976 I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS ˆ T =− ¯ h 2 2m 1 1 − ¯ h 2 2m 2 2 =− ¯ h 2 2m 1 m 1 M 2 ∂ 2 ∂X 2 CM + ∂ 2 ∂x 2 −2 ∂ 2 ∂X CM ∂x +(similarly for y and z) − ¯ h 2 2m 2 m 2 M 2 ∂ 2 ∂X 2 CM + m 1 M 2 ∂ 2 ∂x 2 +2 m 1 m 2 M 2 ∂ 2 ∂X CM ∂x +(similarly for y and z) =− ¯ h 2 2M CM − ¯ h 2 2m 1 m 1 M 2 xyz − ¯ h 2 2m 2 m 1 M 2 xyz − ¯ h 2 2m 1 m 1 M 2 −2 ∂ 2 ∂X CM ∂x +··· − ¯ h 2 2m 2 2 m 1 m 2 M 2 ∂ 2 ∂X CM ∂x +··· =− ¯ h 2 2M CM − ¯ h 2 2m 1 m 1 M 2 xyz − ¯ h 2 2m 2 m 1 M 2 xyz =− ¯ h 2 2M CM − ¯ h 2 2 m 1 m 2 M xyz It is seen that once again we have reached a situation allowing us to separate the motion of the centre of mass in the Schrödinger equation. This time, however, the form of the operator ˆ H is different (e.g., xyz has only formally the same form as ), only because the variables are different (the operator remains the same). Once again this is the kinetic energy of a point-like particle 9 with coordinates xyz (defined in this example) and mass equal to m 2 M m 1 9 Let us first denote the nucleus as particle 1 and the electron as particle 2. Then, R CM almost shows the position of the nucleus, and x y z are almost the coordinates of the electron measured from the nu- cleus, while m 2 M m 1 is almost equal to the mass of the electron. Thus we have a situation which resembles Example 1. If the particles are chosen the other way (the electron is particle 1 and the nucleus is particle 2), the same physical situation looks completely different. The values of x y z are very close to 0, while the mass of the effective point-like particle becomes very large. Note, that the new coordinates describe the potential energy in a more complex way. We need differences of the kind x 2 −x 1 , to insert them into Pythagoras’ formula for the distance. We have x 1 =X CM m 1 +m 2 m 1 − m 2 m 1 x 2 =X CM m 1 +m 2 m 1 − m 2 m 1 (x +X CM ) =X CM − m 2 m 1 x x 1 −x 2 =X CM − m 2 m 1 x −x −X CM =−x 1 + m 2 m 1 This gives immediately (r stands for the electron-centre of mass distance): V(new) =− Ze 2 (1+ m 2 m 1 )r J. ORTHOGONALIZATION 1 SCHMIDT ORTHOGONALIZATION Two vectors Imagine two vectors u and v, each of length 1 (i.e. normalized), with the dot prod- uct u|v=a.Ifa = 0, the two vectors are orthogonal. We are interested in the case a = 0. Can we make such linear combinations of u and v, so that the new vectors, u and v , will be orthogonal? We can do this in many ways, two of them are called the Schmidt orthogonalization: Case I: u =u, v =v −uu|v, Case II: u =u −vv|u, v =v. It is seen that Schmidt orthogonalization is based on a very simple idea. In Case I the first vector is left unchanged, while from the second vector, we cut out its component along the first (Fig. J.1). In this way the two vectors are treated differently (hence, the two cases above). In this book the vectors we orthogonalize will be Hilbert space vectors (see Appendix B), i.e. the normalized wave functions. In the case of two such vectors φ 1 and φ 2 having a dot product φ 1 |φ 2 we construct the new orthogonal wave Fig. J.1. The Schmidt orthogonalization of the unit (i.e. normalized) vectors u and v. The new vectors are u and v . Vector u ≡u, while from vector v we subtract its component along u. The new vectors are orthogonal. 