258

CHAPTER 12

Quantum Mechanical Model Systems

SECTION 12.1

12.1 Half the particles here have twice the momentum of the other half and thereforealso have twice the wavevector of the others. The analog of the wavefunctionin Eq. (12.8) is thus

ψ ∝ 12

1/2eikx +

12

1/2ei2kx .

The probability distribution is

ψ 2 ∝ 12

1/2eikx +

12

1/2ei2kx

12

1/2e–ikx +

12

1/2e–i2kx

= 12

+ 12

e–ikx + 12

eikx + 12

= 1 + 12

eikx + e–ikx = 1 + cos kx .

This is an oscillatory function that oscillates between 0 and 2 with a spacingbetween maxima of 2π/k. The particles are localized to regions around thesemaxima (much as a standing waves of water form a regular pattern of maximaand minima), but these regions of maximum probability extend from –∞ to ∞.

12.2 We wish to evaluate Eq. (12.9) using the w(k) function of Eq. (12.10):

ψ(x) ∝ w(k)e–ikx dk–∞

∞ = e–(k – k0)2/2(δk)2

e–ikx dk–∞

∞ .

This integral is known. It has the form of the general definite integral

259

e–p2z 2 ± qz dz–∞

∞ =

πp

eq2/4p 2 , p > 0 ,

and we can identify p2 = 1/2(δk)2, q = ix, q2 = –x2, and z = k – k0. This gives

ψ(x) ∝ e–(k – k0)2/2(δk)2e–ikx dk

–∞

∞

= e–ik0x e–(k – k0)2/2(δk)2 – i(k – k0)x dk –∞

∞

= δk 2π e–ik0x e–x 2(δk)2/2 ,

which, except for the factor δk 2π , is the function shown in the text on page392 and plotted (as the real part) in Figure 12.2(b). (Whether one has a factore–ik0x, as we have here, or a factor eik0x, as in the text, is unimportant. Bothhave the same real part, cos(k0x), and the imaginary part of one is the negativeof that of the other.) The definite integral above is a special case of the moregeneral definite integral we have seen before (see Problem 11.22):

e–αy 2 dy

–∞

∞ =

πα

.

We “complete the square” in the argument of the exponential:

e–p2z 2 ± qz dz–∞

∞ = e–p2z 2 ± qz – q2/4p 2 + q2/4p 2

dz–∞

∞

= eq2/4p 2 e–p2z 2 ± qz – q2/4p 2

dz–∞

∞

= eq2/4p 2 e– pz +– q/2p 2

dz–∞

∞ ,

then define y = pz +– q/2p so that dy = p dz. Making these substitutions (with α= 1) completes the derivation.

260

12.3 (a) If the source of particles is at x = –∞, then the D coefficient must be zero,since the term in ψ< containing D, e–ik<x, represents particles in the region x >0 moving toward x = –∞. Since there is no source at x = ∞ directing particlestoward x = –∞ and since particles moving toward x = ∞ are not reflected backat x = ∞, this coefficient must be zero.

(b) The requirement ψ<(0) = ψ>(0) means 1 + B = C, and the requirement thatthe first derivatives are equal at x = 0 means ik< – ik<B = ik>C. Solving thesetwo simultaneous equations for B and C gives

B = k< – k>

k< + k> and C =

2k<

k< + k> .

(c) The product B*B is

B*B = k< – k>

k< + k>

2

since the wavevectors k< and k> are both real numbers. As the total energyincreases, the potential step becomes an ever-decreasing perturbation; the par-ticle’s kinetic energy in the positive x region is almost the same as it was in thenegative x region. This means k> is approaching k< as the energy increases,making B*B go to zero. Note as well that C approaches 1 in this high-energylimit, as it should.

12.4 (a) Here C must be zero in order to keep the wavefunction finite as x → ∞.

(b) Equality of the wavefunctions at x = 0 leads to 1 + B = D, and equality oftheir first derivatives leads to ik< – ik<B = –κ>D. Simultaneous solution ofthese equations gives

B = ik< + κ>

ik< – κ>

and D = 2ik<

ik< – κ>

.

(c) Here, B*B is

B*B = ik< + κ>

ik< – κ>

*ik< + κ>

ik< – κ>

= –ik< + κ>

–ik< – κ>

ik< + κ>

ik< – κ>

= k<

2 + κ>2

k<2 + κ>

2 = 1 .

261

(d) The probability of barrier penetration depends on total energy E through thequantity D*D:

D*D = 4k<

2

k<2 + κ>

2 .

Since the wavevectors k< and κ> depend on E through

k< = 2mE

¨ and κ> =

2m(V – E)

¨ ,

we can write

D*D = 4k<

2

k<2 + κ>

2 =

4EE + (V – E)

= 4EV

.

Since E ranges from 0 to V (because E > V returns us to the energy regime ofthe previous problem), the probability is maximal when E = V.

12.5 Here, the wavefunctions can be written

ψ< = C e–ik<x and ψ> = e–ik>x + Beik>x .

Note that ψ< has no term corresponding to particles moving towards x = ∞(i.e., no eik<x term), since particles originate at x = ∞ and, while they can bereflected from the potential step at x = 0 and move back to x = ∞ in the positivex region, they can only move toward x = –∞ in the negative x region. We findexpressions for B and C from the requirements that the wavefunctions and theirfirst derivatives are equal at x = 0:

1 + B = C and –k> + k>B = –k<C .

Simultaneous solution of these two equations gives

B = k> – k<

k> + k< and C =

2k>

k> + k< .

262

The probability for reflection back to the source at the potential step is B*B,since B is the coefficient of the term representing particles in the positive xregion moving towards x = ∞. It is simply

B*B = k> – k<

k> + k<

2 ,

and since k> approaches k< as E increases, B*B approaches 0 as E → ∞.

SECTION 12.2

12.6 The zero-point energy of mass me, the electron mass, in a 1-D box is

E1 = ¨ 2

π2

2meL2.

Equating this to the electron’s mass-equivalent energy, mec2, and solving for Lgives

L = h

2 2 mec =

2.43 × 10–12 m

2 2 = 8.58 × 10–13 m .

As an aside, we can repeat this line of reasoning for the zero-point energy of aharmonic oscillator and get another answer of the same magnitude. We write,with X representing the classical turning point of the motion,

E0 = ¨ 2

2meX 2 = mec2 or X =

¨2mec

= 1.93 × 10–13 m .

Thus, the full classical motion of the electron, from –X to X, spans 1/2π timesthe Compton wavelength: ¨/mec = (h/mec)/2π = 3.86 × 10–13 m.

12.7 Usually, one thinks of classical behavior as the “large quantum number limit” ofa quantum-mechanical system. Here, we will find classical behavior (in thesense defined in this problem—the position uncertainty is greater than the dis-tance between successive maxima in the probability distribution) at a surpris-ingly small quantum number. First, we find ∆x for a particle in a box. Theresult of Problem 11.23 gives us an easy route, if we recall that <x> = L/2:

263

(∆x)2 = <x2> – <x>2 = L2 1

3 –

1

2n2π2 –

L2

4 or

∆xL

= 112

– 1

2n2π2

1/2 .

(As an aside, note that ∆x/L approaches the constant limit 1/ 12 = 0.289 as n→ ∞.) We wish to compare ∆x/L to 1/n (the spacing between maxima dividedby L) and look for that n = n* for which ∆x/L > 1/n. We make the followingtable:

n 1 2 3 4 5 10

∆x/L 0.181 0.266 0.279 0.283 0.285 0.2881/n 1 0.500 0.333 0.250 0.200 0.100

We find that ∆x > L/n for n ≥ 4, which is perhaps a surprisingly small number.

12.8 We write

P(x) dx = dt

τ1/2

= dx

τ1/2(dx/dt) ,

and with x(t) = X cos ωt, dx/dt = –Xω sin ωt. Since τ1/2 = π/ω, we have

P(x) dx = dt

τ1/2 =

dxτ1/2(dx/dt)

= dx ω

Xω π sin ω t =

dxπX sin ω t

.

(We can neglect the minus sign in dx/dt, since probabilities must be positive.)Now we write cos ωt = x/X, and since sin2 ωt + cos2 ωt = 1, we have sin ωt =(X2 – x2)1/2/X so that

P(x) dx = dx

πX sin ω t =

dx

π X 2 – x2 1/2

as we wished to show. The normalization integral follows from

P(x) dx–X

X

= 1π

dx

X 2 – x2 1/2–X

X

.

