.
where m identifies the various excited states, and we will use i and j (a and b) to denote occupied (unoccupied) spin-orbitals. Inserting this into the Schrödinger equation and left-projecting onto a particular singly excited determinant gives
If we recognize that we have one of these equations for every combination of i and a spin-orbitals, then this equation may be viewed as a matrix eigenvalue problem:
To solve this equation, we need an expression for the matrix elements in terms of things we already know, i.e. Fock matrix elements and two-electron integrals. This can be done using either algebraic or diagrammatic techniques to obtain (in the spin-orbital notation of previous projects):
Our task is then relatively simple: Build the Hamiltonian matrix (expressed in the basis of all singly excited determinants) using the above expression and diagonalize it. Note that the CIS Hamiltonian matrix is symmetric (check for yourself by swapping i/a and j/b), with dimensions of the number of occupied orbitals times the number of unoccupied orbitals. For our STO-3G/H2O test case, with its ten occupied and four unoccupied spin-orbitals, the matrix will be 40 x 40.
Make sure you can compute the correct CIS excitation energies for each of the four test cases provided at the end of the page. Before you move on to the next section, can you explain the degeneracies appearing in the excitation-energy list?
If you carefully examine the list of eigenvalues you computed above, you will notice a pattern: some of the eigenvalues are unique (i.e., they occur only once), while others appear in groups of three. The former correspond to spin-singlet excited states and the latter to spin-triplets, and all are eigenfunctions of the S2 and Sz operators by virtue of the fact that the spin-restricted Hartree-Fock reference wave function is such an eigenfunction (by design, of course) and that the single-excitation operators must yield identical amplitudes (to within a sign) for both α- and β-spin excitations.
Consider the structure of the singlet and triplet determinantal wave functions from a simple two-electron/two-orbital example (such as the 1s 2s excited state configuration of the He atom). One can easily show that the four possible determinants arising from this configuration,
are components of one singlet and one triplet in the following combinations:
where the superscript is the spin multiplicity (2S+1) and the subscript is the MS value of the wave function. So, if we wanted to compute only the eigenvalues and eigenfunctions corresponding spin singlets in our CIS calculation, we could introduce the restriction on our CIS coefficients that α and β excitations involving the same spatial orbitals must be identical (including the sign). Similarly, if we wanted only triplets, we could require that the α and β excitations have the same magnitude but opposite signs.
Let's begin with the singlets. Starting from the spin-orbital eigenvalue expression and the equation for the CIS Hamiltonian matrix elements in the previous section, we may write a spin-factored equation for the α coefficients as
Note that the mix-spin cases (where j=α and b=β or vice versa) do not contribute since the Fock matrix elements and two-electron integrals must all give zero. If we then carry out spin integration on the integrals in the above expression and assume that the α and β CI coefficients are identical for the same spatial orbitals, i.e.,
we obtain the spatial orbital expression
The part in brackets above is an expression for the spatial-orbital CIS Hamiltonian, spin-adapted for singlet excited states, and diagonalization of this matrix will yield only the singlet eigenvalues you obtained from your spin-orbital matrix earlier.
How about the triplets? We use exactly the same spin-factorization, but instead require
This yields a slightly simpler Hamiltonian:
which, upon diagonalization, will yield only the triplet eigenvalues (but each only occurring once) from your earlier diagonalziation.
Why should we worry about spin adaptation? Because the dimension of the spatial-orbital Hamiltonian matrix is half that of the spin-orbital Hamiltonian, which is a factor of four savings in the storage of the matrix and a factor of eight in the diagonalization. For large basis sets and/or large molecules, this is a considerable computational savings and essential for production-level codes.
Make sure you can implement the spin-adapted CIS approach entirely in spatial orbitals and obtain the correct excitation energies so that you can take advantage of this greater efficiency available in this formulation. (NB: For the CH4 test case below, you still get groups of three excitation energies in the spin-adapted formulation. Do you understand why this degeneracy arises?)
The TDHF/RPA approach is closely related to CIS in that only singles are involved in the wave function expansion, except that both excitation and “de-excitation” operators are involved. (It is probably better to view the TDHF/RPA wave function expansion in terms of orbital rotations instead of Slater determinants, but that's a discussion for another day.) The TDHF/RPA eigenvalue equations take the form
The definition of the A matrix is just the CIS matrix itself, viz.
while X and Y are the parameters of single excitations and de-excitations, respectively, and the B matrix is simply
Thus, the row/column dimension of the TDHF/RPA Hamiltonian is twice that of the CIS Hamiltonian, and the matrix is non-symmetric (so you must be careful about the diagonalization function you choose). Do you obtain twice as many excitation energies?
Instead of solving the full-dimensional TDHF/RPA equations, which, as noted above, is twice the size of the CIS problem (and thus four times the Hamiltonian storage cost), one can rearrange the eigenvalue equations. First write eigenvalue equation two separate equations, each in terms of the submatrices A and B:
Now take +/- combinations of these equations to obtain
Solve for (X+Y) in the second equation:
Insert this result into the first equation, rearrange a bit, and finally obtain:
This is an eigenvalue equation of the same dimension as the CIS eigenvalue equation (number of occupied orbitals times number of unoccupied orbitals), where the eigenvalue is the square of the excitation energy and the eigenvector is X-Y.
Make sure you can get the same set of excitation energies using the full-dimensional TDHF/RPA approach above, as well as this reduced-dimension approach, for all four test cases below.