977 978 J. ORTHOGONALIZATION functions ψ 1 = φ 1 , ψ 2 = φ 2 − φ 1 φ 1 |φ 2 or ψ 1 = φ 1 − φ 2 φ 2 |φ 1 , ψ 2 = φ 2 by analogy to the previous formulae. More vectors In case of many vectors the procedure is similar. First, we decide the order of the vectors to be orthogonalized. Then we begin the procedure by leaving the first vector unchanged. Then we continue, remembering that from a new vector we have to cut out all its components along the new vectors already found. Of course, the final set of vectors depends on the order chosen. 2 LÖWDIN SYMMETRIC ORTHOGONALIZATION Imagine the normalized but non-orthogonal basis set wave functions collected as the components of the vector φ. By making proper linear combinations of the wave functions, we will get the orthogonal wave functions. The symmetric orthogonaliza- tion (as opposed to the Schmidt orthogonalization) treats all the wave functions on an equal footing. Instead of the old non-orthogonal basis set φ, we construct a new basis set φ by a linear transformation φ =S − 1 2 φ where S is the overlap matrix with the elements S ij =φ i |φ j , and the square matrix S − 1 2 and its cousin S 1 2 are defined in the following way. First, we diagonalize S using a unitary matrix U,i.e. U † U =UU † =1 (for real S the matrix U is orthogonal, U T U =UU T =1), S diag =U † SU The eigenvalues of S are always positive, therefore the diagonal elements of S diag can be replaced by their square roots, thus producing the matrix denoted by the symbol S 1 2 diag . Using the latter matrix we define the matrices S 1 2 =US 1 2 diag U † and S − 1 2 = S 1 2 −1 =US − 1 2 diag U † Their symbols correspond to their properties: S 1 2 S 1 2 =US 1 2 diag U † US 1 2 diag U † =US 1 2 diag S 1 2 diag U † =US diag U † =S similarly S − 1 2 S − 1 2 =S −1 . Also, a straightforward calculation gives 1 S − 1 2 S 1 2 =1. 1 The matrix S − 1 2 is no longer a symbol anymore. Let us check whether the transformation φ = S − 1 2 φ indeed gives orthonormal wave functions (vectors). Remembering that φ represents a vertical vector with components φ i (being functions): φ ∗ φ T dτ = S,while φ ∗ φ T dτ = S − 1 2 φ ∗ φ T S − 1 2 dτ =1 This is what we wanted to show. 2 Löwdin symmetric orthogonalization 979 An important feature of symmetric orthogonalization is 2 that among all possible orthogonalizations it ensures that i φ i −φ i 2 =minimum where φ i −φ i 2 ≡φ i −φ i |φ i −φ i . This means that the symmetrically orthogonalized functions φ i are the “least distant” from the original functions φ i . Thus symmetric orthogonalization means a gentle pushing the directions of the vectors in order to get them to be orthogonal. Example Symmetric orthogonalization will be shown taking the example of two non- orthogonal vectors u and v (instead of functions φ 1 and φ 2 ), each of length 1, with a dot product 3 u|v=a =0 We decide to consider vectors with real components, hence a ∈ R. First we have to construct matrix S − 1 2 .Hereishowwearrivethere. Matrix S is equal to S = 1 a a 1 , and as we see it is symmetric First, let us di- agonalize S. To achieve this, we apply the orthogonal transformation U † SU (thus, in this case U † = U T ), where (to ensure the orthogonality of the transformation matrix) we choose U = cosθ sinθ −sin θ cosθ and therefore U † = cosθ −sinθ sinθ cosθ with angle θ to be specified. After the transformation we have: U † SU = 1 −a sin 2θacos2θ a cos 2θ 1 +a sin 2θ Weseethatifwechoseθ =45 ◦ ,thematrixU † SU will be diagonal 4 (this is what we would like to have): S diag = 1 −a 0 01+a We then construct S 1 2 diag = √ 1 −a 0 0 √ 1 +a 2 G.W. Pratt, S.P. Neustadter, Phys. Rev. 101 (1956) 1248. 3 −1 a 1. 