Laplace expansion

The derivation of the electrostatic properties from the multipole coefficients given below follows the method of Su and Coppens (1992). It employs the Fourier convolution theorem used by Epstein and Swanton (1982) to evaluate the electric field gradient at the atomic nuclei. A direct-space method based on the Laplace expansion of 1/ RP — r has been described by Bentley (1981). 

Inserting the linear combination (1) into (2), after the Laplace-expansion of the determinants the two-matrix has the form ... 

Using the Laplace expansion in terms of the first row of the determinant and changing over the right side of (15.16) again to the wave function of... 

The relation between the CFP with a detached electrons and the reduced matrix elements of operator q>(lNfiLS generating [see (15.4)] the <7-electron wave function is established in exactly the same way as in the derivation of (15.21). Only now in the appropriate determinants we have to apply the Laplace expansion in terms of a rows. The final expression takes the form... 

Note that the output of roots(p) and eig(compan(p)) each is a complex column vector of length three, i.e., each output lies in C3 and the two solution vectors are identical. Our trial polynomial p(x) = a 3 — 2a 2 + 4 has one pair of complex conjugate roots 1.5652 1.0434 i and one real root -1.1304. The (first row) companion matrix P of a normalized nth degree polynomial p (normalized, so that the coefficient an of xn in p is 1) is the sparse n by n matrix P = C(p) as described in formula (1.2). Note that our chosen p is normalized and has zero as its coefficient ai for a = a 1, i.e., the (1, 2) entry in P is zero. For readers familiar with determinants and Laplace expansion, it should be clear that expanding det(P — xI) along row 1 establishes that our polynomial p(x) is the characteristic polynomial of P. Hence P s eigenvalues are precisely the roots of the given polynomial p. 

Multiplying a row by k and adding it (adding corresponding elements) to another row causes no further change in D (in the Laplace expansion the terms without k cancel). 

The suffixes a and b refer to the two possible inelastic paths. Now by replacing summation by integration and approximating the Legendre polynomials P(cos 0) by their Laplace expansion (valid for 1.0 1), the following expression is obtained for the inelastic scattering amplitude (19)... 

To make the relation between the charge density distribution and the potential more explicit, we first take the Laplace expansion [14,11]... 

Using the Laplace expansion (see Appendix A available at http //booksite.elsevier.com/978-0-444-59436-5 on p. el), we get... 

We start from the Slater determinant built of N molecular spinorbitals. Any of these is a linear combination of the spinorbitals of the donor and acceptor. We insert these combinations into the Slater determinant and expand the determinant according to the first row (Laplace expansion, see Appendix A available at booksite. elsevier.com/978-0-444-59436-5 on p. el). As a result, we obtain a linear combination of the Slater determinants, all having the donor or acceptor spinorbitals in the first row. For each of the resulting Slater determinants, we repeat the procedure, but focusing on the second row, then the third row, etc. We end up with a linear combination of the Slater determinants that contain only the donor or acceptor spinorbitals. We concentrate on one of them, which contains some particular donor and acceptor orbitals. We are interested in the coefficient C d) that multiplies this Slater determinant number i. 

For any square matrix A = /4,y, we may calculate a number called its determinant and denoted by det A or A. The determinant is computed by using the Laplace expansion... 

By repeating (i.e., expanding Aij, ete.) flie Laplace expansion again and again, we arrive finally at a linear combination of products of the elements ... 

From the Laplace expansion, it follows that if one of the spinorbitals is composed of two functions i = - - f, then the Slater determinant is a sum of the two Slater determinants, one with instead of i, and the second with f instead of... 

From the Laplace expansion, it follows that multiplying the determinant by a number is equivalent to multiplying an arbitrary row (column) by this number. Therefore, det (cA) = det A, where N is die matrix dimension. ... 

Indeed, the Laplace expansion (Appendix A) along the row corresponding to the first new virtual spinorbital leads to the linear combination of the determinants containing new virtual, which means that the rank of excitation is not changed by this) orbitals in this row. Continuing this procedure with the Slater determinants obtained, we finally get a linear combination of n-tuply excited Slater determinants expressed in old spinorbitals. 

After this is done the Laplace expansion gives... 

Equations (A. 17) and (A. 18) are known as the Laplace expansion formulas they imply that to obtain A, one must do the following ... 

Fig. 4.1 Number of terms needed for Laplace expansion in order to calculate R + r to within 10 accuracy. Two cases are shown where one of the two radii, R, is fixed to R = 2 R = 6. Number of maximum terms was set to be 160. We can see that convergence improves as the radial separation of the two vectors increases (Angular variables were chosen at random and held constant)...

For the more complicated three- and four-center integrals, direct substitution of the Laplace expansion (Eq.4.10), will clearly result in a multiple open sum expressions such that, numerical calculations will be computationally too slow to be of any realistic usage. However, the Poisson equation technique [9,10] will avoid several of the open sums and allow for the efficient calculation of the three- and four-center integrals. Using these techniques, the calculation of the three- and four-center integrals is under investigation. 

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