15-22. What degeneracy would you anticipate for energies associated with points
Chapter 15 Molecular Orbital Theory of Periodic Systems15-20. Figure out the (k1, k2) coordinates and sketch the COs for s-type basis functions for the following points in the RFBZ of Fig. 15-27c. a) Halfway between X and M. b) Halfway between X and M. () Halfway between and X. d) Halfway between and M. Rank these in order of energy.
15-21. Sketch COs based on a 2px AO at each lattice point for the RFBZ of Fig. 15-27c, at points , X, and X. Rank according to predicted energy.
15-22. What degeneracy would you anticipate for energies associated with points , X, M, and X of Fig. 15-27c?.
15-23. The rectangular lattice of Fig. 15-27a leads to a rectangular RFBZ. What shape RFBZ would result for a square lattice?
15-24. It is stated in the text that mixing π(M) and π ∗(M) of Fig. 15-31b will decrease the difference in their energies. Sketch the functions π(M) ± π∗(M) and comment on their energies.
15-25. For the FBZ of Fig. 15-30f, locate the points (0, π/a) and (π/a, −π/a).
Sketch the π CO corresponding to the first of these and compare it with the π CO for (π/a, 0), sketched in Fig. 15-31b. How are these COs related?
15-26. What physical situation allows us to use the half-range of 0 ≤ k3 ≤ π/a3 for the RFBZ pictured in Fig. 15-33a?
15-27. Give a general argument for there being no interlayer-induced splitting of π bands for any point at the top of the RFBZ of Fig. 15-33a. What does your argument predict for σ bands?
15-28. For two-dimensional graphite, sketch the Bloch sums at and M resulting from a bonding pair of 2px AOs in each unit cell. Let x be the bisector of a1 and a2. Which of these COs would you expect to have higher energy?
References [1] R. Hoffmann, C. Janiak, and C. Kollmar, Macromolecules 24, 3725 (1991).
[2] J.-M. Andr´e and G. Leroy, Int. J. Quantum Chem. 5, 557 (1971).
[3] J. P. Lowe and S. A. Kafafi, J. Amer. Chem. Soc. 106, 5837 (1984).
[4] J. P. Lowe, S. A. Kafafi, and J. P. LaFemina, J. Phys. Chem. 90, 6602 (1986).
[5] D. J. Chadi and M. L. Cohen, Phys. Rev. B 8, 5747 (1973).
[6] G. S. Painter and D. E. Ellis, Phys. Rev. B 1, 4747 (1970).
[7] J. P. LaFemina and J. P. Lowe, Int. J. Quantum Chem. 30, 769 (1986).
[8] J. C. Slater, Quantum Theory of Molecules and Solids, Vol. 2. McGraw-Hill, New York, 1965.
Section 15-17 Summary [9] R. Hoffmann, Solids and Surfaces: A Chemist’s View of Bonding in ExtendedStructures. VCH Publishers, New York, 1988.
[10] N. W. Ashcroft and N. D. Mermin, Solid State Physics. Holt, Rinehart and Winston, New York, 1976.
[11] C. Kittel, Introduction to Solid State Physics, 7th ed. Wiley, New York, 1996.
Appendix 1
Useful Integrals
xneax dx = (xneax/a) − (n/a) xn−1eax dx ∞
xne−ax dx = (n!/an.+1) = n+1(a), n > −1, a > 0
0
∞
e−ax2 dx = 1 π/a
0
2
∞
xe−ax2 dx = 1/2a
0
∞
x2e−ax2 dx = 1 π/a3
0
4
∞
x3e−ax2 dx = 1/2a2
0
∞
π
x2ne−ax2 dx = 1 · 3 · · · · · (2n − 1)
0
2n+1
a2n+1
∞
x2n+1e−ax2 dx = n!
0
2an+1
∞
e−ax dx = e−a
1
a
1
e−ax dx = (1/a)(1 − e−a)
0
∞
xe−ax dx = (e−a/a2)(1 + a)
1
1
xe−ax dx = (1/a2)[1 − e−a(1 + a)]
0
∞
x2e−ax dx = (2e−a/a3)(1 + a + a2/2)
1
1
x2e−ax dx = (2/a3)[1 − e−a(1 + a + a2/2)]
0
∞
n
xne−ax dx = (n!e−a/an+1)
ak/k! ≡ An(a)
1
k=0
582
Appendix 1
∞
n
xne−ax dx = (n!e−ay/an+1) (ay)k/k!
y
k=0
+1
e−ax dx = (1/a)(ea − e−a)
−1
+1
xe−ax dx = (1/a2)[ea − e−a − a(ea + e−a)]
−1
+1
xne−ax dx = (−1)n+1An(−a) − An(a)
−1
+1
0, n = 1, 3, 5, . . .
xndx = −1
2/(n + 1), n = 0, 2, 4, . . .
sin x dx = − cos x
cos x dx = sin x
sin2 x dx = x − sin 2x
2
4
cos2 x dx = x + sin 2x
2
4
x sin x dx = sin x − x cos x x cos x dx = cos x + x sin x x sin2 x dx = x2 − x sin 2x − cos 2x
4
4
8
x cos2 x dx = x2 + x sin 2x + cos 2x
4
4
8
Appendix 2
Determinants
A determinant is a scalar calculated from an ordered set of elements according to a specific evaluation recipe. The elements are ordered in a square array of rows and columns, bounded at left and right by straight vertical lines. For instance, (A2-1) is a 2 × 2 determinant:
x
−i
2
y2
(A2-1)
The recipe for evaluating a 2 × 2 determinant is: From the product of the elements on the principal diagonal (upper left to lower right) subtract the product of the other two elements. Thus, (A2-1) has the value xy2 + 2i.