4 In such a case the transformation matrix is U = ⎛ ⎝ 1 √ 2 1 √ 2 − 1 √ 2 1 √ 2 ⎞ ⎠ = 1 √ 2 11 −11 980 J. ORTHOGONALIZATION Next, we form 5 S 1 2 =US 1 2 diag U † = 1 2 √ 1 −a + √ 1 +a √ 1 +a − √ 1 −a √ 1 +a − √ 1 −a √ 1 −a + √ 1 +a and the matrix S − 1 2 needed for the transformation is equal to S − 1 2 =US − 1 2 diag U † =U 1 √ 1−a 0 0 1 √ 1+a U † = 1 2 ⎛ ⎝ 1 √ 1−a + 1 √ 1+a 1 √ 1+a − 1 √ 1−a 1 √ 1+a − 1 √ 1−a 1 √ 1−a + 1 √ 1+a ⎞ ⎠ Now we are ready to construct the orthogonalized vectors: 6 u v = 1 2 ⎛ ⎝ 1 √ 1−a + 1 √ 1+a 1 √ 1+a − 1 √ 1−a 1 √ 1+a − 1 √ 1−a 1 √ 1−a + 1 √ 1+a ⎞ ⎠ u v u = Cu +cv v = cu +Cv where the “large” coefficient C = 1 2 1 √ 1 −a + 1 √ 1 +a and there is a “small” admixture c = 1 2 1 √ 1 +a − 1 √ 1 −a As we can see the new (orthogonal) vectors are formed from the old ones (non- orthogonal) by an identical (hence the name “symmetric orthogonalization”) admix- ture of the old vectors, i.e. the contribution of u and v in u isthesameasthatofv and u in v . The new vectors are obtained by correcting the directions of the old ones, each by the same angle. This is illustrated in Fig. J.2. 5 They are symmetric matrices. For example, S 1 2 ij = US 1 2 diag U † ij = k l U ik S 1 2 diag kl U jl = k l U ik S 1 2 diag kl δ kl U jl = k U ik S 1 2 diag kk U jk = S 1 2 ji 6 We see that if the vectors u and v were already orthogonal, i.e. a =0, then u = u and v = v.Of course,welikethisresult. 2 Löwdin symmetric orthogonalization 981 Fig. J.2. The symmetric (or Löwdin’s) orthogonal- ization of the normalized vectors u and v.Thevec- tors are just pushed away by the same angle in such a way as to ensure u and v become orthogonal. The new vectors automatically have length 1, the same as the starting vectors. K. DIAGONALIZATION OF A MATRIX In quantum chemistry we often encounter the following mathematical problem. We have a Hermitian 1 matrix A (of dimension n), i.e. A † =A, and are interested in all numbers λ (called “eigenvalues” 2 ) and the corresponding column vectors (“eigenvectors” of dimension n) L, that satisfy the following equation (A −λ1)L =0 (K.1) where 1 is the unit matrix (of dimension n). There are n solutions to the last equa- tion: n eigenvalues of λ and also n eigenvectors L. Some eigenvalues λ may be equal (degeneracy), i.e. two or more linearly independent eigenvectors L corre- spond to a single eigenvalue λ. From (K.1) it is shown that any vector L is deter- mined only to the accuracy of a multiplicative factor. 3 This is why, in future, there will be justification for normalizing them to unity. In quantum chemistry the eigenvalue problem is solved in two ways: one is easy for n 2, but more and more difficult for larger n, the second (using computers) treats all cases uniformly. • The first way sees the eigenvalue equation as a set of linear homogeneous equa- tions for the unknown components of vector L. Then the condition for the non- trivial solution 4 to exist is: det(A −λ1) = 0. This condition can be fulfilled only for some particular values of λ, which are to be found by expanding the determi- nant and solving the resulting n-th degree polynomial equation for λ.Theneach solution λ is inserted into eq. (K.1) and the components of the corresponding vector L are found using any method applicable to linear equations. Thus, we end up with λ k and L k for k =1 2 3n. • The second way is based on diagonalization of A. First,letusshowthatthe same λ’s satisfy the eigenvalue equation (K.1), but with a much simpler matrix. To this end let us multiply (K.1) by (at the moment) 1 In practice, matrix A is usually real, and therefore satisfies (A T ) ∗ =A T =A,i.e.A is symmetric. 