Let x = yX so that dx = X dy. Then the integral above becomes

264

1π

dx

X 2 – x2 1/2–X

X

= 1π

X dy

X 2 – y2X 2 1/2–1

1

= 1π

dy

1 – y2 1/2–1

1

= ππ

= 1 .

12.9 At the classical turning point of a harmonic oscillator, x = Xv where Xv is theclassical turning point for state v. Moreover, the energy of state v is Ev =kX v

2/2 for force constant k. Thus, the Schrödinger equation becomes

– ¨ 2

2µ d2ψv

dx2 + kx2

2 – Ev ψv = –

¨ 2

2µ d2ψv

dx2 + kx2

2 –

kXv2

2ψv = 0 .

If x = Xv, the terms in parentheses cancel each other, and we are left withd2ψv/dx2 = 0. The wavefunction has an inflection point at the classical turningpoint, as Figure 12.8 (and Figure 11.2 for the ground state) indicate. This mustbe a general result, since the definition of a classical turning point is that point atwhich the total energy equals the potential energy and the terms in parenthesesin the Schrödinger equation as written above are always (potential energy – totalenergy).

12.10 The coordinate q used to express the harmonic oscillator wavefunctions in Eq.(12.19) is shown on page 406 in the text to equal 2v + 1x/Xv so that q1 =

3x/X1 and q2 = 5x/X2. The Hermite polynomials for these two states are inTable 12.1: H1(q) = 2q and H2(q) = 4q2 – 2. The normalization constantsgiven by Eq. (12.20) are for the wavefunction written in terms of the dimen-sionless variable q. We want the normalization constants for the wavefunctionwritten in terms of the real distance variable x, and thus we use the expressionslisted in the Summary to Chapter 12 on page 440. (See also the comments atthe end of Example 12.4 on page 406.) We have

N1 = k

π¨ω12

1/2 and N2 = k

π¨ω18

1/2 .

Since q can also be written as k/¨ω x, we have k/¨ω = 2v + 1 /Xv.This makes the normalization constants equal to

N1 = 1

2 π3

X1

1/2 and N2 =

1

8 π5

X2

1/2 .

265

Thus, the entire wavefunctions are

ψ1 = 1

2 π3

X1

1/2 2

3xX1

e–3x2/2X12

ψ2 = 1

8 π5

X2

1/2 4

5x2

X22

– 2 e–5x2/2X22 .

12.11 As we used in the previous problem, there is a simple relationship between theclassical turning point of any harmonic oscillator state and its parameters m andk. For the ground state, the relationship is

X0 = ¨ωk

.

Since ω = k/m , this can also be written as

X0 = ¨ω

k =

¨ kk m

= ¨km

.

Using the values for k and m quoted in the problem, we find X0 = 0.123 Å forH2 and X0 = 4.98 × 10–2 Å for I2. Thus, X0/Re = 0.166 for H2 and X0/Re =1.87 × 10–2 for I2. Hydrogen vibrates over a total distance (classically) that isabout 33% of its bond length while I2 vibrates over a much smaller 4% or so.

12.12 There is a clear symmetry we can exploit here. We want the ground-state wave-function to have its maximum at x = 0. If the potential was a particle-in-a-boxextending over –L/2 ≤ x ≤ L/2, we know the ground-state wavefunction wouldbe symmetric about and maximal at x = 0. Likewise, a harmonic potential cen-tered at x = 0 has a ground-state wavefunction symmetric about and maximal atx = 0. Thus, since what we have here is half of each type of potential, we wanta continuous wavefunction that is the particle-in-a-box wavefunction for –L/2 ≤x ≤ 0 and a harmonic oscillator wavefunction for x ≥ 0. Our usual coordinatesystem for a particle-in-a-box extends from 0 to L; since the potential we are im-agining here has been shifted towards smaller x by L/2, the ground-state wave-function has the sine function of Eq. (12.11) replaced by cosine:

ψ(x ≤ 0) = 2L

cos πxL

266

The harmonic oscillator ground-state wavefunction is

ψ(x ≥ 0) = 1π1/4 X0

e–x 2/2X02 = k

π¨ω

1/4e–k 2x 2/2¨ω .

For these to join smoothly at x = 0, they must equal each other and have equalfirst derivatives. Equal derivatives is assured: both wavefunctions have maxima(and thus zero first derivatives) at x = 0. Continuity of the wavefunctions(along with ω = k/m ) gives us the relationship among m, L, and k that weseek:

2L

1/2 = k

π¨ω

1/4 = mk

π¨1/4

or L = 2π¨mk

.

12.13 Here, we can exploit the eigenvalue expressions for the creation and annihila-tion operators given on page 405 in the text:

q – ddq

ψv = [2(v + 1)]1/2ψv + 1 , v = 0, 1, 2, …

q + ddq

ψv = (2v)1/2ψ v – 1 , v = 1, 2, 3, …

We write the matrix element qnm and expand it as follows:

qnm = ψnqψm dq = 12

ψn q – ddq

+ q + ddq

ψm dq

= 12

ψn q – ddq

ψm dq + ψn q + ddq

ψm dq

= 12

2(m + 1) ψnψm + 1 dq + 2m ψnψm – 1 dq .

Since the wavefunctions are orthonormal, the first integral in the last line abovevanishes unless n = m + 1 (in which case it equals 1), and the second integralvanishes unless n = m – 1 (in which case it, too, equals 1). Thus, if n = m + 1,qnm = qn,n – 1 = [ 2(m + 1)]/2 = 2n/2, and if n = m – 1, qnm = qn,n + 1 =

2m /2 = [ 2(n + 1)]/2. See also Problem 13.15 for another representation ofthese integrals.

267

12.14 The matrix representation of the square if the q operator is the square of thematrix representation of the q operator alone:

qnm2 = qniqim∑

i = 1

∞ .

The previous problem gives the pattern of nonzero values for qnm, and armedwith this pattern and the equation above, we can deduce that the pattern of non-zero values for qnm

2 must be

q2 =

0 q01 0q10 0 q120 q21 0

2

=

q01q10 0 q12q01 0 0 q12q21 + q01q10 0 q23q12

q21q10 0 q23q32 + q12q21 0 0 q32q21 0 q34q43 + q23q32

.

In particular, we see that q002 is nonzero and equals

q002 = q01q10 =

22

22

= 12

.

Problem 11.34 found <x2> = X2/2 where X is the classical turning point. Ourresult here, q00

2 = <q2>00 = 1/2, is the dimensionless variant of <x2> for theground state for which q = x/X. Thus, <q2>00 = <x2>00/X2 = 1/2.

12.15 Differentiating the generating function gives

dHv

dq =

ddq

(–1)v eq2 dv e–q2

dqv = (–1)v 2qeq2

dv + 1 e–q2

dqv + 1 = 2qHv – Hv + 1 .

Differentiating this gives

268

d2Hv

dq2 =

d(2qHv – Hv + 1)

dq = 2Hv + 2q

dHv

dq –

dHv + 1

dq

= 2Hv + 2q(2qHv – Hv + 1) – 2qHv + 1 + Hv + 2

= (2 + 4q2)Hv – 4qHv + 1 + Hv + 2 .

If we substitute these expressions for the first and second derivatives of Hv intoHermite’s differential equation, we find

d2Hv

dq2 – 2q

dHv

dq + 2vHv = (2 + 2v)Hv + Hv + 2 – 2qHv + 1 = 0 .

To arrive at the recurrence relation shown in the text, define w = v + 1, so thatv = w –1. This gives

[2 + 2(w – 1)]Hw – 1 + Hw + 1 – 2qHw = 2wHw – 1 + Hw + 1 – 2qHw = 0 .

Changing the symbol that indexes the Hermite polynomial in this equation fromw to v lets us write the recurrence relation as it is shown in the text, Hv + 1 =2qHv – 2vHv – 1. Finally, if we substitute this expression into our originalexpression for the first derivative of Hv, we find

dHv

dq = 2qHv – Hv + 1 = 2qHv – 2qHv + 2vHv – 1 = 2vHv – 1 .

