Quantum chemistry (chap 2)

CHAPTER 2

In 1925, Erwin Schrodinger and Werner Heisenberg independently formulated a general quantum theory At first sight, the two methods appeared to be different because Heisenberg's method is formulated in terms of matrices, whereas Schrodinger's method is formulated in terms of partial differential equations Just a year later, however, Schrodinger showed that the two formulations are mathematically equivalent Because most students of physical chemistry are not familiar with matrix algebra, quantum theory is customarily presented according to Schrodinger's formulation, the central feature of which is a partial differential equation now known as the Schrodinger equation Partial differential equations may sound no more comforting than matrix algebra, but fortunarely we require only elementary calculus to treat the problems in this book We learned in Chapter I that matter can behave as a wave, so it's not surprising that the Schrodinger equation (sometimes called the Schrodinger wave equation) describes wavelike behavior The wave equation of classical physics describes various wave phenomena such as a vibrating string, a vibrating drum head, ocean waves, and acoustic waves Not only does the classical wave equation provide a physical background to the Schrodinger equation, but, in addition, the mathematics involved in solving the classical wave equation are central to any discussion of quantum mechanics Because most students of physical chemistry have little experience with classical wave equations, this chapter discusses this topic In particular, we will solve the standard problem of a vibrating string because not only is the method of solving this problem similar to the method we will use to solve the Schrodinger equation, but it also gives us an excellent opportunity to relate the mathematical solution of a problem to the physical nature of the problem Many of the problems at the end of the chapter illustrate the connection between physical problems and the mathematics developed in the chapter 53

54 Chapter 2 I The Classical Wave Equation

I -~ x

0

FIGURE 2.1

A vibrating string whose ends are fixed at 0 and l The displacement of the vibration at position

x and time t is u(x, t)

2.1 The One-Dimensional Wave Equalion Describes lhe Motion of a Vibrating String

Consider a uniform string stretched between two fixed points, as shown in Figure 2.1 The maximum displacement of the string from its equilibrium horizontal position is called its amplitude If we let u(x, t) be the displacement of the string, then u(x, t)

satisfies the equation

ox2 v2 a12

where u is the speed with which a disturbance moves along the string Equation 2.1 is the classical wave equation Equation 2 l is a partial differential equation because the unknown, u(x, 1) in this case, occurs in partial derivatives The variables x and / are

said to be the independent variables and u(x, 1), which depends upon x and t, is said to be the dependent variable Equation 2 I is a linear partial differential equation because

u (x, t) and its derivatives appear only to the first power and there are no cross terms In addition to having to satisfy Equation 2.1, the displacement u (x, t) must satisfy certain physical conditions as well Because the ends of the string are held fixed, the displacement at these two points is always zero, and so we have the requirement that

u(O, 1) = O and u(l,t)=O (for all t) (2.2) These two conditions are called boundary conditions because they specify the behavior of u(x, t) at the boundaries Generally, a partial differential equation must be solved subject to certain boundary conditions, the nature of which will be apparent on physical grounds

2.2 The Wave Equation Can Be Solved by the Method of Separation of Variables

The classical wave equation, as well as the Schrodinger equation and many other partial differential equations that arise in physical chemistry, can be solved readily by a method

2.2 The Wave Equation Can Be Solved by the Method of Separation of Variables called separation of variables We shall use the problem of a vibrating string to illustrate this method

The key step in the method of separation of variables is to assume that u(x, t) factors into a function of x, X (x ), times a function oft, T (t), or that

Ifwe substitute Equation 2.3 into Equation 2.1, we obtain

d2X(x) l d2T(t) T(l) = - X(x)- -

The left side of Equation 2.5 is a function of x only and the right side is a function

oft only Because x and tare independent variables, each side of Equation 2.5 can be varied independently The only way for the equality of the two sides to be preserved under any variation of x and l is for each side to be equal to a constant If we let this constant be K, we can write

55

56 Chapter 2 I The Classical Wave Equation

equations are called linear differential equations with constant coefficients and are quite easy to solve, as we shall see

The value of K in Equations 2.8 and 2.9 is yet to be determined We do not know right now whether K is positive, negativ·e, or even zero Let's first assume that K = 0 In this case, Equations 2.8 and 2.9 can be integrated immediately to find

