In the case of two such vectors φ1and φ2 having a dot product φ1|φ2 we construct the new orthogonal wave Fig.. 2 LÖWDIN SYMMETRIC ORTHOGONALIZATION Imagine the normalized but non-orthogo
Trang 1976 I SPACE- AND BODY-FIXED COORDINATE SYSTEMS
ˆT = − ¯h2 2m1 1− ¯h
2 2m2 2
= − ¯h 2 2m1
m1 M
2
∂2
∂XCM2 + ∂2
∂x2− 2 ∂2
∂XCM∂x
+ (similarly for y and z)
− ¯h2 2m2
m2 M
2
∂2
∂XCM2 +
m1 M
2
∂2
∂x2+ 2m1m2
M2
∂2
∂XCM∂x
+ (similarly for y and z)
= − ¯h2 2M CM− ¯h2
2m1
m1 M
2 xyz
− ¯h 2 2m2
m1 M
2 xyz− ¯h
2 2m1
m1 M
2
−2 ∂2
∂XCM∂x
+ · · ·
− ¯h2 2m22m1m2
M2
∂2
∂XCM∂x+ · · ·
= − ¯h2 2M CM− ¯h2
2m1
m1 M
2 xyz− ¯h2 2m2
m1 M
2 xyz
= − ¯h2 2M CM− ¯h2
2
m1
m2M
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 ˆH 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 particle9 with coordinates x y z
(defined in this example) and mass equal to m2 M
m1
9 Let us first denote the nucleus as particle 1 and the electron as particle 2 Then, RCMalmost shows the position of the nucleus, and x y z are almost the coordinates of the electron measured from the nu-cleus, whilem2 M
m1 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− x 1 , to insert them into Pythagoras’ formula for the distance We have
x1= X CM
m1+ m 2
m1 −m2
m1x2= X CM
m1+ m 2
m1 −m2
m1(x+ X CM ) = X CM −m2
m1x
x1− x 2 = X CM −m2
m1x− x − X CM = −x1 +m2
m1
This gives immediately (r stands for the electron-centre of mass distance): V (new) = − Ze2
(1+m2m1)r
Trang 2J 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 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, uand 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 − u u|v ,
Case II: u= u − v v|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
φ1and φ2 having a dot product φ1|φ2 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
Trang 3978 J ORTHOGONALIZATION
functions ψ1= φ1, ψ2= φ2− φ1 φ1|φ2 or ψ1= φ1− φ2 φ2|φ1 , 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 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 orthogonalizaorthogonaliza-tion) treats all the wave funcorthogonaliza-tions 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−1φ where S is the overlap matrix with the elements Sij= φi|φj , and the square matrix S−1 and its cousin S1 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
the symbol S
1 diag Using the latter matrix we define the matrices
S1 = USdiag1 U† and S− 1
=S1−1
= US−diag1 U†
Their symbols correspond to their properties:
S1S1 = USdiag1 U†US
1 diagU†= USdiag1 S
1 diagU†= USdiagU†= S
similarly S−1S− 1
= S−1 Also, a straightforward calculation gives1S− 1
S1 = 1.
1 The matrix S − 1
is no longer a symbol anymore Let us check whether the transformation
φ = S−1φ indeed gives orthonormal wave functions (vectors) Remembering that φ represents
a vertical vector with components φi (being functions):
φ ∗φT dτ = S, while φ ∗φTdτ=
S − 1
φ ∗φT S − 1
dτ= 1 This is what we wanted to show.
Trang 4An important feature of symmetric orthogonalization is2that among all possible
orthogonalizations it ensures that
i
::φi− φ
i::2= minimum whereφi− φ
i2≡ φi− φ
i|φi− φ
i This means that the symmetrically orthogonalized functions φiare 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 φ1and φ2), each of length 1, with
a dot product3 u|v = a = 0 We decide to consider vectors with real components,
hence a∈ R First we have to construct matrix S− 1
Here is how we arrive there
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†= UT), where (to ensure the orthogonality of the transformation
matrix) we choose
U=
cos θ sin θ
− sin θ cos θ
and therefore U†=
cos θ − sin θ sin θ cos θ
with angle θ to be specified After the transformation we have:
U†SU=
1− a sin 2θ a cos 2θ
a cos 2θ 1+ a sin 2θ
We see that if we chose θ= 45◦, the matrix U†SUwill be diagonal4(this is what
we would like to have):
Sdiag=
1− a 0
0 1+ a
We then construct
S
1 diag=
√
1+ a
2G.W Pratt, S.P Neustadter, Phys Rev 101 (1956) 1248.