12.16 If we expand the Lennard-Jones potential function in a Taylor’s series with theaim of finding its effective force constant, k we see that we need the secondderivative of V, since k = (d2V/dx2)x = xe. We can write V as

V(x) = D 1 – 2xe6x–6 + xe

12x–12

so that the first derivative is

dV(x)dx

= D 12xe6x–7 – 12xe

12x–13

and the second derivative is

269

d2V(x)

dx2 = D –12·7xe

6x–8 + 12·13xe12x–14 .

Evaluating this at x = xe gives the answer we seek:

k = d2V(x)

dx2x = xe

= D –12·7xe6xe

–8 + 12·13xe12xe

–14 = 72D

xe2

.

SECTION 12.3

12.17 A particle moving freely in 3-D space has the same Hamiltonian whether thespace is confined to a finite region (the 3-D particle-in-a-box potential) orextends to infinity. The Hamiltonian is given by Eq. (12.21):

H = –¨ 2

2m

∂2

∂x2 +

∂2

∂y2 +

∂2

∂z2 = –

¨ 2

2m ∇2

.

Since the three directions in space x, y, and z are equivalent and independent ofeach other, the most general 3-D free particle wavefunction has the form

ψ(x,y , z) = Aei(kxx + kyy + kzz) + Be–i(kxx + kyy + kzz) = Aeik⋅r + Be–ik⋅r

where kx, ky, and kz are the wavevectors for motion along the x, y, and z direc-tions, respectively, and where k is the vector with components kx, ky, and kz, ris the vector with components x, y, and z, and k·r represents the vector dotproduct of k and r. The three quantum numbers needed to describe the state ofthe system (including its energy) are the wavevector components since

kx = px

¨ =

mvx

¨ , ky =

py

¨ =

mvy

¨ , kz =

pz

¨ =

mvz

¨ , and E =

¨ 2kx

2 + ky2 + kz

2

2m.

Each wavevector can vary continuously from –∞ to ∞ (although ±∞ implies aninfinite velocity, which relativity does not allow).

12.18 The 2-D particle-in-a-box quantum numbers are nx and ny, and each can rangefrom 1 to ∞ in integral steps. Since the box is a square (all sides of length L),the total energy expression is

270

E = Ex + Ey = ¨ 2

π2

2mL2 nx

2 + ny2 .

To find all the states with energy E ≤ 9¨2π2/mL2, we can define the dimen-sionless energy ε = E/(¨2π2/mL2) = (nx

2 + ny2)/2, which we can tabulate:

nx\ny 1 2 3 4

1 1 5/2 5 17/2

2 5/2 4 13/2 10

3 5 13/2 9 25/2

4 17/2 10 25/2 32

Entries in italics have ε > 9 and thus correspond to states we are not interestedin here. This shows us that the state with ε = 1 (or E = ¨2π2/mL2) is nonde-generate, as are those with ε = 4 (E = 4¨2π2/mL2) and those with ε = 9 (E =9¨2π2/mL2). The states with ε = 5/2 (E = 5¨2π2/2mL2), with ε = 13/2 (E =13¨2π2/2mL2), and with ε = 17/2 (E = 17¨2π2/2mL2) are doubly degenerate.To locate the most probable position of the particle in any one state, we look atthe square of the wavefunction. If we define the x and y coordinates so that theedges of the box extend over the ranges 0 ≤ x, y ≤ L, the square of the wave-function is

ψ2(x,y) = ψnx

2 (x)ψny

2 (y) = 2L

1/2 sin2

nxπx

L

2L

1/2 sin2

nyπy

L

= 2L

sin2 nxπx

L sin2

nyπy

L .

Wherever this function is a maximum for any (nx, ny) choice locates the mostprobable position or positions. An easy way to see this probability is as densityplot (an analog of a contour plot) that plots ψ2 in two dimensions as a gradationin shade that is proportional to the value of ψ2. Such plots are shown at the topof the next page for the states (nx, ny) = (1, 1), (3, 1), (1, 3), and (2, 2), drawnso that the darkest shade locates the maximum or maxima in ψ2.

271

00

L

L

y

x

00

L

L

y

x00

L

L

y

x

00

L

L

y

x

(nx, ny) = (1, 1) (nx, ny) = (1, 3)

(nx, ny) = (2, 2)(nx, ny) = (3, 1)

The probability peaks in the center of the box (at x = y = L/2) for the (1, 1)state. The states (1, 3) and (3, 1) are related by simple symmetry: rotate thepicture of one by 90° to get the picture of the other. State (1, 3) peaks at x = L/2and y = L/6, L/2, and 5L/6, while state (3, 1) peaks at Y = L/2 and x = L/6,L/2, and 5L/6. State (2, 2) peaks at the four positions (x, y) = (L/4, L/4),(3L/4, L/4), (L/4, 3L/4), and (3L/4, 3L/4).

12.19 For an isotropic 3-D harmonic oscillator, ω = k/m is the only parameterneeded to characterize the energy of the system. There are three independentquantum numbers that specify a particular state and energy of such a system:vx, vy, and vz, and each can range from 0 to ∞. The energy expression is

E = Ex + Ey + Ez

= ¨ω vx + 12

+ vy + 12

+ vz + 12

= ¨ω vx + vy + vz + 32

,

272

and we can use it to tabulate ε = E/¨ω for various combinations of quantumnumbers:

v x v y v z ε v x v y v z ε v x v y v z ε v x v y v z ε0 0 0 3/2 1 0 1 7/2 1 1 1 9/2 0 1 2 9/2

1 0 0 5/2 0 1 1 7/2 2 0 1 9/2 1 0 2 9/2

0 1 0 5/2 2 0 0 7/2 2 1 0 9/2 3 0 0 9/2

0 0 1 5/2 0 2 0 7/2 0 2 1 9/2 0 3 0 9/2

1 1 0 7/2 0 0 2 7/2 1 2 0 9/2 0 0 3 9/2

We see that the lowest energy state is nondegenerate, at E = 3¨ω/2, the nexthighest state is three-fold degenerate at E = 5¨ω/2, the next is six-fold degen-erate at E = 7¨ω/2, and the fourth state is ten-fold degenerate at E = 9¨ω/2.

12.20 We first convert the C–C bond energy, 200 kJ mol–1, to joule per moleculeunits:

200 kJ mol–1

6.022 × 1023 molecules mol–1 = 3.32 × 10–19 J molecule–1.

Now we seek a quantum number n that, when substituted in the energyexpression in Example 12.5 with m = mass of one He atom = 6.65 × 10–27 kgand rm = radial distance the He can move inside the C60 cage = 0.45 Å =4.5 × 10–11 m, equals or exceed this energy. Thus, we solve the followingexpression for n:

En = 3.32 × 10–19 J = ¨ 2

π2

2mrm2

n2 = (4.08 × 10–21 J)n2 ,

and find n = 9.02, so that, since n must be an integer, any state with n ≥ 9 or sogives the He atom sufficient energy to escape the C60 cage. (Of course, the 200kJ mol–1 bond energy is not a very accurately known number here, and sinceour answer n = 9.02 is so close to 9, it may well be that the state with n = 9 isnot really sufficiently energetic. On the other hand, the model itself is quitecrude, and we should expect the motion of a He atom inside C60 to be morecomplex than what we have indicated here. As mentioned in Example 12.5,there are other states for this system in which the He atom has angular momen-tum about the center of the cage. These should also be considered.)

273

SECTION 12.4

12.21 We start with the Cartesian representation of Lz given by Eq. (12.31):

Lz = ¨i

x∂∂y

– y∂∂x

and follow the steps outlined in the problem to transform to plane polar coor-dinates. Using the relations r = (x2 + y2)1/2 and θ = tan–1(y/x), we can derive

∂r∂x y

= x

x 2 + y2 1/2 =

xr

∂r∂y x

= y

x 2 + y2 1/2 =

yr

∂θ∂x y

= – y

x 2 1 + (y/x)2 = –

y

x 2 + y2 = –

y

r2

∂θ∂y x

= 1

x 1 + (y/x)2 =

x

x 2 + y2 =

x

r2 .