Larger determinants are evaluated by a process that reduces them step by step to a linear combination of smaller determinants until, finally, they are all 2 × 2’s, which are then evaluated as above. The process of reduction involves the concept of a cofactor.
As our example, we use the 4 × 4 determinant (A2-2), symbolized |M|, where M is the array of elements within the vertical bars:
(A2-2)
The elements are numbered so that the first index tells which row, and the second index which column, the element is in. The cofactor of element a11 is defined as the determinant obtained by removing the row and column containing a11. We see in (A2-2) that striking out row 1 and column 1 gives us a 3 × 3 determinant (dashed outline) as cofactor of a11. Symbolize this cofactor as |A11|.
To evaluate the determinant |M|, we expand in terms of cofactors. We begin by choosing any row or column of M. (We will choose row 1.) Then we write a linear combination containing every element in this row or column times its cofactor: |M| = a11|A11| − a12|A12| + a13|A13| − a14|A14|
(A2-3)
The sign of each term in the linear combination is determined as follows. lf the sum of row and column indices is even, the sign is plus. If the sum is odd, the sign is minus.
Since the indices of a12 and a14 sum to odd numbers, they are minus in (A2-3).
584
Appendix 2
The method of expanding in cofactors is successively applied until a large determi nant is reduced to 3 × 3’s or 2 × 2’s that can be evaluated directly (see Problem A2-1).
Thus, a 5 × 5 is first expanded to five 4 × 4’s and each 4 × 4 is expanded to four 3 × 3’s giving a total of 20 3 × 3’s. This method becomes extremely clumsy for large determinants.
Some useful properties of determinants, symbolized |M|, are stated below without proof. The reader should verify that these are true using 2 × 2 or 3 × 3 examples, or by examining Eq. (A2-3).
1. Multiplying every element in one row or one column of M by the constant c multi plies the value of |M| by c.
2. If every element in a row or column of M is zero, then |M| = 0.
3. lnterchanging two rows or columns of M to produce M results in M = − |M|; i.e., it reverses the sign of |M|.
4. Adding to any row (column) of M the quantity c times any other row (column) of M does not affect the value of the determinant.
5. If two rows or columns of M differ only by a constant multiplier, then |M| = 0.
A2-1 Use of Determinants in Linear Homogeneous Equations
Suppose that we seek a nontrivial solution for the following set of linear homogeneous equations: a1x + b1y + c1z = 0
(A2-4)
a2x + b2y + c2z = 0
(A2-5)
a3x + b3y + c3z = 0
(A2-6)
Here, x, y, and z are unknown and the coefficients ai, bi, ci are given. Let us collect the coefficients into a determinant |M|, a1 b1 c1
|
M| = a2 b2 c2
(A2-7)
a
3
b3 c3
As before, let |A1| be the cofacter of a1, etc. Now, multiply Eq. (A2-4) by |A1| (since |A1| is a determinant, it is just a number, and so this is a scalar multiplication), Eq. (A2-5) by −|A2|, and Eq. (A2-6) by |A3| and add the results to get x(a1|A1| − a2|A2| + a3|A3|)+y(b1|A1| − b2|A2| + b3|A3|) +z(c1|A1| − c2|A2| + c3|A3|) = 0
(A2-8)
The coefficient of x is just |M|. The coefficients of y and z correspond to determinants having two identical rows and hence are zero. Therefore, |M| x = 0
(A2-9)
Determinants
In order for x to be nonzero (i.e., nontrivial) it is necessary that |M| = 0. This is a result that is very useful. The condition that must be met by the coefficients of a set of linear
homogeneous equations in order that nontrivial solutions exist is that their determinant
vanish.
A2-2 ProblemsA2-1. Expand the 3 × 3 determinant (Fig. PA2-1), by cofactors and show that this result is equivalent to the direct evaluation of the 3 × 3 by summing the three products parallel to the main diagonal (solid arrows) and subtracting the three products parallel to the other diagonal (dashed arrows).
Figure PA2-1
1 2 0 1 1 1 0
0 1
3 0 1 4 x 2
A2-2. Evaluate (a)
=
0 1 1 (c)
(d)
0 for x 2 1 (b) 1 1 0 1 1 x 1 1 1 0 2 1 1 A2-3. Verify that the coefficient of y in Eq. (A2-8) is zero.
