Section 9-4
Section 9-4 Solving the Matrix Equation
column in C refers to a different MO. The first column refers to the MO having energy E1. In multiplying E by C from the left, each coefficient in column 1 becomes multiplied by E1. This would not occur if we multiplied E by C1 from the right. Therefore, HC = SCE is correct, whereas HC = ESC is incorrect.
9-4 Solving the Matrix Equation
Since we know the basis functions and the effective hamiltonian (in principle, at least), we are in a position to evaluate the elements in H and S. How do we then find C and E?
Let us first treat the simplified situation where our basis set of functions is orthonor mal, either by assumption or design. Then all the off-diagonal elements of S (which correspond to overlap between different basis functions) are zero, and all the diagonal elements are unity because of normality. In short, S = 1. Therefore, Eq. (9-17)
becomes
HC = CE
(9-18)
and our problem is, given H, find C and E.
Now, we want a set of coefficients that correspond to normalized MOs. We have seen earlier that, for an orthonormal basis set, this requires each MO to have coefficients satisfying the equation (assuming real coefficients) c2 + + · · · +
=
1i
c22i
c2ni
1
(9-19)
We can write this as a vector equation
c1i
c
˜
2i
c
i ci ≡ ci1 ci2 · · · cin .
= 1
(9-20) .
.
cni
Furthermore, we know that any two different MOs must be orthogonal to each other.
That is ˜cicj = 0, i = j . All this may be summarized in the matrix equation ˜CC = 1
(9-21)
Hence, the coefficient matrix is orthogonal. In the more general case in which coefficients may be complex, C is unitary; i.e., C†C = 1. Our problem then is, given H, find a unitary matrix C such that HC = CE with E diagonal.
We can multiply both sides of a matrix equation by the same matrix and preserve the equality. However, because matrices do not necessarily commute, we must be careful to carry out the multiplication from the left on both sides, or from the right on both sides. Thus, multiplying Eq. (9-18) from the left by C†, we obtain C†HC = C†CE = 1E = E
(9-22)
where we have used the fact that C is unitary. Now our problem may be stated as, given H, find a unitary matrix C such that C†HC is diagonal.4 Several techniques 4Not every matrix (not even every square matrix) can be diagonalized by a unitary transformation, but every hermitian matrix can be so diagonalized.
Chapter 9 Matrix Formulation of the Linear Variation Method exist for finding such a matrix C. These are generally much more suitable for machine computation than are determinantal manipulations.
We can illustrate that the allyl radical energies and coefficients already found by the HMO method do in fact satisfy the relations C†C = 1 and C†HC = E. The matrix C can be constructed from the HMO coefficients and is
1
1
√
1
2
2
2
C = 1 √
0
− 1 √
2
2
(9-23)
1
− 1 √
1
2
2
2
Therefore
1
1
√
1
1
1
√
1
2
2
2
2
2
2
C†C = 1
1
√
0
− 1 √ √
0
− 1 √
2
2
2
2
1
− 1 √
1
1
− 1 √
1
2
2
2
2
2
2
1
0
0
=
0
1
0 = 1
(9-24)
0
0
1
The matrix H for the allyl radical is, in HMO theory,
α
β
0
H = β α β
(9-25)
0
β
α
The reader should verify that
√
α + 2β
0
0
C†HC =
0
α
0√ = E
(9-26)
0
0
α − 2β
The diagonal elements can be seen to correspond to the HMO energies. Note that the energy in the (1 1) position of E corresponds to the MO with coefficients appearing in column 1 of C, illustrating the positional correlation of eigenvalues and eigenvectors referred to earlier.