We use these expressions to build the operator through the chain rule:

x∂∂y

= x∂r∂y x

∂∂r

+ ∂θ∂y x

∂∂θ

= xyr

∂∂r

+ x

r2

∂∂θ

y∂∂x

= y∂r∂x y

∂∂r

+ ∂θ∂x y

∂∂θ

= yxr

∂∂r

– y

r2

∂∂θ

so that

x∂∂y

– y∂∂x

= xyr

∂∂r

+ x

r2

∂∂θ

– yxr

∂∂r

– y

r2

∂∂θ

= xyr

∂∂r

+ x 2

r2

∂∂θ

– xyr

∂∂r

+ y 2

r2

∂∂θ

= x 2 + y2

r2

∂∂θ

= ∂

∂θ .

This proves that Lz in plane polar coordinates equals (¨/i)(∂/∂θ).

274

12.22 We note first that the four first derivatives among the two coordinate systems’variables that we derived in the previous problem can also be written

∂r∂x y

= xr = cos θ

∂r∂y x

= yr = sin θ

∂θ∂x y

= –y

r2 = –

sin θr

∂θ∂y x

= x

r2 =

cos θr

.

Using these expressions in the chain rule lets us write

∂F∂x y

= cos θ ∂F∂r θ

– sin θ

r

∂F

∂θ r

∂F∂y x

= sin θ ∂F∂r θ

+ cos θ

r

∂F

∂θ r

.

We substitute from these six expressions as called for into the general expres-sions for the second derivatives:

∂2F

∂x2 =

∂r∂x y

∂∂r

∂F∂x y

+ ∂θ∂x y

∂∂θ

∂F∂x y

= cos θ ∂∂r

cos θ ∂∂r

– sin θ

r

∂∂θ

– sin θ

r

∂∂θ

cos θ ∂∂r

– sin θ

r

∂∂θ

∂2F

∂y2 =

∂r∂y x

∂∂r

∂F∂y x

+ ∂θ∂y x

∂∂θ

∂F∂y x

= sin θ ∂∂r

sin θ ∂∂r

+ cos θ

r

∂∂θ

+ cos θ

r

∂∂θ

sin θ ∂∂r

+ cos θ

r

∂∂θ

and expand each expression:

275

∂2F

∂x2 = cos2 θ

∂2F

∂r2 +

sin2 θr

∂F∂r

– 2 cos θ sin θ

r ∂2F∂r∂θ

+ sin2 θ

r2 ∂2F

∂θ2 +

2 cos θ sin θr2

∂F∂θ

∂2F

∂y2 = sin2 θ

∂2F

∂r2 +

cos2 θr

∂F∂r

+ 2 cos θ sin θ

r ∂2F∂r∂θ

+ cos2 θ

r2 ∂2F

∂θ2 –

2 cos θ sin θr2

∂

∂θ .

Adding these and recalling that sin2 θ + cos2 θ = 1 gives us

∇2 = ∂2F

∂x2 +

∂2F

∂y2

= ∂2F

∂r2 +

1r

∂F∂r

+ 1

r2 ∂2F

∂θ2

= 1r

∂∂r

1r

∂F∂r

+ 1

r2 ∂2F

∂θ2 .

12.23 We start with the expressions for the components of the angular momentumoperator written in spherical polar coordinates as given in Eq. (12.32):

Lx = ¨i

–sin φ ∂

∂θ – cot θ cos φ

∂∂φ

Ly = ¨i

cos φ ∂

∂θ – cot θ sin φ

∂∂φ

Lz = ¨i

∂∂φ

,

then we square each operator:

LxLx = –¨ 2 –sin φ

∂∂θ

– cot θ cos φ ∂∂φ

–sin φ ∂

∂θ – cot θ cos φ

∂∂φ

276

LyLy = –¨ 2 cos φ

∂∂θ

– cot θ sin φ ∂∂φ

cos φ ∂

∂θ – cot θ sin φ

∂∂φ

LzLz = –¨ 2

∂∂φ

∂∂φ

,

and expand the products:

LxLx = –¨ 2 –cos φ sin φ cot2 θ + csc2 θ

∂∂φ

+ cos2 φ cot2 θ ∂2

∂φ2

+ 2 cos φ sin φ cot θ ∂2

∂θ∂φ + sin2 φ

∂2

∂θ2 + cos2 φ cot θ

∂∂θ

LyLy = –¨ 2 cos φ sin φ cot2 θ + csc2 θ

∂∂φ

+ sin2 φ cot2 θ ∂2

∂φ2

– 2 cos φ sin φ cot θ ∂2

∂θ∂φ + cos2 φ

∂2

∂θ2 + sin2 φ cot θ

∂∂θ

LzLz = –¨ 2

∂2

∂φ2 .

We add these to form the operator we seek, L2 = LxLx + LyLy + LzLz. Fortu-

nately, several terms cancel, and the identity sin2 φ + cos2 φ = 1 simplifiesseveral others:

L2 = –¨ 2

cot2 θ + 1 ∂2

∂φ2 +

∂2

∂θ2 + cot θ

∂∂θ

= –¨ 2

∂2

∂θ2 + cot θ

∂∂θ

+ 1sin2 θ

∂2

∂φ2

= –¨ 2 1

sin θ

∂∂θ

sin θ ∂

∂θ + 1

sin2 θ ∂2

∂φ2 .

To express L2 in Cartesian coordinates, we use Eq. (12.31) for each compo-

nent operator:

277

Lx = ¨ i

y ∂∂z

– z ∂∂y

Ly = ¨ i

z ∂∂x

– x ∂∂z

Lz = ¨ i

x ∂∂y

– y ∂∂x

,

then we square and expand each component. Since all three operators are sosimilar, once we square and expand one in detail, we can immediately write theother two by analogy:

LxLx = –¨ 2 y

∂∂z

– z ∂∂y

y ∂∂z

– z ∂∂y

= –¨ 2 y

∂∂z

y ∂∂z

– y ∂∂z

z ∂∂y

– z ∂∂y

y ∂∂z

+ z ∂∂y

z ∂∂y

= –¨ 2 y 2

∂2

∂z2 – y

∂∂y

– yz ∂2

∂z∂y – z

∂∂z

– zy ∂2

∂y∂z + z2

∂2

∂y2

= –¨ 2 y 2

∂2

∂z2 – 2yz

∂2

∂z∂y + z2

∂2

∂y2 – y

∂∂y

– z ∂∂z

.

Thus, the squares of the other two components are

LyLy = –¨ 2 z2

∂2

∂x2 – 2zx

∂2

∂x∂z + x2

∂2

∂z2 – z

∂∂z

– x ∂∂x

and

LzLz = –¨ 2 x 2

∂2

∂y2 – 2xy

∂2

∂y∂x + y2

∂2

∂x2 – x

∂∂x

– y ∂∂y

.

We add these three expressions to form the operator for the square of the totalangular momentum, L

2 = LxLx + LyLy + LzLz, shown at the top of the next

page. Note that this version of the operator is considerably more complicatedthan the version written in spherical polar coordinates earlier in this problem.Consequently, the Cartesian representation is rarely used.

278

L2 = –¨ 2

y 2 ∂2

∂z2 – 2yz

∂2

∂z∂y + z2

∂2

∂y2 – y

∂∂y

– z ∂∂z

+ z2 ∂2

∂x2 – 2zx

∂2

∂x∂z + x2

∂2

∂z2 – z

∂∂z

– x ∂∂x

+ x2 ∂2

∂y2 – 2xy

∂2

∂y∂x + y2

∂2

∂x2 – x

∂∂x

– y ∂∂y

= –¨ 2 x 2 ∂2

∂y2 +

∂2

∂z2 + y2 ∂2

∂x2 +

∂2

∂z2 + z2 ∂2

∂x2 +

∂2

∂y2

– 2 yz ∂2

∂z∂y + zx

∂2

∂x∂z + xy

∂2

∂y∂x – 2 x

∂∂x

+ y ∂∂y

+ z ∂∂z

.

12.24 The expression in part (a), [Lx,Ly] = i¨Lz, is easiest to prove using Cartesiancoordinates:

[Lx,Ly] = LxLy – LyLx

= ¨ i

y ∂∂z

– z ∂∂y

¨ i

z ∂∂x

– x ∂∂z

– ¨ i

z ∂∂x

– x ∂∂z

¨ i

y ∂∂z

– z ∂∂y

= – ¨ 2 y

∂∂x

+ yz ∂2

∂z∂x – yx

∂2

∂z2 – z2

∂2

∂y∂x + zx

∂2

∂z∂y

+ ¨ 2 zy

∂2

∂x∂z – z2

∂2

∂x∂y – xy

∂2

∂z2 + x

∂∂y

+ xz ∂2

∂z∂y

= ¨ 2 x

∂∂y

– y ∂∂x

= i¨¨i

x ∂∂y

– y ∂∂x

= i¨Lz .

