MODULE 8 APPROXIMATION METHODS I Once we move past the two particle systems, the Schrödinger equation cannot be solved exactly. The electronic inter-repulsion term in the hamiltonian prevents us from separating it into the sum of one-electron operators, and thus exact solutions for systems of three or more particles elude us. Fortunately approximation methods have been developed which can generate solutions that are very accurate. Variation theory and Perturbation theory

MODULE 8 The Variation Method The ground state of an arbitrary system defined by a wavefunction  0 with eigenvalue E 0. When the wavefunction is normalized the denominator is unity, but this is not necessarily the case

MODULE 8 Now substitute any other function  for  0 and calculate the corresponding energy E  Then according to the variation theorem and the equality holds only if   

MODULE 8 Thus the energy E  of the new system will never be less than that of the ground state. From this we see that we can calculate an upper limit (bound) for the value of E 0 by using some trial function. The closer we can get the wavefunction  to be like  0, the closer will E  approach the true value E 0. The way to proceed is to choose a trial function that depends on one or more arbitrary parameters (variational parameters).

At this point we minimize E   with respect to the parameters and thus obtain the best possible ground state energy that our trial function can provide. The success of this approach clearly depends on our ability to select a good trial wavefunction ( clever guesswork). MODULE 8 For example choose  ( , , , …). The energy will also depend on the same parameters

MODULE 8 We already know that we can determine the energy of the ground (1s) state of a hydrogen atom exactly without recourse to approximations. So this is a good test case for the variation approach-how well does it recover the known value? In the 1s state, l = 0, and the hamiltonian can be simplified to

MODULE 8 Now we decide on the trial function. From our knowledge of the exact solution we can expect that the wavefunction will decay to zero as r increases. One such function is a Gaussian of the form  (r) = exp(-  r 2 ) FIGURE 8.1where  can be the variational parameter. Now substitute trial function and the operator into the numerator and denominator of the basic equation

MODULE 8 Now we want to minimize E  with respect to  by differentiating and setting the result equal to zero.

MODULE 8 This value can be compared to the exact value of E 1s from above As the theorem predicts, E min > E 1s, i.e. it is less negative. In the normalized trial wavefunction it is convenient to put  in terms of the Bohr radius, a 0. Thus:

MODULE 8 The Gaussian is a very poor approximation to the true exponential, particularly at values of r less than the Bohr radius. Nevertheless the trial wavefunction leads to a value for the state energy which is within 80% of the exact result. This was with a single variational parameter (  ). It might be anticipated that by using more flexible trial functions with more parameters, we could approach the true value even closer.

MODULE 8 The Helium Atom Now we use the variational method to find the upper bound of the energy for He, with one nucleon (Z = 2) and two electrons. We do not know the exact value of this. The Hamiltonian operator for this is given by: The resulting Schrödinger equation cannot be solved exactly because of the potential energy term involving r 12 If we ignore the inter-electronic repulsion term the ground state wavefunction becomes separable

MODULE 8 This equation can be used as a trial function in which Z is a variational parameter. Because the variational parameter is part of the exponential function this is an example of a non-linear variation The 1s wavefunctions in the equation are of the form (remember for 1s, l = 0)

MODULE 8 or, in au laborious math Minimizing E(Z) with respect to Z we find Z min = 27/16

MODULE 8 Substitution back into equation for E(Z) The experimental result is – and the best-refined calculation provides – Thus the result for E min is reasonable (it is less negative, i.e. greater than the experimental value) considering the simplicity of the trial function we have employed. au

MODULE 8 The use of nuclear charge (Z) as a variational parameter implies that Z min = 27/16 (or ) can be interpreted as an effective nuclear charge. That the value is significantly less than 2 reflects the fact that each electron is partially screening the nucleon from the other electron, thereby reducing the net charge seen by the nucleon. This lessening of the nuclear attraction means that the 1s orbital in helium will be larger than a hydrogenic 1s with Z = 2.

MODULE 8 Linear Combinations and Secular Determinants improved accuracy can be achieved by increasing the number of variational parameters employ a trial function that is a linear combination of known functions with variable coefficients. Such a trial function can be written as We opt for simplicity by choosing N = 2 and require that c n and f n are real (not necessary, but keeps it simple).

MODULE 8 where

MODULE 8 the operator has to be Hermitian in order to guarantee real eigenvalues, hence Following the same routine, the denominator of the variational equation becomes

MODULE 8 multiplying out the LHS and taking the partial derivative of E wrt c 1, we get Thus we have E as a function of the variational parameters c i Prior to differentiating it is better to write it differently

MODULE 8 Now put the derivative term equal to zero since we are minimizing, and rearrange Differentiate equation again, but now wrt c 2 These last two equations are a pair of linear homogeneous algebraic equations, containing the two unknowns c 1 and c 2. The trivial solution is that both coefficients equal zero, but then the wavefunction would equal zero.

MODULE 8 Non-trivial solutions to such a set of equations do exist If, and only if, the determinant of the coefficients vanishes, i.e. if This is called the secular determinant, and expansion of it yields a polynomial (here a quadratic) secular equation in E. Finding the roots of the quadratic yields two values of E, of which we select the lowest as the approximate energy of our ground state entity.

MODULE 8 An example of the linear combination (LC) procedure: The particle in the one-dimensional box with vertical walls where the potential energy becomes infinitely large, but within the box, V(x) = 0. The Hamiltonian for this situation is that for kinetic energy only The trial functions we choose are from the set

MODULE 8 These functions have zero amplitude at the walls and a single maximum at the center of the box, as needed for the n = 1 wavefunction. As usual to keep matters simple we use the first two terms only as our BASIS SET, thus

MODULE 8 Using the outlined procedure we evaluate the H and S integrals to be:

MODULE 8 E’ = and This is to be compared to the exact energy Thus the LC technique with a basis set of only two members leads to an energy value within 2 ppm of the exact value - a remarkable achievement.

MODULE 8 It is often useful to operate with a larger basis set than two In the general case for a basis set of N-terms we write the set of linear equations as:

MODULE 8 In the determinant the elements have been inserted using the convention M rc where r = row number and c = column number. It can be simplified using the Hermitian conditions that H ij = H ji and S ij = S ji. Expansion of the determinant leads to a secular N-order polynomial in E. For N > 2 we need to use computational methods to obtain the roots. we select the smallest of the roots for the value of E min.

MODULE 8 From the root taken we can substitute back to find the values of c i. Only N-1 of the equations are independent and we can evaluate only the ratios c N /c 1. If we require that the function be normalized to unity viz. then the value of c 1 can be extracted and so all the rest.

MODULE 8 An extension of the LC method is to use a trial function of the form where f n also contain variational parameters, e.g. Now two sets of variational parameters, c n and  n. The minimization of E with respect to the two sets of parameters involves complicated math and needs to be done numerically. Fortunately algorithms are available for this.

MODULE 8 N E min / au exact The results obtained by taking different numbers of terms for the ground state of the hydrogen atom are shown in Table