If the basis functions are not orthogonal, S = 1 and the procedure is slightly more complicated. Basically, one first transforms to an orthogonal basis to obtain an equation of the form HC = CE. One diagonalizes H as indicated above to find E and C, where C is the matrix of coefficients in the orthogonalized basis. Then one back transforms C into the original basis set to obtain C. There are many choices available for the orthogonalizing transformation. The Schmidt transformation, based on the Schmidt orthogonalization procedure described in Chapter 6, is popular because it is very rapidly performed by a computer. Here we will simply indicate the matrix algebra involved. Let the matrix that transforms a nonorthonormal basis to an orthonormal one be symbolized A. This matrix satisfies the relation A†SA = 1
(9-27)
Section 9-4 Solving the Matrix Equation Furthermore, |A| = 0 and so A−1 exists. We can insert the unit matrix (in the form AA−1) wherever we please in the matrix equation HC = SCE without affecting the equality. Thus,
HAA−1C = SAA−1CE
(9-28)
Multiplying from the left by A† gives A†HAA−1C = A†SAA−1CE
(9-29)
By Eq. (9-27), this reduces to (A†HA)(A−1C) = (A−1C)E
(9-30)
where the parentheses serve only to make the following discussion clearer. If we define
A†HA to be H, and A−1C to be C, Eq. (9-30) becomes
HC = CE
(9-31)
Since we know the matrix H and can compute A from knowledge of S, it is possible to write down an explicit H matrix for a given problem. Then, knowing H (which is just the hamiltonian matrix for the problem in the orthonormal basis), we can seek the unitary matrix C such that C†HC is diagonal. These diagonal elements are our orbital energies. (Note that E in Eq. (9-31) is the same as E in HC = SCE.) To find the coefficients for the MOs in terms of the original basis (i.e., to find C), we use the relation AC = A(A−1C) = 1C = C
(9-32)
One nice feature of this procedure is that, even though we use the inverse matrix A−1 in our formal development, we never need to actually compute it. (A and A† are used to find H, and A is used to find C.) This is fortunate because calculating inverse matrices is a relatively slow process.
A few more words should be said about the process of diagonalizing a hermitian matrix H with a unitary transformation. Two methods are currently in wide use. The older, slower method, known as the Jacobi method, requires a series of steps on the starting matrix. In the first step, a matrix O1 is constructed that causes the largest off-diagonal pair of elements of H to vanish in the transformation H1 = ˜ O1HO1. Now a second transformation matrix O2 is constructed to force the largest off-diagonal pair of elements in H
˜
1 to vanish in the transformation ˜ O2H1O2 = ˜ O2O1HO1O2. This procedure is continued. However, since each transformation affects more elements in the matrix than just the biggest pair, we eventually “unzero” the pair that was zeroed in forming H1 or H2, etc. This means that many more transformations are required than there are off-diagonal pairs. Eventually, however, the off-diagonal elements will have been nibbled away (while the diagonal elements have been building up) until they are all smaller in magnitude than some preselected value, and so we stop the process. The transformation matrix C corresponds to the accumulated product O1O2O3 . . . .
column in C refers to a different MO. The first column refers to the MO having energy E1. In multiplying E by C from the left, each coefficient in column 1 becomes multiplied by E1. This would not occur if we multiplied E by C1 from the right. Therefore, HC = SCE is correct, whereas HC = ESC is incorrect.
9-4 Solving the Matrix Equation
Since we know the basis functions and the effective hamiltonian (in principle, at least), we are in a position to evaluate the elements in H and S. How do we then find C and E?
Let us first treat the simplified situation where our basis set of functions is orthonor mal, either by assumption or design. Then all the off-diagonal elements of S (which correspond to overlap between different basis functions) are zero, and all the diagonal elements are unity because of normality. In short, S = 1. Therefore, Eq. (9-17)
becomes
HC = CE
(9-18)
and our problem is, given H, find C and E.
Now, we want a set of coefficients that correspond to normalized MOs. We have seen earlier that, for an orthonormal basis set, this requires each MO to have coefficients satisfying the equation (assuming real coefficients) c2 + + · · · +
=
1i
c22i
c2ni
1
(9-19)
We can write this as a vector equation
c1i
c
˜
2i
c
i ci ≡ ci1 ci2 · · · cin .
= 1
(9-20) .
.
cni
Furthermore, we know that any two different MOs must be orthogonal to each other.
That is ˜cicj = 0, i = j . All this may be summarized in the matrix equation ˜CC = 1
(9-21)
Hence, the coefficient matrix is orthogonal. In the more general case in which coefficients may be complex, C is unitary; i.e., C†C = 1. Our problem then is, given H, find a unitary matrix C such that HC = CE with E diagonal.
We can multiply both sides of a matrix equation by the same matrix and preserve the equality. However, because matrices do not necessarily commute, we must be careful to carry out the multiplication from the left on both sides, or from the right on both sides. Thus, multiplying Eq. (9-18) from the left by C†, we obtain C†HC = C†CE = 1E = E
(9-22)
where we have used the fact that C is unitary. Now our problem may be stated as, given H, find a unitary matrix C such that C†HC is diagonal.4 Several techniques 4Not every matrix (not even every square matrix) can be diagonalized by a unitary transformation, but every hermitian matrix can be so diagonalized.