(2.10) and

(2.11) where the a 's and b 's are just integration constants, which can be determined by using the boundary conditions given in Equations 2.2 In terms of X(x) and T(l), the boundary conditions are

u(O, t) = X(O)T(t) = 0 and u(l, t) = X(l)T(t) =0 Because T (t) certainly does not vanish for all t, we must have that

Now let's assume that K > 0 in Equation 2.8 To this end, write K as k2 where k

is real This assures that K is positive because it is the square of a real number In this case, Equation 2.8 becomes

(2.13) Experience shows that solutions to a linear differential equation with constant coeffi-cients whose right side is equal to zero are of the form X (x) = e°'x, where a is a constant

to be determined

EXAMPLE 2-1

Solve the equaton

d2y ~dy - -5-+2y=0

dx2 dx

2.2 The Equation Can Be Solved by the Method of Separation of Variables

SOLUTION: Ifwe substitute y(x) = e<>x into this differential equation, we obtain

a2y - 3ay + 2y = 0

a2 - 3a + 2 = 0 (a- 2)(a- 1)= 0 or that a= I and 2 The two solutions are y(x) =ex and y(x) = e2x and

is also a solution Prove this by substituting this solution back into the original equation

We now look for a solution to Equation 2.13 by letting X (x) = eax and get

(a2 - k2)X (x) = 0

Therefore, either (a2 - k2) or X (x) must equal zero The case X (x) = 0 is a trivial

solution, and so a2 - k2 must equal zero Therefore,

being a linear differential equation (Problem 2-9) Note that the highest derivative in Equation 2.13 is a second derivative, which implies that in some sense we are performing two integrations when we find its solution When we do two integrations,

we always obtain two constants of integration The solution we have found has two constants, c1 and c2, which suggests that it is the most general solution

Applying the boundary conditions given by Equations 2.12 to Equation 2.14 gives

58 Chapter 2 I The Classical Wave Equation

2.3 Some Differential Equations Have Oscillatory Solutions

Let's hope that assuming K to be negative gives us something interesting If we set

K = -{32 then K is negative if f3 is real In this case Equation 2.8 is

Equation 2.16 using Euler's formula (Equation A.6):

e±ifJ = cos e ± i sin e

If we substitute Euler's formula into Equation 2.16, we find

X (x) = c1(cos f3x + i sin f3x) + c2(cos f3x -i sin f3x)

= (c 1 + c2) cos f3x + (ic1 -ic2) sin f3x

But c1 + c2 and ic1 - ic2 are also just constants, and if we call them c3 and c4, respectively, we can write

X (x) = c3 cos f3x + c4 sin f3x

instead of

These two forms for X (x) are equivalent

2.3Some Differential Equations Have Oscillatory Solutions

EXAMPLE 2-2

Prove that

y(x) =A cos fJx + B sin fJx

(where A and Bare constants) is a solution to the differential equation

SOLUTION: Thefirstderivativeofy(x) is

-= -A{J s1n {Jx + B{J cos {Jx dx

and the second derivative is

Therefore, we see that

-= -AfJ cos {Jx - B{J s1n {Jx dx2

or that y (x) = A cos {Jx + B sin {Jx is a solution of the differential equation

The next example is important and one whose general solution should be learned

subject to the initial conditions x (0) = A and dx / dt = 0 at t = 0

SOLUTION: In this case, we find a= ±iw and

59

60 Chapter 2 I The Classical Wave Equation

2.4 The General Solution to the Wave Equation Is a Superposition of Normal Modes

Equation 2.18 can be satisfied in two ways One is that B = 0, but this along with the fact that A = 0 yields a trivial solution The other way is to require that sin {JI = 0 Because sine = 0 when e = 0, rr, 2rr, 3rr, , Equation 2.18 implies that

where Equation 2.19 says that f3 = nn: / l Referring to the result obtained in Example

