CHAPTER 12
Quantum Mechanical Model Systems
SECTION 12.1
12.1
Half the particles here have twice the momentum of the other half and therefore
also have twice the wavevector of the others. The analog of the wavefunction
in Eq. (12.8) is thus
ψ∝
1 1/2 ikx 1 1/2 i2kx
e +
e
.
2
2
The probability distribution is
1 1/2 ikx 1 1/2 i2kx 1 1/2 –ikx 1 1/2 –i2kx
ψ2∝
e +
e
e
+
e
2
2
2
2
1 1
1
1
1
= + e –ikx + e ikx + = 1 + e ikx + e –ikx = 1 + cos kx .
2 2
2
2
2
This is an oscillatory function that oscillates between 0 and 2 with a spacing
between maxima of 2π/k. The particles are localized to regions around these
maxima (much as a standing waves of water form a regular pattern of maxima
and 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 =
2
2
e –(k – k0) /2(δk) e –ikx dk .
–∞
This integral is known. It has the form of the general definite integral
258
∞
2 2
π q2/4p 2
e
e –p z ± qz dz =
, p>0,
p
–∞
and we can identify p2 = 1/2(δk)2, q = ix, q2 = –x2, and z = k – k0. This gives
ψ(x) ∝
∞
2
2
e –(k – k0) /2(δk) e –ikx dk
–∞
= e –ik0 x
∞
2
2
e –(k – k0) /2(δk) – i(k – k0) x dk
–∞
2
2
= δk 2π e –ik0 x e –x (δk) /2 ,
which, except for the factor δk 2π, is the function shown in the text on page
392 and plotted (as the real part) in Figure 12.2(b). (Whether one has a factor
e –ik0 x , as we have here, or a factor e ik 0 x , as in the text, is unimportant. Both
have the same real part, cos(k0x), and the imaginary part of one is the negative
of that of the other.) The definite integral above is a special case of the more
general definite integral we have seen before (see Problem 11.22):
∞
2
e –αy dy =
–∞
π
α
.
We “complete the square” in the argument of the exponential:
∞
–∞
2 2
e –p z ± qz dz =
∞
2 2
2
2
2
2
e –p z ± qz – q /4p + q /4p dz
–∞
2
2
= e q /4p
∞
2 2
2
2
e –p z ± qz – q /4p dz
–∞
2
2
= e q /4p
∞
2
–
e – pz + q/2p dz ,
–∞
then define y = pz –+ q/2p so that dy = p dz. Making these substitutions (with α
= 1) completes the derivation.
259
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 particles
toward x = –∞ and since particles moving toward x = ∞ are not reflected back
at x = ∞, this coefficient must be zero.
(b) The requirement ψ<(0) = ψ>(0) means 1 + B = C, and the requirement that
the first derivatives are equal at x = 0 means ik< – ik<B = ik>C. Solving these
two simultaneous equations for B and C gives
k – k>
B= <
k < + k>
and
C=
2k<
.
k < + k>
(c) The product B*B is
B*B =
k < – k> 2
k < + k>
since the wavevectors k< and k> are both real numbers. As the total energy
increases, the potential step becomes an ever-decreasing perturbation; the particle’s kinetic energy in the positive x region is almost the same as it was in the
negative 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-energy
limit, 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 of
their first derivatives leads to ik< – ik<B = –κ>D. Simultaneous solution of
these equations gives
B=
ik< + κ>
ik< – κ>
and
D=
2ik<
ik< – κ>
.
(c) Here, B*B is
B*B =
ik< + κ>
ik< – κ>
*
2
k <2 + κ>
–ik< + κ> ik< + κ>
=
=
=1.
2
ik< – κ>
–ik< – κ> ik< – κ>
k <2 + κ>
ik< + κ>
260
(d) The probability of barrier penetration depends on total energy E through the
quantity D*D:
D*D =
4k<2
2
k <2 + κ>
.
Since the wavevectors k< and κ> depend on E through
k< = 2mE
¨
κ> =
and
2m(V – E)
,
¨
we can write
D*D =
4k<2
2
k <2 + κ>
=
4E
4E
=
.
E + (V – E) V
Since E ranges from 0 to V (because E > V returns us to the energy regime of
the previous problem), the probability is maximal when E = V.
12.5
Here, the wavefunctions can be written
ψ< = C e –ik<x
ψ> = e –ik>x + Be ik> x .
and
Note that ψ< has no term corresponding to particles moving towards x = ∞
(i.e., no e ik< x term), since particles originate at x = ∞ and, while they can be
reflected from the potential step at x = 0 and move back to x = ∞ in the positive
x region, they can only move toward x = –∞ in the negative x region. We find
expressions for B and C from the requirements that the wavefunctions and their
first derivatives are equal at x = 0:
1+B=C
–k> + k>B = –k<C .
and
Simultaneous solution of these two equations gives
k – k<
B= >
k > + k<
and
261
C=
2k>
.
k > + k<
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 x
region moving towards x = ∞. It is simply
B*B =
k > – k< 2
,
k > + k<
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
2
E1 =
¨ π2
2meL 2
.