Chapter 9 Matrix Formulation of the Linear Variation Method exist for finding such a matrix C. These are generally much more suitable for machine computation than are determinantal manipulations.
We can illustrate that the allyl radical energies and coefficients already found by the HMO method do in fact satisfy the relations C†C = 1 and C†HC = E. The matrix C can be constructed from the HMO coefficients and is
1
1
√
1
2
2
2
C = 1 √
0
− 1 √
2
2
(9-23)
1
− 1 √
1
2
2
2
Therefore
1
1
√
1
1
1
√
1
2
2
2
2
2
2
C†C = 1
1
√
0
− 1 √ √
0
− 1 √
2
2
2
2
1
− 1 √
1
1
− 1 √
1
2
2
2
2
2
2
1
0
0
=
0
1
0 = 1
(9-24)
0
0
1
The matrix H for the allyl radical is, in HMO theory,
α
β
0
H = β α β
(9-25)
0
β
α
The reader should verify that
√
α + 2β
0
0
C†HC =
0
α
0√ = E
(9-26)
0
0
α − 2β
The diagonal elements can be seen to correspond to the HMO energies. Note that the energy in the (1 1) position of E corresponds to the MO with coefficients appearing in column 1 of C, illustrating the positional correlation of eigenvalues and eigenvectors referred to earlier.
If the basis functions are not orthogonal, S = 1 and the procedure is slightly more complicated. Basically, one first transforms to an orthogonal basis to obtain an equation of the form HC = CE. One diagonalizes H as indicated above to find E and C, where C is the matrix of coefficients in the orthogonalized basis. Then one back transforms C into the original basis set to obtain C. There are many choices available for the orthogonalizing transformation. The Schmidt transformation, based on the Schmidt orthogonalization procedure described in Chapter 6, is popular because it is very rapidly performed by a computer. Here we will simply indicate the matrix algebra involved. Let the matrix that transforms a nonorthonormal basis to an orthonormal one be symbolized A. This matrix satisfies the relation A†SA = 1
(9-27)
Section 9-4 Solving the Matrix Equation Furthermore, |A| = 0 and so A−1 exists. We can insert the unit matrix (in the form AA−1) wherever we please in the matrix equation HC = SCE without affecting the equality. Thus,
HAA−1C = SAA−1CE
(9-28)
Multiplying from the left by A† gives A†HAA−1C = A†SAA−1CE
(9-29)
By Eq. (9-27), this reduces to (A†HA)(A−1C) = (A−1C)E
(9-30)
where the parentheses serve only to make the following discussion clearer. If we define
A†HA to be H, and A−1C to be C, Eq. (9-30) becomes
HC = CE
(9-31)
Since we know the matrix H and can compute A from knowledge of S, it is possible to write down an explicit H matrix for a given problem. Then, knowing H (which is just the hamiltonian matrix for the problem in the orthonormal basis), we can seek the unitary matrix C such that C†HC is diagonal. These diagonal elements are our orbital energies. (Note that E in Eq. (9-31) is the same as E in HC = SCE.) To find the coefficients for the MOs in terms of the original basis (i.e., to find C), we use the relation AC = A(A−1C) = 1C = C
(9-32)
One nice feature of this procedure is that, even though we use the inverse matrix A−1 in our formal development, we never need to actually compute it. (A and A† are used to find H, and A is used to find C.) This is fortunate because calculating inverse matrices is a relatively slow process.
A few more words should be said about the process of diagonalizing a hermitian matrix H with a unitary transformation. Two methods are currently in wide use. The older, slower method, known as the Jacobi method, requires a series of steps on the starting matrix. In the first step, a matrix O1 is constructed that causes the largest off-diagonal pair of elements of H to vanish in the transformation H1 = ˜ O1HO1. Now a second transformation matrix O2 is constructed to force the largest off-diagonal pair of elements in H
˜
1 to vanish in the transformation ˜ O2H1O2 = ˜ O2O1HO1O2. This procedure is continued. However, since each transformation affects more elements in the matrix than just the biggest pair, we eventually “unzero” the pair that was zeroed in forming H1 or H2, etc. This means that many more transformations are required than there are off-diagonal pairs. Eventually, however, the off-diagonal elements will have been nibbled away (while the diagonal elements have been building up) until they are all smaller in magnitude than some preselected value, and so we stop the process. The transformation matrix C corresponds to the accumulated product O1O2O3 . . . .