2-2 again, the general solution to Equation 2.21 is

T(1) = D cos w,,r + E sin w11t (2.22) where u>11 = {Jv = mr v/ I We have no conditions to specify D and E, so the amplitude

u(x, t) is (cf Equation 2.3)

where we have let F = DB and G = EB Because there is a u (x, l) for each integer n

and because the values of F and G may depend on n, we should write u(x, t) as

) ( G ) nn: x u11(.x,1 = F11cosw111+ 11smw111 sm- -

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62 Chapter 2 I The Classical Wave Equation

EXAMPLE 2-4

Show that Equation 2.23 is a solution to Equation 2.1

SOLUTION: The second partial derivatives of u,, (x, t) are

Equa-The quantity A is the amplitude of the wave and</> is called the phase angle Using this relation, we can write Equation 2.24 in the form

Equation 2.25 has a nice physical interpretation Each u11 (x, t) is called a normal

mode, and the time dependence of each normal mode represents harmonic motion of frequency

2.4 The General Solution to the Wave Equation Is a Superposition of Normal Modes

The first term, u 1(x, t), called the fundamental mode or.first harmonic, represents a

cosinusoidal (harmonic) time dependence of frequency v /2l of the motion depicted in

Figure 2.3a The second harmonic or first overtone, u2(x, t), vibrates harmonically with frequency v / l and looks like the motion depicted in Figure 2.3b Note that the midpoint of this harmonic is fixed at zero for all t Such a point is called a node, a concept that arises in quantum mechanics as well Notice that u(O) and u(l) are also equal to zero

These terms are not nodes because their values are fixed by the boundary conditions

Note that the second harmonic oscillates with twice the frequency of the first harmonic

Figure 2.3c shows that the third harmonic or second overtone has two nodes It is easy to continue and show that the number of nodes is equal to n -l (Problem 2-11 ) The

waves shown in Figure 2.3 are called standing waves because the positions of the nodes are fixed in time Between the nodes, the string oscillates up and down

Consider a simple case in which u(x, t) consists of only the first two harmonics

and is of the form (cf Equation 2.25)

de-into a sum or superposition of normal modes is a fundamental property of oscillatory behavior and follows from the fact that the wave equation is a linear equation

Our path from the wave equation to its solution was fairly long because we had to learn to solve a certain class of ordinary differential equations on the way The

overall procedure is actually straightforward, and to illustrate this procedure, we will

solve the problem of a vibrating rectangular membrane, a two-dimensional problem, in

Section 2.5

63

64 Chapter 2 I The Classical Wave Equation

An illustration of how two standing waves can combine to give a traveling wave In both parts, time increases downward The left portion shows the independent motion of the first two harmonics Both hannonics are standing waves; the first harmonic goes through half a cycle and the second harmonic goes through one complete cycle in the time shown The right side

shows the sum of the two harmonics The sum is not a standing wave As shown, the sum is a traveling wave that travels back and forth between the fixed ends The traveling wave has gone through one-half of a cycle in the time shown

2.5 A Vibrating Membrane Is Described by a Two-Dimensional

to the geometry in Figure 2.5, we see that the bounda1y conditions that u(x, y, t) must

satisfy (because its four edges are clamped) are

u(O, y) = u(a, y) = 0 }

(for all t)

By applying the method of separation of variables to Equation 2.28, we assume

that u (x, y, t) can be written as the product of a spatial part and a temporal part or that

u(x, y, t) = F(x, y)T(t) (2.30)

by

y

b 1 - - -

a x A rectangular membrane clamped along its perimeter

We substitute Equation 2.30 into Equation 2.28 and divide both sides by F(x, y)T(t)

to find

(2.31)

The right side of Equation 2.31 is a function of x and y only and the left side is a function oft only The equality can be true for all t, x, and y only if both sides are equal to a constant Anticipating that the separation constant will be negative, as it was in the previous sections, we write it as -{32 and obtain the two separate equations

66 Chapter 2 I The Classical Wave Equation

where p2 and q2 are separation constants, which according to Equation 2 34 must satisfy

which imply that

X(O)Y(y) = X(a)Y(y) = 0

2.5 A Vibrating Membrane Is Described by a Two-Dimensional Wave Equation

Recalling that p2 + q2 = {32 we see that

(n2 m2)1;2

f311111 =rr a2 + f;i m= I, n = l, 2, 2, (2.45) where we have subscripted f3 to emphasize that it depends on the two integers n and m