The expression in part (b), [L2,Lz] = 0, is easiest to prove in spherical polar

coordinates, and the calculus is perhaps a bit easier and more transparent if wewrite the operator L

2 in the following expanded way, which we first encoun-

tered in the previous problem:

L2 = –¨ 2

1sin θ

∂

∂θ sin θ

∂∂θ

+ 1sin2 θ

∂2

∂φ2

= –¨ 2

∂2

∂θ2 + cot θ

∂∂θ

+ 1sin2 θ

∂2

∂φ2 .

279

This lets us write L2Lz as

L2Lz = –¨ 2

∂2

∂θ2 + cot θ

∂∂θ

+ 1sin2 θ

∂2

∂φ2 ¨i

∂∂φ

= – ¨ 3

i

∂3

∂θ2∂φ + cot θ

∂2

∂θ∂φ + 1

sin2 θ ∂3

∂φ3

and LzL2 as

LzL2 =

¨i

∂∂φ

–¨ 2

∂2

∂θ2 + cot θ

∂∂θ

+ 1sin2 θ

∂2

∂φ2

= – ¨ 3

i

∂3

∂θ2∂φ + cot θ

∂2

∂θ∂φ + 1

sin2 θ ∂3

∂φ3 .

Subtracting these to form the commutator clearly gives zero.

For part (c), we substitute the definition L+ = Lx + iLy into the expanded com-mutator of interest:

[Lz,L+] = Lz Lx + iLy – Lx + iLy Lz

= LzLx – LxLz + i LzLy – LyLz

= [Lz,Lx] + i[Lz,Ly]

= i¨Ly + i(–i¨Lx)

= ¨ Lx + iLy = ¨L+ .

For part (d), we again substitute the definition L+ = Lx + iLy in the commutatorof interest here and write

[L+,L2] = [ Lx + iLy , L

2] = [Lx,L

2] + i[Ly,L

2] .

(Note how commutator algebra is distributive. This lets us jump directly andconfidently from the second to the third step above.) Since L

2 commutes with

every angular momentum component operator, both of the final two commuta-

280

tors equal zero, as, therefore, does the operator [L+,L2]. Thus, these two

operators commute.

To see if L+ and L– commute, we write

[L+,L–] = [ Lx + iLy , Lx – iLy ]

= Lx + iLy Lx – iLy – Lx – iLy Lx + iLy

= Lx2 – iLxLy + iLyLx + Ly

2 – Lx

2 – iLxLy + iLyLx – Ly

2

= 2i LyLx – LxLy = 2i[Ly,Lx]

= 2i –i¨Lz = 2¨Lz .

Since the final result is not identically zero, these operators do not commute.

12.25 Let ψm be an eigenfunction such that Lzψm = m¨ψm. Then we argue as fol-lows:

Lz L–ψm = LzL–ψm

= LzL– – L–Lz + L–Lz ψm

= [Lz,L–] + L–Lz ψm

= –¨L– + L–Lz ψm

= L–Lzψm – ¨L–ψm

= L–(m¨)ψm – ¨L–ψm

= (m – 1)¨ L–ψm .

Thus, the function L–ψm has an eigenvalue for the Lz operator that is one ¨unit lower than ψm itself.

12.26 If l = 1 and m = 1, the eigenfunction ψlm = ψ1,1 must satisfy the followingequation:

L2 – Lz

2 ψ1,1 = L2ψ1,1 – Lz

2ψ1,1 = 2¨ 2ψ1,1 – ¨ 2

ψ1,1 = ¨ 2ψ1,1

so that Lx2 +Ly

2 = ¨. This, along with Lz = ¨ and |L| = 2¨, lets us con-struct the diagram at the top of the next page for the classical vector L. The L

281

vector lies somewhere on the surface of the shaded cone in this figure; theUncertainty Principle does not allow us to say exactly where.

45°

¨

x

y

z

¨

L=

2¨

For the general case of l = m, we note that the length of L is l (l + 1)¨ and thelength of its z component is m¨ = l¨. Thus, we can make the following dia-gram of the general classical vector geometry:

z

θl¨

l (l + 1) ¨

We see that cos θ = l/ l (l + 1), and thus as l → ∞, cos θ → 1, or θ → 0.

12.27 To write L+ in spherical polar coordinates, we substitute the spherical polarforms of Lx and Ly into the definition of L+ . Since we will need L– later on inthis problem, we can derive both at once. We also take advantage of the iden-tity eiφ = cos φ + i sin φ and find

282

L± = Lx ± iLy

= ¨i

–sin φ ∂

∂θ – cot θ cos φ

∂∂φ

± i cos φ ∂

∂θ – cot θ sin φ

∂∂φ

= ¨i

(±i cos φ – sin φ)∂

∂θ – cot θ(cos φ ± i sin φ)

∂∂φ

= ¨i

ie±iφ ∂

∂θ – cot θ e±iφ

∂∂φ

= ¨ e±iφ ∂

∂θ + i cot θ

∂∂φ

.

Next, we take l = m, Θl,l(θ) = sinl θ, and Φl(φ) = eimφ = eilφ and operate onΘΦ with L+. Since L+ raises the z component one ¨ unit, but for m = l, thiscomponent is already as large as it can be, the operator should yield zero:

L+Θl, l(θ)Φl(φ) = L+ sinl θ eilφ

= ¨ eiφ ∂

∂θ + i cot θ

∂∂φ

sinl θ eilφ

= ¨ eiφ leilφ cos θ sinl – 1 θ – l eilφ cot θ sinl θ

= ¨ eiφ leilφ cos θ sinl – 1 θ – l eilφ cos θsin θ

sinl θ

= ¨ eiφ leilφ cos θ sinl – 1 θ – l eilφ cos θ sinl – 1 θ= 0 .

Finally, we set l = 2 and operate on (sin2 θ)Φ2(φ) = (sin2 θ)e2iφ with L–:

L–Θ2,2(θ)Φ2(φ) = L– sin2 θ e2iφ

= ¨e–iφ i cot θ ∂∂φ

– ∂

∂θ sin2 θ e2iφ

= ¨e–iφ –2e2iφ cot θ sin2 θ – 2e2iφ cos θ sin θ

= ¨e–iφ –2e2iφ cos θsin θ

sin2 θ – 2e2iφ cos θ sin θ

= ¨e–iφ –2e2iφ cos θ sin θ – 2e2iφ cos θ sin θ= –4¨eiφ cos θ sin θ .

283

Since we have not included normalization constants in our wavefunctions, thisresult is only proportional to ψ2,1, but we can see that it has the correct func-tional dependence on θ and φ for this wavefunction: for m = 1, we should finda factor eiφ, and we do; for l = 2 and m = 1, we should find the factors(cos θ sin θ) that represent the θ dependence of the Y2,1(θ,φ) spherical har-monic function (see Table 12.2), and we do.

12.28 With the help of Table 12.2, which lists the spherical harmonic functions weneed and shows us that the φ part of each term in the Theorem’s sum vanishes,since φ appears in the spherical harmonic functions as eimφ and this term timesits complex conjugate gives simply 1, we write Unsöld’s Theorem for theparticular case of l = 1 as

Y1m(θ,φ)2∑

m = –1

1

= Y1,–1 2 + Y1,0 2 + Y1,1 2

= 1

2π34

sin2θ + 32

cos2θ + 34

sin2θ

= 3

4πsin2θ + cos2θ =

34π

= constant.