3
4 In such a case the transformation matrix is
U =
⎛
⎝
1
√ 2 1
√ 2
− √ 1 2 1
√ 2
⎞
⎠ = √1 2
−1 1
Trang 5
980 J ORTHOGONALIZATION
Next, we form5
S1 = USdiag1 U†=1
2
!√
1− a +√1+ a √1+ a −√1− a
√
1+ a −√1− a √1− a +√1+ a
"
and the matrix S−1 needed for the transformation is equal to
S− 1
= US−diag1 U†= U
√
1 +a
"
U†=1 2
⎛
⎝
1
√
1 −a+√ 1
1 +a √11+a−√ 1
1 −a 1
√
1 +a−√ 1
1 −a √11−a+√ 1
1 +a
⎞
⎠
Now we are ready to construct the orthogonalized vectors:6
u
v
=1 2
⎛
⎝
1
√
1 −a+√ 1
1 +a √11+a−√ 1
1 −a 1
√
1 +a−√ 1
1 −a √11−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 uis the same as that of v 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,
S1
ij =US 1
diag U†
ij =
k
l
Uik S 1
diag
kl Ujl=
k
l
Uik S 1
diag
kl δklUjl
k
Uik S 1
diag
kk Ujk=S1
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, we like this result.
Trang 6Fig J.2.The symmetric (or Löwdin’s)
orthogonal-ization of the normalized vectors u and v The
vec-tors are just pushed away by the same angle in such
a way as to ensure u and vbecome orthogonal.
The new vectors automatically have length 1, the same as the starting vectors
Trang 7K DIAGONALIZATION OF A MATRIX
In quantum chemistry we often encounter the following mathematical problem
We have a Hermitian1matrix 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
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.3This 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
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 solution4to 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 Then each 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 λkand Lkfor k= 1 2 3 n
• The second way is based on diagonalization of A
First, let us show that the 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 (AT) ∗ = 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
Trang 8the arbitrary non-singular5 square matrix6 B We obtain the following chain of
transformations B−1(A− λ1)L = B−1(ABB−1− λ1)L = (B−1AB− λ1)B−1L=
( ˜A− λ1) ˜L = 0, where7 ˜A = B−1AB, and ˜L= B−1L 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 ˜Awere diagonal, then
det ˜A− λ1
= n
@
k =1
˜Akk− λ= 0
which gives the solution λk= ˜Akk Then, it is easy to find the corresponding
vec-tor ˜Lk For example, ˜L1we find from equation ( ˜A− λ11) ˜L1= 0 in the following
way:9
⎛
⎜
⎜
⎝
0 A˜22− λ1 0 0
0 0 A˜33− λ1 0
⎞
⎟
⎟
⎠
⎛
⎜
⎜
⎝
˜L1 1
˜L1 2
˜L1 3
˜L1 n
⎞
⎟
⎟
⎠=
⎛
⎜
⎜
⎝
0 0 0 0 0
⎞
⎟
⎟
⎠
which means that to get 0 on the right side, we have to have an arbitrary ˜L1 1, while
the other ˜L1 j= 0 for j = 2 3 n
To have the length of ˜L1equal to 1, it is sufficient to put ˜L1 1= 1 Similarly, we
easily find that the vectors ˜Lk corresponding to λk simply represent the column
vectors with all components equal to 0 except component k, which equals 1 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 Lkis nothing else but the k-th
column of matrix B Since B 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.
7Such a unitary matrix B (i.e satisfying (BT) ∗ = B −1) can be found, that B−1ABis 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 , because we are interested in getting ˜ L
Trang 9L SECULAR EQUATION
A typicalε 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 equation1(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: Hij= φi| ˆHφj , Sij= φi|φj we will use the orthogonal basis set φ= S−1φ, where S−1 is calculated as described in Appendix J Then we multiply the secular equation (H− εS)c = 0 from the left by S−1 and make the following transformations
S− 1
H− εS−1S
c= 0
S− 1
H1− εS− 1
S
c= 0
S− 1
HS− 1
S1 − εS−1S
c= 0
S− 1
HS− 1
S1 − εS1c= 0
S− 1
HS− 1
− ε1S1c= 0
˜H− ε1˜c = 0 with ˜H= S−1HS− 1
and˜c = S1c
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, we have to diagonalize S to calculate S1 and S−1
1 See Chapter 5.
984
Trang 10Secular 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 The eigenvalues
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 ai Then a
new overlap integral and the corresponding matrix element of the Hamiltonian ˆH
would be Sij= aiajSij, Hij = aiajHij The new secular determinant det(H− εS)
may be expressed by the old secular determinant times a number.2This 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 Hidentical, the answer
would be: no.3However, in quantum chemistry we do not calculate the
eigenval-ues4 of H , but solve the secular equation (H− εS)c= 0 If H changes with
respect to H , there is a corresponding change of S when compared to S This
guarantees that the εs do not change
2 We divide the new determinant by a1, which means dividing the elements of the first row by a1and
in this way removing from them a1, both in H 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.
... data-page="7">K DIAGONALIZATION OF A MATRIX
In quantum chemistry we often encounter the following mathematical problem
We have a Hermitian1matrix A (of dimension n), i.e...
admix-ture of the old vectors, i.e the contribution of u and v in uis the same as that of v and u in v
The new vectors are obtained by correcting the directions of the... eigenvalues of the matrices H are Hidentical, the answer
would be: no.3However, in quantum chemistry we not calculate the
eigenval-ues4 of H , but