Finally, now we solve Equation 2.32 for the time dependence:

(2.46) where

and the general solution is given by

As in the one-dimensional case of a vibrating string, we see that the general

vibrational motion of a rectangular drum can be expressed as a superposition of normal modes, u11111(x, y, 1) Some of these modes are shown in Figure 2.6 Note that in this two-dimensional problem we obtain nodal lines In two-dimensional problems, the nodes are lines, as compared with points in one-dimensional problems Figure 2.6 shows the normal modes for a case in which a -I= b The case in which a = b is an interesting

FIGURE 2.6

The first few normal modes of a rectangular membrane The positive values of u are in orange

and the negative values are in grey

67

68 Chapter 2 I The Classical Wave Equation

one The frequencies of the normal modes are given by Equation 2.47 When a= bin

The normal modes of a square membrane, illustrating the occurrence of degeneracy in this

system The normal modes u 12 and u21 have different 01ientations but the same frequency, given by Equation 2.50 As in Figure 2.6, the positive values of u are in orange and the negative values

of u are in grey Notice that the nodal line for u 12 is parallel to the x axis while that for u21 is parallel to they axis The same is true for the normal modes u 13 and u31 •

2.6 A Characteristic Property of Waves Is Thal They Lead to Interference

We saw in Section 2.4 that a traveling wave can be expressed as a superposition of standing waves To see this more expliciil:ly, start with the first term in Equation 2.27,

J'(X u(x, t) =cos wit sin -

2.6 A Characteristic Property of Waves Is That They Lead to Interference

, _ _ _ _ _ ' l.?;i"'_ - _,,

FIGURE 2.8

An illustration of the fact that a wave of the form sin[rr(x - vt)/ l] moves uniformly to the right as r increases The time, r2, labeling the dashed curve is greater than the time, ti labeling the solid curve

to write u(x, t) in the form

u(x, t) =~sin [ T(x + vr)J +~sin [ T(x -ur)] (2.51) Let's look at the second term in this expression Figure 2.8 shows sin[rr(x - vi)/ I]

plotted against x for increasing values of t Note that the shape does not change with time; the curve simply moves uniformly to the right In fact, the speed with

which it moves is v, which you can see by setting rr(x - vt)/l =constant, and then differentiating with respect to l to find that dx /dt = v Similarly, the first term in Equation 2.51 is a waveform moving uniformly to the left with a velocity v

We can write sin[ rr (x - vi)/ I] in a different notation Notice from Figure 2.3a that A = 21 Therefore, we can write

Equation 2.53 expresses a traveling wave in very common notation The quantity k is

called the wave vector In one dimension, k is just a scalar, but in two or three dimensions

k is a vector that describes the direction of propagation of the wave The magnitude of

k, k = lkl, is still equal to 2rr/A Equation 2.53 represents a traveling wave of wave vector k and frequency u>

traveling in the positive x direction with a velocity w/k = 2rrv/(2rr/A) =AV= v Of course, we could also have cos(kx - wt) instead of sin(kx - w1) When dealing with

69

70 Chapter 2 I The Classical Wave Equation

traveling waves, it is usually more convenient to use a complex exponential notation and to express f (x, l) above as

I (x, t) = ei<kx-wr> (2.54) and take either the real or the imaginary part A wave of the form described by Equation 2.54 is called a plane wave because it is uniform in the y-z plane

If, for example, the wave is an electromagnetic wave, then /(x, l) represents the electric field at a point (x, t) and we wriif:e

where E0 is the amplitude of the wave We simply take the real part of f(x, l) to obtain cos(kx - wt) or the imaginary part to obtain sin(kx - wt) One reason that the complex exponential notation is more convenient than using either cos(kx - wt) or sin(kx - wt)

is that every derivative of Equation 2.54 gives a constant times f (x, t), but you have to cycle through an even number of derivatives in Equation 2.53

EXAMPLE 2-5

SOLUTION: The second derivatives of f(x, t) are

Let's use Equation 2.55 to derive the interference pattern of a two-slit experiment

like that discussed in Section 1.13 Figure 1.15, reproduced as Figure 2.9 here, illustrates the geometry associated with the experiment If we let x0 be the distance S 1 P in Figure 2.9 and use Equation 2.55, we can write the electric field at the point P as