For l = 2, we follow the same argument, using two trigonometric identities tosimplify our result, sin2 θ = 1 – cos2 θ and sin4 θ – cos4 θ = sin2 θ – cos2 θ:

Y2m(θ,φ)2∑

m = –2

2

= Y2,–2 2 + Y2,–1 2 + Y2,0 2 + Y2,1 2 + Y2,2 2

= 1

2π 2

1516

sin4θ +2154

sin2θ cos2θ + 58

3cos2θ – 12

= 5

16π9cos4θ – 6cos2θ + 3sin4θ + 12cos2θ sin2θ + 1

= 5

16π9cos4θ – 6cos2θ + 3sin4θ + 12cos2θ 1 – cos2θ + 1

= 5

16π6cos2θ + 3 sin4θ – cos4θ + 1

= 5

16π6cos2θ + 3 sin2θ – cos2θ + 1

= 5

16π3sin2θ + 3cos2θ + 1 =

54π

= constant.

284

One can show that, in general, the constant is (2l + 1)/4π.

12.29 We follow the logic used in Example 12.7 here, using the energy expression inEq. (12.42) with R = a0, the Bohr radius of Eq. (12.45) with µ = me. We find

E = l(l + 1)¨ 2

2mea02

= l(l + 1)¨ 2

2me4πε0¨

2

mee2

2 = l(l + 1)

e4me

2(4πε0)2¨ 2

= l(l + 1) × (2.18 × 10–18 J) = l(l + 1) × (13.6 eV) .

Note that the energy constant e4me/2(4πε0)2¨2 = e2/2(4πε0)a0 = –E1, which isthe negative of the usual H atom ground state energy (see Eq. (12.46b)),appears here.

12.30 Equating the attractive force between the proton and electron to the centrifugalforce of an orbiting electron in the Bohr model gives us

e2

4πε0a2 = meaω2

.

The Bohr quantization condition, mea2ω = n¨, gives us an expression for theorbiting frequency ω:

ω = n¨

mea2 ,

that, when substituted in the first equality, gives

e2

4πε0a2 = mea

n¨mea2

2

, or a = n24πε0¨2

mee2 = n2a0 .

If E = –T = –mea2ω2/2, we can use the expressions above for ω to write anexpression for a2ω2 that, when substituted in the energy expression, givessomething we can arrange in the form of Eq. (12.52):

285

E = – mea2ω2

2 = –me

n2¨ 2

2me2a2

= – n2¨ 2

2mea2

= – n2¨ 2

2me

e4me2

n4¨ 4(4πε0)2

= – e4me

2¨ 2(4πε0)2

1n2

= – e2

2(4πε0)a0 1n2

,

where, in the last step, we have collected those constants that define the Bohrradius, a0 = (4πε0)¨2/mee2.

12.31 If we scale a 1.2 × 10–15 m proton radius up to 1 mm (a magnification factorof 8.33 × 1011), and if we define the proton distance from the center of massto be rp and the electron distance re, then we can write two expressions relatingrp and re. The first is that they add to give our measure of a 1s H atom radius,which we are taking to be <r> so that rp + re = <r>. The second is the defini-tion of center of mass: mprp = mere. The 1s state value for <r> is 3a0/2 =7.94 × 10–11 m, which, when scaled by our magnification factor, becomesabout 66 m. Solving the two expressions for rp and re gives

rp = me

mp + me <r> and re =

mp

mp + me <r> .

The magnified numerical values are rp = 36 mm and re = 66 m. (All currenttheories and experiments on electrons, by the way, indicate that the electron hasno radius; it is a true point.) In terms familiar to American readers, if we placethe center of mass on the goal line of an American football field, the proton isabout 1.4 in. into the end zone and the electron is on the opponent’s 28 yardline.

12.32 From the general expression ∆E = hc/λ for the emission wavelength λ accom-panying an energy change ∆E, we can write

λ = hc∆E

= hc

En – Em =

hcE1

n2 –

E1

m2

= hcn2m2

E1 m2 – n2 =

hc–E1

n2m2

n2 – m2 .

With E1 = –e2/2(4πε0)a0, we identify the constant in the general expression,Eq. (11.3): 91.127 nm = 2(4πε0)a0hc/e2. Setting m = 2 gives the constant inEq. (11.2), the Balmer formula.

286

12.33 The differences in size and ionization energy for the three isotopic forms ofhydrogen are due to the different reduced masses µ of each isotope. The re-duced mass enters the Bohr radius expression and the ground-state energy ex-pression (the negative of which is the ionization potential):

a0 = 4πε0¨

2

µe 2 and IP = –E1 = e2

2(4πε0)a0 .

Using the full number of significant figures in the mass of the electron alongwith the accurate nuclear masses given in the problem and the conversion factorbetween atomic mass units and kg, we can calculate reduced masses:

µH = memH

me + mH = 9.104 434 × 10–31 kg

µD = memD

me + mD = 9.106 909 × 10–31 kg

µT = memT

me + mT = 9.107 733 × 10–31 kg .

From these, the following radii and ionization potentials follow:

a0(H) = 4πε0¨

2

e2µH

= 5.294 653 × 10–11 m

IPH = e2

2(4πε0)a0(H) = 2.178 688 × 10–18 J

a0(D) = 4πε0¨

2

e2µD

= 5.293 214 × 10–11 m

IPD = e2

2(4πε0)a0(D) = 2.179 281 × 10–18 J

287

a0(T) = 4πε0¨

2

e2µT

= 5.292 735 × 10–11 m

IPT = e2

2(4πε0)a0(T) = 2.179 478 × 10–18 J .

The size differences are inconsequential and unobservable, but the ionizationenergy differences are experimentally observable.

12.34 The general rules that guide us here are: the total number of nodes = n – 1 andthe total number of spherical nodes = n – l – 1. For wavefunction (a), “spheri-cally symmetric” tells us l = 0 (an s state), and “three spherical nodes” tells us 3= n – 0 – 1 or n = 4. It also follows that m = 0. For wavefunction (b), thethree planar nodes (and thus no spherical nodes) again tell us 3 = n – 1 or n = 4.Since there are no spherical nodes, 0 = n – l – 1 = 4 – l – 1 or l = 3 (an f state).This is the 4fxyz wavefunction shown in Figure 12.16(e), and we cannotuniquely identify m. Wavefunction (c) has one spherical and one planar nodeso that 2 = n – 1 or n = 3. The one spherical node tells us 1 = n – l – 1 = 3 – l –1 or l = 1 (a p state). Since the nodal plane is the xy plane, the wavefunction iscylindrically symmetric about the z axis. This is the 3pz wavefunction (m = 0).Wavefunction (d) has two spherical nodes and two nonspherical (conical)nodes: 4 = n – 1 or n = 5 and 2 = n – l – 1 = 5 – l – 1 or l = 2 (a d state).Again, m = 0 because conical nodes are cylindrically symmetric about z. Thisis the 5dz2 wavefunction. For (e), the symmetry tells us m = 0, and the singlenodal plane (which must be the xy plane) tells us l = 1 so that n = 2. This is the2pz wavefunction of in Figure 12.14(d).

12.35 From Table 12.4, we see that the angular portion of the 2pz wavefunction issimply cos θ. For the 3dz2 wavefunction, it is 3 cos2 θ – 1. Polar plots of r =cos θ and r = 3 cos2 θ – 1 are shown below.

288

3dz22pz

These plots show the symmetry and rough spatial extent of the electron prob-ability density plots shown in Figures 12.14(d) and 12.15(a).

12.36 For the 1s state, <r> = 3a0/2 = 7.94 × 10–11 m, while a0 = 5.29 × 10–11 m.The radius r* that encloses 90% of the electron probability in this state is givenimplicitly by the equation

0.90 = dφ0

2π sin θ dθ

0

π ψ1s

2 r2dr

0

r*

= 4

ao3 e–2r/a0r2dr

0

r*

.

If we let x = r/a0, this integral becomes

0.90 = 4 e–2xx 2dx0

x*

,

289

which cannot be evaluated in closed form. Numerical integration gives x* =2.661, or r* = 2.661a0 = 1.41 × 10–10 m, a larger measure of the 1s atomic“size” than either a0 or <r>. It is interesting to look at a graph of probabilityversus r for this state just to see how rapidly the probability is approaching 1:

2 4

0.2

0.4

0.6

0.8

10 3 5 60

1.0

r/a0

Prob

abili

ty

At the <r> value, which is 1.5 on the scale of this figure, the probability is only0.6 or so.

12.37 From Example 12.8, the radial distribution function, rdf, is defined as 4πr2Rnl2

where Rnl is the radial part of the wavefunction for the state with quantum num-bers n and l. These radial factors are listed in Table 12.3, and thus we can write

rdf1s = 4πr2 4

a03 e–2r/a0 =

16π

a03

r2e–2r/a0

rdf2pz = 4πr2

1

24a03 r2

a02 e–r/a0 =

π

6a05

r3 e–r/a0 .

