A numerical method for computing dispersion constants

Abstract

We reformulate and discuss a previously proposed variational numerical technique for the computation of dispersion coefficients. The method extends the Full CI idea to the perturbation equation for the intermolecular interaction, by expanding the perturbative solution in a small number of tensor products of suitably chosen Full CI vectors. Some new expansion vectors are proposed and their convergence properties are tested by performing computations on HF and H₂O. Last, a natural state analysis of the solution is performed via an orthogonal transformation of the original expansion vectors and it is found that a single couple of natural states strongly dominates the expansion.

Appendices

Appendix 1

Strictly speaking, Eq. 1 has an infinity of solutions, since to any given solution Φ ^AB_1ab we can add products of the type Φ ^A₀ f ^B and g ^A Φ ^B₀ , where f ^B and g ^A are arbitrary functions, and get another solution. The dispersion solution is characterized by a kind of strong orthogonality to the ground state eigenvectors of molecule A and B, as clearly shown by the London sum-over-states representation given in Eq. 2:

$$ \int\limits_A \Upphi^{AB}_{1 a b} \Upphi_{0}^{A} dV_A =\int\limits_B \Upphi^{AB}_{1 a b} \Upphi_{0}^{B} dV_B=0 $$

Equivalently, we can define Φ ^AB_1ab to be the solution of minimal norm of Eq. 1. As far as Eq. 11 is concerned, one has the conditions:

$$ {{\mathbf{v}}}_{0}^{A} {{\mathbf{X}}}={{\mathbf{0}}}^{B}={{\mathbf{0}}}, \quad {{\mathbf{X}}}({{\mathbf{v}}}_{0}^{B} )^T= ({{\mathbf{0}}}^A)^T $$

where v ^A₀ , v ^B₀ are the Full CI ground state eigenvectors of molecule A, B, respectively, and 0 ^A, 0 ^B are zero vectors. Consequently, the expansion vectors should fulfill the orthogonality requirements:

$$ ({{\mathbf{z}}}_{i}^{A})^T {{\mathbf{v}}}_{0}^{A}=({{\mathbf{z}}}_{j}^{B})^T {{\mathbf{v}}}_{0}^{B}=0. $$

Appendix 2

For reader’s convenience we report here the explicit formulae used to compute the dispersion coefficients from the cartesian matrix elements:

$$ \begin{aligned} C_{6}^{00000}& = {\frac{2}{3}} \left[\langle xx|xx \rangle_\otimes + \langle xy|xy \rangle_\otimes + \langle xz|xz \rangle_\otimes \right. \\ &\quad\left. + \langle yx|yx \rangle_\otimes + \langle yy|yy \rangle_\otimes + \langle yz|yz \rangle_\otimes\right.\\ &\quad\left. + \langle zx|zx \rangle_\otimes + \langle zy|zy \rangle_\otimes + \langle zz|zz \rangle_\otimes \right] \\ C_{6}^{20002} = & -{\frac{\sqrt{5}}{3}} \left[\langle xx|xx \rangle_\otimes + \langle xy|xy \rangle_\otimes + \langle xz|xz \rangle_\otimes \right. \\ &\quad\left. + \langle yx|yx \rangle_\otimes + \langle yy|yy \rangle_\otimes \langle yz|yz \rangle_\otimes \right. \\ &\quad \left.-2( \langle zx|zx \rangle_\otimes + \langle zy|zy \rangle_\otimes + \langle zz|zz \rangle_\otimes) \right] \\ C_{6}^{22002} = & \sqrt{{\frac{5}{6}}} \left[\langle xx|xx \rangle_\otimes + \langle xy|xy \rangle_\otimes + \langle xz|xz \rangle_\otimes \right.\\ &\quad \left. - (\langle yx|yx \rangle_\otimes + \langle yy|yy \rangle_\otimes + \langle yz|yz \rangle_\otimes) \right] \cr C_{6}^{22224} = & {\frac{27}{\sqrt{70}}} \left[ \langle xx|xx \rangle_\otimes - \langle xy|xy \rangle_\otimes - \langle yx|yx \rangle_\otimes + \langle yy|yy \rangle_\otimes \right] \end{aligned} $$

in the notation defined by Eq. 5.