E(B) = Eoei(kxo-wt) + Eoei[k(x0+d sin 8)-wt]

2.6 A Characteristic Property of Waves Is That They Lead to Interference

p

FIGURE 2.9

A schematic diagram ofa two-slit interference experiment (right) The distance I is large enough

compared to the distance between the slits, d, so that the two light waves are essentially parallel The angle fJ is the angle that the beams make with the perpendicular line between the two

screens The diagram on the left shows a magnification of the circled region on the right

the radiation described by Equation 2.56 is given by

Figure 2.10 shows I (8) plotted against 8 ford= 0.010 mm and A.= 6000 A Note that

rrd sine

Note that this is just the relation (d sine= nA.) that we derived in Section 1.13

71

A.= 6000 A

Suppose the second screen is placed l = 1.0 m beyond the slits What is the separation

of adjacent maxima along the second screen? Take d = 0.0 I 0 mm and A = 6000 A

SOLUTION: The distance z along the second screen from the perpendicular line from

halfway between the slits to the second screen is given by z = l sine (see Figure 2.9)

Therefore, the distance between successive maxima is given by

[ / A.] A ~z =I (n + 1) n- = i-

= (1.0 m)(6.0 x ~0-7

m) = 60 mm 1.0 x 10-) m

Problem 2-19 has you generalize Equation 2 58 to the case of N slits This problem nicely illustrates the convenience of using Equation 2.54 instead of Equation 2.53

This chapter has presented a discussion of the wave equation and its solutions In Chapter 3, we will use the mathematical methods developed here, and so we recommend

doing many of the problems at the end ofithis chapter before going on Several problems involve physical systems and serve as refreshers or introductions to classical mechanics

Problems

d2y dy (b)dx2-5dx+6y=0 y(0)= - 1

2-3 Prove that x(t) = cos wt oscillates with a frequency v = w/2rc Prove that x(t) =

A cos wt + B sin wt oscillates with the same frequency, w/2rc

2-4 Solve the following differential equations:

(a)dt2+wx(t)=0 x(O)= O dr(att= O)= vo

(b) 2 + w x(t) = 0 x(O) =A - (at t = 0) = v0

Prove in both cases that x(t) oscillates with frequency w/2rc

2-5 The general solution to the differential equation

Show that all three of these expressions for x (t) are equivalent Derive equations for A and</>

in terms of Ct and c2, and for B and >./! in terms of Ct and c2 Show that au three fo1ms of

x(t) oscillate with frequency w/2rc Hint: Use the trigonometric identities

sin(a + {3) = sin a cos fJ + cos a sin fJ and cos( a+ {3) = cos a cos f3 - si a sin fJ

2-6 In all the differential equations that we have discussed so far, the values of the exponents a

that we have found have been either real or purely imaginary Let us consider a case in which

a turns out to be complex Consider the equation

d2y dy

dx2 dx

If we substitute y (x) = eax into this equation, we find that a2 + 2a + l 0 = 0 or that

a = - 1 ± 3i The general solution is

73

74 Chapter 2 I The Classical Wave Equation

m

A body of mass m connected to a wall by a spring

Show that y(x) can be written in the equivalent form

y(x) = e-x (c3 cos 3x + c4 sin 3x) Thus we see that complex values of the et 's lead to trigonometric solutions modulated by an exponential factor Solve the following equations:

2-7 This problem develops the idea of a classical harmonic oscillator Consider a mass m

attached to a spring, as shown in Figure 2.11 Suppose there is no gravitational force acting on m so that the only force is from the spring Let the relaxed or undistorted length of the spring be x0 Hooke's law says that the force acting on the mass m is f = -k(x - x0), where

k is a constant characteristic of the spring and is called the force constant of the spring.Note that the minus sign indicates the direction of the force: to the left if x > x0 (extended) and to the right if x < x0 (compressed) The momentum of the mass is

Replacing f(x) by Hooke's law, show that

Upon letting e = x - Xo be the displacement of the spring from its undistorted length, then