We find the maxima in these functions by differentiation:

drdf1s

dr =

16π

a03

2re–2r/a0 – r2 2a0

e–2r/a0

drdf2pz

dr =

π

6a05

3r2e–r/a0 – r3

a03

e–r/a0 .

290

When these derivatives equal zero, we are at the maximum. For 1s, this hap-pens when (2r – 2r2/a0) = 0 or when r = a0. For 2pz, we want 3r2 – r3/a0 = 0or r = 3a0. Finally, we find the 2s radial distribution function:

rdf2s = 4πr2 1

8a03

2 – ra0

2 e–r/a0 =

πr2

2a03

2 – ra0

2 e–r/a0 .

This function is zero at r = 0 and ∞ (neither of which counts as a node) and at r= 2a0, which is the node location. The graph in Example 12.8 on page 432 inthe text shows this node as well as the maximum in the 1s rdf at r = a0.

12.38 The general expression for the ionization potential of a one-electron atom ofnuclear charge Z is IP(Z) = –Z2E1 where E1 is the H atom ground-state energy.For U91+, Z = 92 and IP = 8464E1 = 115 093. eV. The general expression for<r> is

<r> = a0

2Z 3n2 – l(l + 1) .

If Z = 92, n = 1, and l = 0, we find <r> = 3a0/184 = 0.0163 a0 = 8.63 × 10–3

Å. If we consider only l = 0 states, for some n = n* the U91+ size will exceedthe H atom 1s size, given by <r> = 3a0/2. Thus, we write

3a0

2 =

3a0

2·92 n*2 or n*2 = 92

so that (since n* must be integral) n* ≥ 10. We can make a plot of <r>1s =3a0/2Z versus IP(Z ) = –Z2E1 for Z = 1 through 92 if we solve each expressionfor Z and equate the results. We find <r> = (3a0/2) × (–E1/IP)1/2, which isgraphed below in a log–log plot.

291

0.1

1.0

10 100 10000.01

10

1 10 000–IP/E1

<r>

/a0

H

U

12.39 The Earth’s speed v in its orbit around the sun is

v = 2πr

τ =

2π(1.5 × 1011 m)

10π × 106 s ≅ 3 × 104 m s–1

so that its angular momentum is

L = mvr = (6 × 1024 kg) × (3 × 104 m s–1) × (1.5 × 1011 m)

≅ 2.4 × 1040 kg m2 s–1 ≅ 2.6 × 1074¨ .

The huge number 2.6 × 1074 represents an angular momentum quantum num-ber l. (The correct quantum expression L = l(l + 1)¨ reduces to L = l¨ in thelarge l limit.) An H atom must have n at least this large in order to have such anangular momentum. Thus, if we take n = l = 2.6 × 1074 and substitute intothe general expression for <r>, we find

<r> = a0

23n2 – l(l + 1) =

a0

23n2 – n2 – n ≅ a0n2

or <r> ≅ 6.6 × 10148 a0 = 3.5 × 10138 m. This huge number, larger than theradius of the known Universe, indicates that while n can increase without limitin the simplest theory of atomic hydrogen, states with very large n do not havephysical significance.

12.40 The Lz operator is (¨/i)(∂/∂φ), and the 2px and 2py hybrid functions depend onφ as shown in the text on page 434: 2px ∝ cos φ and 2py ∝ sin φ. Since we

292

know ∂cos φ/∂φ = sin φ and ∂sin φ/∂φ = –cos φ, these hybrid functions are noteigenfunctions of Lz. In fact, the Lz operator turns one function into somethingat least proportional to the other.

12.41 Table 12.5 lists the general form of the hybrid H-atom wavefunctions, and forthe 3px and 3py functions, we find

3px = 1

2 Ψ311 + Ψ31–1 and 3py =

–i

2 Ψ311 – Ψ31–1 .

Table 12.4 in turn gives the hydrogenic wavefunctions Ψ31±1 as

Ψ31±1 = 1

π1

a03/2

e–r/3a06ra0

– r2

a02

sin θe±iφ .

Substitution and simplification (using 2 cos φ = eiφ + e–iφ and 2i sin φ = eiφ –e–iφ) gives

3px = 2

π a03/2

6ra0

– r2

a02

cos φ sin θ e–r/3a0

3px = 2

π a03/2

6ra0

– r2

a02

sin φ sin θ e–r/3a0 .

The one radial node in each of these hybrids occurs when 6r/a0 – r2/a02 = 0, or

at r = 6a0. The 3px hybrid has a nodal plane wherever cos φ = 0, which is they-z plane with x = 0, exactly like the 2px hybrid, and the 3py hybrid has a nodalplane wherever sin φ = 0, which is the x-z plane with y = 0, exactly like the 2pyhybrid.

12.42 We follow the logic of the previous problem again here, starting with the gen-eral forms of the hybrids in Table 12.5:

293

3dxz = 1

2 Ψ321 + Ψ32–1

3dyz = –i

2 Ψ321 – Ψ32–1

3dxy = –i

2 Ψ322 – Ψ32–2 .

The hydrogenic wavefunctions we need are in Table 12.4:

Ψ32±1 = 1

81 π a03/2

e–r/3a0 r2

a02 sin θ cos θ e±iφ

Ψ32±2 = 1

162 π a03/2

e–r/3a0 r2

a02 sin2 θ e±2iφ .

Substitution and simplification again gives the hybrids:

3dxz = 2

81 π a03/2

r2

a02 e–r/3a0 cos φ cos θ sin θ

3dyz = 2

81 π a03/2

r2

a02 e–r/3a0 sin φ cos θ sin θ

3dxy = 2

81 π a03/2

r2

a02 e–r/3a0 cos φ sin φ sin2 θ .

Using the relationships between Cartesian and spherical polar coordinates

x = r sin θ cos φ , y = r sin θ sin φ , and z = r cos θ ,

we can see that the 3dxz hybrid contains xz/r2 = cos φ cos θ sin θ, the 3dyzhybrid contains yx/r2 = sin φ cos θ sin θ, and the 3dxy hybrid contains xy/r2 =cos φ sin φ sin2 θ. Thus, 3dxz is zero wherever xz = 0 and similarly for theother two hybrids.

SECTION 12.5

294

12.43 Equation (12.53b) gives us the eigenvalues for the Sz operator: Szα = (¨/2)αand Szβ = –(¨/2)β. The eigenvalue expressions for the square of Sz are there-fore

Sz2α = SzSzα = Sz

¨2

α = ¨2

Szα = ¨ 2

4 α

Sz2β = SzSzβ = Sz –

¨2

β = – ¨2

Szβ = ¨ 2

4 β .

Since the expectation value of an operator that gives an exact eigenvalue is justthe eigenvalue, we can write, for either the α or β spin state, <Sz

2> = ¨2/4.

Thus, <S2> = 3<Sz

2> = 3¨2/4 = (1/2)(1/2 + 1)¨2, in accord with Eq.

(12.53a).

GENERAL PROBLEMS

12.44 For scenario (a), the H2O bending angle, we note first that the H–O–H bendingangle θ varies between 0° (the two H atoms overlap) and 180° (the molecule islinear). Moreover, we know that water has an equilibrium (lowest energy)structure with θ ≅ 105°. As θ → 0°, the potential energy should rise rapidly asthe two H atoms bump into and repel each other; the θ = 0° limit should not bereachable. As θ → 180°, the potential should reach a maximum, since, bysymmetry, the linear configuration divides configurations that are bent to oneside or the other of linear. This leads to the qualitative sketch shown below.

