Ideas of Quantum Chemistry P102

976 I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS 2 2 T = −2m1 11 − 2m2 12 2 ·µ ¶2µ 2 2 2 ¶¸ = −2m1 M ∂X2M + ∂x2 −2∂XCM∂x +(similarly for y and z) ¯2 ·µm2 ¶2 ∂2 µm1 ¶2 ∂2 m1m2 ∂2 ¸ 2m2 M ∂X2M M ∂x2 M2 ∂XCM∂x +(similarly for y and z) 2 2 µ ¶2 = −2M1CM − 2m1 M 1xyz 2 µ ¶2 2 µ ¶2µ 2 ¶ − 2m2 M 1xyz − 2m1 M −2∂XCM∂x +··· − 2m2 2m1m2 ∂XCM∂x +··· 2 2 µ ¶2 2 µ ¶2 = −2M1CM − 2m1 M 1xyz − 2m2 M 1xyz 2 2 µ ¶ = −2M1CM − 2 m2M 1xyz: 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 H is different (e.g., 1xyz has only formally the same form as 1), only because the variables are different (the operator remains the same). Once again this is the kinetic energy of a point-like particle9 with coordinates x;y;z (defined in this example) and mass equal to m2M : 9Let us first denote the nucleus as particle 1 and the electron as particle 2. Then, RCM almost shows the position of thenucleus, and x;y;z arealmost the coordinates ofthe electron measuredfromthe nu-cleus, while m2M 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 x2 −x1, to insert them into Pythagoras’ formula for the distance. We have x1 =XCM m1 +m2 − m2 x2 =XCM m1 +m2 − m2 (x+XCM)=XCM − m2 x; x1 −x2 =XCM − m1 x−x−XCM =−xµ1+ m1 ¶: This gives immediately (r stands for the electron-centre of mass distance): V (new)=−(1+ e2 )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 hu|vi = a. If a = 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, u0 and v0, will be orthogonal? We can do this in many ways, two of them are called the Schmidt orthogonalization: Case I: u0 =u, v0 =v −uhu|vi, Case II: u0 =u−vhv|ui, v0 =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 φ1 and φ2 having a dot product hφ1|φ2i 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 u0 and v0. Vector u0 ≡u, while from vector v we subtract its component along u. The new vectors are orthogonal. 977 978 J. ORTHOGONALIZATION functions ψ1 = φ1, ψ2 = φ2 − φ1hφ1|φ2i or ψ1 = φ1 − φ2hφ2|φ1i, ψ2 = φ2 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 vectorunchanged.Thenwecontinue,rememberingthatfromanewvectorwehave 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 componentsofthevector φ.Bymakingproperlinearcombinationsofthewave functions, we will get the orthogonal wave functions. The symmetricorthogonaliza-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 φ0 by a linear transformation φ0 = S−1 φ; where S is the overlap matrix with the elements Sij =hφi|φji, and the square matrix S−2 ; and its cousin S2 ; 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, UTU =UUT =1), Sdiag =U†SU: The eigenvalues of S are always positive, therefore the diagonal elements of Sdiag can be replaced by their square roots, thus producing the matrix denoted by 1 the symbol Sdiag. Using the latter matrix we define the matrices S1 =USdiagU† and S−1 =¡S1 ¢−1 =USdiagU†: Their symbols correspond to their properties: S1 S1 =USdiagU†USdiagU† =USdiagSdiagU† =USdiagU† =S; similarly S−1 S−1 =S−1. Also, a straightforward calculation gives1 S−1 S1 =1. 1The matrix S−2 is no longer a symbol anymore. Let us check whether the transformation φ0 = S−1 φ indeed gives orthonormal wave functions (vectors). Remembering that φ represents a vertical vector with components φi (being functions): φ∗φT dτ = S, while φ0∗φ0T dτ = S−2 φ∗φTS−2 dτ =1: This is what we wanted to show. 2 Löwdin symmetric orthogonalization 979 Animportantfeatureofsymmetricorthogonalizationis2 thatamongallpossible orthogonalizations it ensures that X°φi −φ0°2 =minimum i where kφi −φ0k2 ≡hφi −φ0|φi −φ0i. This means that the symmetrically orthogonalized functions φ0 are the “least distant” from the original functions φi. 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 φ1 and φ2), each of length 1, with a dot product hu|vi=a=0: We decide to consider vectors with real components, hence a ∈ R. First we have to construct matrix S−2 . Here is how we arrive there. Matrix S is equal to S = 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† = UT), where (to ensure the orthogonality of the transformation matrix) we choose U =µ−sinθ cosθ¶; and therefore U† =µcosθ −sinθ¶ with angle θ to be specified. After the transformation we have: † µ1−asin2θ acos2θ acos2θ ¶ 1+asin2θ We see that if we chose θ=45◦, the matrix U†SU will be diagonal4 (this is what we would like to have): 1−a 0 diag 0 1+a We then construct 1 µ√1−a 0 ¶ diag 0 1+a 2G.W. Pratt, S.P. Neustadter, Phys. Rev. 101 (1956) 1248. 3−1 6a61. 4In such a case the transformation matrix is ⎞ √ √ U =⎝ 2 √2 ⎠= √ −1 2 2 1¶ 1 980 J. ORTHOGONALIZATION Next, we form5 1 1 1 Ã√1−a+√1+a 2 diag 2 1+a− 1−a √1+a−√1−a! 1−a+ 1+a and the matrix S−2 needed for the transformation is equal to S−2 =USdiagU† =U à √1−a 0 !U† = 1 ⎛ √1−a + √1+a 1+a 1+a 1−a ⎞ √1+a − √1−a ⎠ √1−a + √1+a Now we are ready to construct the orthogonalized vectors:6 µu0 ¶ 1 ⎛ √1−a + √1+a √1+a − √1−a ⎞µu¶ v0 2 √1+a − √1−a √1−a + √1+a v u0 = Cu+cv; v0 = cu+Cv: where the “large” coefficient µ ¶ C = 2 √1−a + √1+a ; and there is a “small” admixture µ ¶ c = 2 √1+a − √1−a : As we can see the new (orthogonal) vectors are formed from the old ones (non-orthogonal) by an identical(hence the name “symmetricorthogonalization”) admix-ture of the old vectors, i.e. the contribution of u and v in u0 is the same as that of v and u in v0. 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. 5They are symmetric matrices. For example, ¡S1 ¢ij = ¡USdiagU†¢ij =XXUik¡Sdiag¢klUjl =XXUik¡Sdiag¢klδklUjl k l k l = XUik¡Sdiag¢kkUjk =¡S1 ¢ji: k 6We see that if the vectors u and v were already orthogonal, i.e. a = 0, then u0 = u and v0 = v. Of course, we like this result. ... - tailieumienphi.vn