2 They are real. 3 In other words, a unnormalized wave function still satisfies the Schrödinger equation, or an arbitrary amplitude can be assigned to any normal mode. 4 The trivial one is obviously L =0, which is however unacceptable, since the wave function cannot vanish everywhere, or atoms have to vibrate, etc. 982 K. DIAGONALIZATION OF A MATRIX 983 the arbitrary non-singular 5 square matrix 6 B. We obtain the following chain of transformations B −1 (A − λ1)L =B −1 (ABB −1 − λ1)L = (B −1 AB − λ1)B −1 L = ( ˜ A − λ1) ˜ L = 0,where 7 ˜ A = B −1 AB,and ˜ L = B −1 L. Thus, another matrix and other eigenvectors, but the same λ’s! Now, let us choose such a special B so as to have the resulting equation as simple as possible, i.e. with a diagonal ˜ A.Then we will know, 8 what the λ values have to be in order to satisfy the equation ( ˜ A −λ1) ˜ L =0. Indeed, if ˜ A were diagonal, then det ˜ A −λ1 = n k=1 ˜ A kk −λ =0 which gives the solution λ k = ˜ A kk . Then, it is easy to find the corresponding vec- tor ˜ L k .Forexample, ˜ L 1 we find from equation ( ˜ A −λ 1 1) ˜ L 1 = 0 in the following way: 9 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 00 0 0 0 ˜ A 22 −λ 1 0 0 00 ˜ A 33 −λ 1 0 00 0 ˜ A nn −λ 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ˜ L 11 ˜ L 12 ˜ L 13 ˜ L 1n ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 0 0 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ which means that to get 0 ontherightside,wehavetohaveanarbitrary ˜ L 11 ,while the other ˜ L 1j =0forj =2 3n. To ha v e t h e l e n g t h of ˜ L 1 equalto1,itissufficienttoput ˜ L 11 =1. Similarly, we easily find that the vectors ˜ L k corresponding to λ k simply represent the column vectors with all components equal to 0 except component k,whichequals1.We are interested in vectors L, rather than ˜ L. We get these vectors from L =B ˜ L,and taking into account the form of ˜ L, this means that L k is nothing else but the k-th column of matrix B.SinceB is known, because this is precisely the matrix which led to the diagonalization, there is therefore no problem with L: the columns of B represent the eigenvectors L of the equation (A − λ1)L =0. This is it. 5 That is, its inverse matrix exists. 6 To be found. 7 Such a unitary matrix B (i.e. satisfying (B T ) ∗ =B −1 ) can be found, that B −1 AB is real and diagonal. When (as is the case in most applications) we have to do with real matrices, then instead of unitary and Hermitian matrices, we have to do with orthogonal and symmetric matrices, respectively. 8 Just by looking. 9 The λ has been replaced by λ 1 , because we are interested in getting ˜ L 1 . L. SECULAR EQUATION (H −εS)c =0 Atypicalε approach for solving an eigenvalue problem is its “algebraization”, i.e. representation of the wave function as a linear combination of the known basis functions with the unknown coefficients. Then instead of searching for a function, we try to find the expansion coefficients c from the secular equation 1 (H −εS)c = 0. Our goal is to reduce this task to the eigenvalue problem of a matrix. If the basis set used is orthonormal, the goal would be immediately achieved, because the secular equation would be reduced to (H −ε1)c =0, i.e. the eigenvalue problem. However, in most cases the basis set used is not orthonormal. We may, however, orthonormalize the basis. We will achieve this using symmetric orthogonalization (see