30° 90° 120°

V(θ

)

0° 60° 180°150°θ

295

We can graph the potential for scenario (b) exactly: it is the Coulombic repulsivepotential energy between two identical charges, V(r12) = e2/(4πε0)r12, with thezero of energy taken at r12 = ∞. It looks like this:

r12/10–10 m

1 2 3 4 50

4

3

2

1

0

V(r

12)/

10–1

8 J

The methyl cyanide, CH3CN, to methyl isocyanide, CH3NC, isomerization inpart (c) is characterized by an isomerization angle θ that we can take to be 0° inthe cyanide configuration and 180° in the isocyanide configuration. Since thecyanide configuration is the more stable of the two, V(θ) must have an absoluteminimum at θ = 0°. But since the isocyanide is also stable, there must be a sec-ondary minimum in V(θ) at 180° that is higher in energy than the absolute mini-mum. For all angles in between, the energy must rise smoothly to a maximumand then fall again, leading to the following qualitative diagram:

30° 90° 120°

V(θ

)

0° 60° 180°150°θ

As the O–H group is rotated around the C–O bond, the alcoholic H alternatelyeclipses and staggers hydrogens on the methyl group. Thus, every 120°, there

296

is a maximum in V(θ) at each eclipse, and 60° past each maximum, half waybetween successive maxima, we encounter a minimum representing each stag-gered conformation. All the maxima have the same height, all the minima havethe same height, and θ ranges from 0° to 360°. These are the characteristics of afunction we can write as V(θ) = V0 [1 + cos(3θ)] where the constant V0 mea-sures the potential difference between the minima and maxima. This function isshown below.

120° 240° 360°θ

300°180°60°0°0

V0

V(θ

)

Note that there is no sound theoretical argument that says the cosine (or sine—our choice is arbitrary) function accurately represents the true V(θ). Anyfunction with the right number of wiggles of uniform height could perhaps doas well or better, but the cosine is the simplest function with a chance at beingcorrect, and experiments have shown that it can do a good job of reproducingsome of the phenomena associated with this sort of internal motion.

12.45 If the new variables are x′ = (x + y)/ 2 and y′ = (y – x)/ 2, we can solve theseequations for the original variables: x = (x′ – y′)/ 2 and y = (x′ + y′)/ 2, andsubstitute these expressions into the potential energy function V(x, y):

V(x, y) = k0(5x2 + 5y2 + 6xy)

= k0 52

x′ – y′ 2 + 52

x′ + y′ 2 + 62

x′ – y′ x′ + y′

= k0 (8x′2 + 2y′2)

= 12

(16k0) x′2 + 12

(4k0) y′2 .

297

Not only does this variable change remove the cross term that precluded a sepa-ration of variables solution, the final form of V shows at a glance that the poten-tial is simply a sum of harmonic oscillator potentials in the two independentvariables x′ and y′. Written this way, we can easily identify the force constantskx′ and ky′ for these two directions:

kx′ = 16k0 and ky′ = 4k0 .

The energy expression for a mass m bound to this 2-D potential is therefore

E = ¨ ωx′ vx′ + 12

+ ¨ ωy′ vy′ + 12

, ωx′ = 4k0m and ωy′ = 2

k0m .

12.46 Classically, the electron and positron in positronium move toward each other,moving symmetrically about a center of mass that, since e– and e+ have thesame mass, is always half way between them. For states of zero orbital angularmomentum, these two particles move along a line. The only difference betweenthe Ps atom quantum mechanical problem and the ordinary H atom problem isthe reduced mass: for H, µH = memp/(me + mp) ≅ me, but for Ps, µPs =meme/(me + me) = me/2 = µH/2. Since the ionization energy of H is the nega-tive of the ground-state energy: IP = –E1, and since E1 ∝ µ (see Eq. (12.46a)),the Ps ionization energy is half that of H. Since the Bohr radius is proportionalto 1/µ (see Eq. (12.45)), the Ps Bohr radius is twice that of H. The energylevel expression for Ps is identical to that for H with the exception of the changein the energy constant:

En(H) = – 2.179 × 10–18 J

n2 , En(Ps) = –

2.179 × 10–18 J

2n2 = –

1.090 × 10–18 J

n2 .

For H, the longest wavelength a ground-state atom can absorb is that whichtakes the atom from n = 1 to n = 2. It is 121.5 nm, the longest wavelength ofthe Lyman series mentioned on page 351 and shown in Figure 11.1 in the text.The same calculation holds for Ps:

λ = hc∆E

= hc

E2(Ps) – E1(Ps)

= hc

– 1.090 × 10–18 J

22 – –

1.090 × 10–18 J

12

= 243 nm ,

298

exactly twice the H atom n = 2 → 1 wavelength. In a classical picture of the Psground state, the positron–electron motion is along a line connecting both parti-cles through the center of mass. The classical turning point of this motion forPs is 4a0, found by writing E1(Ps) = the total energy = V(r = classical turningpoint) = –e2/(4πε0)r and solving for r. The time τ it takes the electron (say) togo from its turning point, in to r = 0, out to the opposite turning point, thenback to r = 0 and on to the original turning point must be four times the time tomove from the first turning point to r = 0. (This time is the same for the posi-tron to make its mirror-image motion.) It can be shown that this total time isfour times as long as the time for the electron in the Bohr model of H to makeone Bohr orbit around the proton, which for Ps is a total time τ ≅ 6 × 10–16 s.Thus, in 0.1 ns = 10–10 s, the Ps mean lifetime, the electron and positron makeabout 170 000 “classical orbits” around their mutual center of mass.

12.47 The density of states for a particle in a 1-D box is worked out as Example 20.1on page 754 in the text. We can follow the logic shown there to derive thedensity of states for a particle-on-a-ring, introduced in Example 12.6 on page419 in the text. The energy expression is E = m2¨2/2MR2 for a particle ofmass M with quantum number m = 0, ±1, ±2, …. One difference between thissystem and the particle-in-a-box is that every state here except m = 0 is doublydegenerate. It introduces little error to derive the general expression assumingall states are doubly degenerate, and thus we write the density of states ρ(E) as

ρ(E) = 2 dmdE

= 2 dEdm

–1 = 2

m¨ 2

MR2

–1

= 2 2MR2

¨ 2

1/2 ¨ 2

MR2

–1

= R

¨ 2M

E .

12.48 If V(r) = (constant)rn, the general form of the Virial Theorem becomes

2<T> = <r∂V∂r

> = <(constant)rnrn – 1> = n<V> .

Since E = <T> + <V>, we can write

E = <T> + <V> = n2

<V> + <V> = n + 2

2 <V> = <T> +

2n

<T> = n + 2

n <T> .

If n = 2 (harmonic), we see that <T> = <V>. For n = –1 (Coulombic), <T> =–<V>/2.

299

12.49 If we equate the total energy of a state of the H atom, En, to the effective po-tential energy (the sum of the centrifugal potential and the Coulombic potential),we can write

En = – e2

2(4πε0)a0 1n2

= l(l + 1)¨ 2

2µr 2 – e2

(4πε0)r , a0 =

(4πε0)¨ 2

µe 2 .

With the definition suggested in the problem, ρ = r/a0, this can be written

0 = e2

(4πε0)r – e2

2(4πε0)a0 1n2

– l(l + 1)¨ 2

2µr 2

= e2

(4πε0)r – e2

2(4πε0)a0 1n2

– l(l + 1)a0e2

2(4πε0)r2

= 1r – 12n2

– l(l + 1)a0

2r2 = 1

ρ – 1

2n2 –

l(l + 1)

2ρ2

or 2n2ρ – ρ2 – l(l + 1)n2 = 0. A quadratic solution for ρ gives

ρ = n2 ± n n2 – l(l + 1) .

The two solutions represent the two classical turning points. Note that for l = 0(i.e., s states), the expression simplifies to ρ = 2n2 and ρ = 0, which simplymeans that r = 0 can be reached in s states. In units of a0, the outer classicalturning points (the values for ρ with the + sign chosen) are 32 for 4s, 30.967for 4p, 28.649 for 4d, and 24 for 4f. These can be compared to the <r> valuesfor the same states: 24, 23, 21, and 18, respectively, using the expression for<r> given on page 431 in the text.

12.50 If we take p = ¨/R, so that T = ¨2/2meR and thus Fout = –dT/dR = ¨2/meR3,we can write the force balance equation

Fout + Fin = ¨ 2

meR3 – e2

(4πε0)R2 = 0 .

Solving this expression for R gives R = ¨2(4πε0)/mee2 = a0, the usual Bohrradius. The Virial Theorem says E = –T = –¨2/2mea0 = E1, the exact H atomground-state energy. Thus, this simple model gives the same energy and sizeparameter as the Bohr theory, which, in turn, gives the same energy and size asthe complete quantum mechanical theory.