A2-4. Consider the following set of linear homogeneous equations: 4x + 2y − z = 0, 3x − y − 2z = 0, 2y + z = 0 Do nontrivial roots exist?
A2-5. Find a value for c that allows nontrivial solutions for the equations cx − 2y + z = 0, 4x + cy − 2z = 0, −8x + 5y − cz = 0 A2-6. Five properties of determinants have been listed in this appendix. (a) Prove statement (5) is true assuming statements (1)–(4) are true. (b) Demonstrate statements (1)–(4) using simple examples.
Appendix 3
Evaluation of the Coulomb Repulsion
Integral Over 1s AOs
Evaluation of 1s(1)1s(2)(1/r12)1s(1)1s(2) dv(1) dv(2)
(A3-1)
where
1s(1) = ζ 3/π exp(−ζ r)
(A3-2)
may be carried out in two closely related ways.1 Each method is instructive and sheds light on the other, and so we will give both of them here.
The first method works from a physical model and requires knowledge of two features of situations governed by the 1/r2 force law (e.g., electrostatics, gravitation). Suppose that there exists a spherical shell in which charge or mass is distributed uniformly, like the soap solution in a soap bubble. The first feature is that a point charge or mass outside the sphere has a potential due to attraction (or repulsion) by the sphere that is identical to the potential produced if the sphere collapsed to a point at its center (conserving mass or charge in the process). Thus, the electrostatic interaction between two separated spherical charge distributions may be calculated as though all the charge were concentrated at their centers. The second feature is that the potential is identical for all points inside the spherical shell; that is, if the core of the earth were hollow, a person would be weightless there. There would be no tendency for that person to drift toward a wall or toward the center.
Armed with these facts, we can evaluate the integral. First, we remark that all the functions in the integrand commute. This enables us to write Eq. (A3-1) as 1s2(1)(1/r12)1s2(2)dv(1)dv(2)
(A3-3)
The functions 1s2(1) and 1s2(2) are just charge clouds for electrons 1 and 2, and the integral is evidently just the energy of repulsion between the clouds. Suppose (see Fig. 3-1) that, at some instant, electron 1 is at a distance r1 from the nucleus. What is its energy of repulsion with the charge cloud of electron 2? The charge cloud of electron 2 can be divided into two parts: the charge inside a sphere of radius r1 and the charge outside that sphere. From what we just said, electron 1 experiences a repulsion due to the cloud inside the sphere that is the same as the repulsion it would feel if that 1Other methods, not discussed here, also exist. See, for example, Margenau and Murphy [1, pp. 382–383].
587
Evaluation of the Coulomb Repulsion Integral Over 1s AOs Charge cloud 1s2 (2)
1
r1
Figure A3-1
Sketch of spherical charge cloud 1s2(2) with electron 1 at a distance r1 from the nucleus.
part of the cloud were collapsed to the center. We can calculate the energy due to this repulsion (call it the “inner repulsion energy”) by dividing the product of charges by the distance between them: inner repulsion energy = (fraction of charge cloud 2 inside r1) × 1/r1
r1
= 4π 1s2(2)r2 2 dr2 × 1/r1
(A3-4)
0
where the factor 4π comes from integrating over θ2, φ2. Electron 1 also experiences repulsion from charge cloud 2 outside the sphere of radius r1. But, from what we said above, electron 1 would experience this same “outer repulsion” no matter where it was inside the inner sphere. Therefore we will calculate the energy due to this repulsion as though electron 1 were at the center, since this preserves spherical symmetry and simplifies the calculation. It follows that all the charge in a thin shell of radius r2 repels electron 1 through an effective distance of r2. Integrating over all such shells gives
∞
outer repulsion energy = 4π (1/r2)1s2(2)r22dr2
(A3-5)
r1
The total energy of repulsion between charge cloud 2 and electron 1 at r1 is the sum of inner and outer repulsive energies. But electron 1 is not always at r1. Therefore, we must finally integrate over all positions of electron 1, weighted by the frequency of their occurrence:
∞
r1 repulsive energy = 16π2 1s2(1)
(1/r1) 1s2(2)r2 2 dr2
0
0
∞
+
1s2(2)r2dr2 r21dr1
(A3-6)
r1
Appendix 3
Figure A3-2
The inner integrals contain but one variable, r2, and can be evaluated with the help of Appendix 1. After they are performed, the integrand depends only on r1 and this is also easily integrated yielding 5ζ/8 as the result. A positive value is necessary since a net repulsion exists between two clouds of like charge.
The second method of evaluation is more mathematical and more general. The func tion 1/r12 is expressible2 as a series of terms involving associated Legendre functions:
∞
+l
1
=
(l − |m|)! rl<
|
|
P m|(cos θ
m|
1)P (cos θ2) exp[im(φ1 − φ2)]
(A3-7)
r
l
l
12
(l + |m|)! rl+1
l=0 m=−l
>
This infinite series will give the distance between particles 1 and 2 located at positions r1, θ1, φ1 and r2, θ2, φ2 (Fig. A3-2). All we need to do is pick the larger of r1 and r2 and call that r>, the other being r