Appendix J, p. 977). Instead of the old basis set (collected in the vector φ), in which the matrices H and S were calculated: H ij =φ i | ˆ Hφ j , S ij =φ i |φ j we will use the orthogonal basis set φ =S − 1 2 φ,whereS − 1 2 is calculated as described in Appendix J. Then we multiply the secular equation (H − εS)c = 0 from the left by S − 1 2 and make the following transformations S − 1 2 H −εS − 1 2 S c = 0 S − 1 2 H1 −εS − 1 2 S c = 0 S − 1 2 HS − 1 2 S 1 2 −εS − 1 2 S c = 0 S − 1 2 HS − 1 2 S 1 2 −εS 1 2 c = 0 S − 1 2 HS − 1 2 −ε1 S 1 2 c = 0 ˜ H −ε1 ˜ c = 0 with ˜ H =S − 1 2 HS − 1 2 and ˜ c =S 1 2 c. The new equation represents the eigenvalue problem, which we solve by diago- nalization of ˜ H (Appendix K, p. 982). Thus, the equation (H −εS)c = 0 is equivalent to the eigenvalue problem ( ˜ H − ε1) ˜ c =0. To obtain ˜ H,wehavetodiagonalizeS to calculate S 1 2 and S − 1 2 . 1 See Chapter 5. 984 L. SECULAR EQUATION (H −εS)c = 0 985 Secular equation and normalization If we used non-normalized basis functions in the Ritz method, this would not change the eigenvalues obtained from the secular equation. The only thing that would change are the eigenvectors. Indeed, imagine we have solved the secular equation for the normalized basis set functions: (H −εS)c = 0.Theeigenvalues ε have been obtained from the secular determinant det(H − εS) = 0. Now we wish to destroy the normalization and take new basis functions, which are the old basis set functions multiplied by some numbers, the i-th function by a i .Thena new overlap integral and the corresponding matrix element of the Hamiltonian ˆ H would be S ij =a i a j S ij , H ij =a i a j H ij . The new secular determinant det(H −εS ) may be expressed by the old secular determinant times a number. 2 This number is irrelevant, since what matters is that the determinant is equal to 0. Thus, whether in the secular equation we use the normalized basis set or not, the eigenvalues do not change. The eigenfunctions are also identical, although the eigenvectors c are different – they have to be, because they multiply different functions (which are proportional to each other). If we ask whether the eigenvalues of the matrices H are H identical, the answer would be: no. 3 However, in quantum chemistry we do not calculate the eigenval- ues 4 of H, but solve the secular equation (H − εS )c = 0.IfH changes with respect to H, there is a corresponding change of S when compared to S.This guarantees that the ε s do not change. 2 We divide the new determinant by a 1 , which means dividing the elements of the first row by a 1 and in this way removing from them a 1 ,bothinH and in S . Doing the same with a 2 and the second row, etc., and then repeating the procedure for columns (instead of rows), we finally get the old determinant times a number. 3 This is evident, just think of diagonal matrices. 4 Although we often say it this way. . starting vectors. K. DIAGONALIZATION OF A MATRIX In quantum chemistry we often encounter the following mathematical problem. We have a Hermitian 1 matrix A (of dimension n), i.e. A † =A, and are. orthogonalization”) admix- ture of the old vectors, i.e. the contribution of u and v in u isthesameasthatofv and u in v . The new vectors are obtained by correcting the directions of the old ones, each. ask whether the eigenvalues of the matrices H are H identical, the answer would be: no. 3 However, in quantum chemistry we do not calculate the eigenval- ues 4 of H, but solve the secular equation