Equating this to the electron’s mass-equivalent energy, mec2, and solving for L
gives
L=
2.43 × 10–12 m
h
=
= 8.58 × 10–13 m .
2 2
2 2 mec
As an aside, we can repeat this line of reasoning for the zero-point energy of a
harmonic oscillator and get another answer of the same magnitude. We write,
with X representing the classical turning point of the motion,
E0 =
¨
2
2m eX 2
= mec 2
or
X=
¨
= 1.93 × 10–13 m .
2mec
Thus, the full classical motion of the electron, from –X to X, spans 1/2π times
the 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” of
a quantum-mechanical system. Here, we will find classical behavior (in the
sense defined in this problem—the position uncertainty is greater than the distance between successive maxima in the probability distribution) at a surprisingly small quantum number. First, we find ∆x for a particle in a box. The
result of Problem 11.23 gives us an easy route, if we recall that <x> = L/2:
262
2
1
1
L2
–
(∆x)2 = <x > – <x>2 = L 2 –
3 2n 2π2
4
1
∆x
1 1/2
–
=
.
12 2n 2π2
L
or
(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 divided
by L) and look for that n = n* for which ∆x/L > 1/n. We make the following
table:
n
∆x/L
1/n
1
2
3
4
5
10
0.181 0.266 0.279 0.283 0.285 0.288
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
dx
dx ω
dx
=
=
=
.
τ 1/2 τ 1/2 (dx/dt) Xω π sin ω t πX 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
dx
=
πX sin ω t π X 2 – x 2 1/2
as we wished to show. The normalization integral follows from
X
X
P(x) dx =
–X
1
π
dx
–X
X2
–
1/2
x2
.
Let x = yX so that dx = X dy. Then the integral above becomes
263
X
1
π
12.9
1
1
X dy
dy
1
1
π
=
=
= =1.
2
2 1/2 π
2
2 2 1/2 π
2 1/2 π
–X X – x
–1 X – y X
–1 1 – y
dx
At the classical turning point of a harmonic oscillator, x = Xv where Xv is the
classical turning point for state v. Moreover, the energy of state v is Ev =
2
kX v /2 for force constant k. Thus, the Schrödinger equation becomes
2
2
¨ d 2 ψv k x 2
¨ d 2 ψ v k x 2 kX v2
–
– Ev ψ v = –
–
ψv = 0 .
+
+
2µ dx 2
2
2µ dx 2
2
2
If x = Xv, the terms in parentheses cancel each other, and we are left with
d2ψv/dx2 = 0. The wavefunction has an inflection point at the classical turning
point, as Figure 12.8 (and Figure 11.2 for the ground state) indicate. This must
be a general result, since the definition of a classical turning point is that point at
which the total energy equals the potential energy and the terms in parentheses
in the Schrödinger equation as written above are always (potential energy – total
energy).
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 + 1 x/Xv so that q1 =
3x/X1 and q2 = 5x/X2. The Hermite polynomials for these two states are in
Table 12.1: H1(q) = 2q and H2(q) = 4q2 – 2. The normalization constants
given by Eq. (12.20) are for the wavefunction written in terms of the dimensionless variable q. We want the normalization constants for the wavefunction
written in terms of the real distance variable x, and thus we use the expressions
listed in the Summary to Chapter 12 on page 440. (See also the comments at
the end of Example 12.4 on page 406.) We have
N1 =
k 1 1/2
π¨ ω 2
and
k 1 1/2 .
π¨ ω 8
N2 =
Since q can also be written as k/¨ω x, we have
This makes the normalization constants equal to
N1 =
3 1/2
2 π X1
1
and
264
N2 =
k/¨ω =
5 1/2
.
8 π X2
1
2v + 1 /X v .
Thus, the entire wavefunctions are
3 1/2
3x –3x 2 /2X 2
2
e
1
X1
2 π X1
2
2
1 5 1/2 5x 2
ψ2 =
4
– 2 e –5x /2X 2 .
8 π X2
X2
ψ1 =
1
2
12.11 As we used in the previous problem, there is a simple relationship between the
classical turning point of any harmonic oscillator state and its parameters m and
k. For the ground state, the relationship is
X0 =
¨ω
.
k
Since ω = k/m , this can also be written as
X0 =
¨ω
=
k
¨ k
=
k m
¨
.
km
Using the values for k and m quoted in the problem, we find X0 = 0.123 Å for
H 2 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 is
about 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 wavefunction to have its maximum at x = 0. If the potential was a particle-in-a-box
extending over –L/2 ≤ x ≤ L/2, we know the ground-state wavefunction would
be symmetric about and maximal at x = 0. Likewise, a harmonic potential centered at x = 0 has a ground-state wavefunction symmetric about and maximal at
x = 0. Thus, since what we have here is half of each type of potential, we want
a 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 coordinate
system for a particle-in-a-box extends from 0 to L; since the potential we are imagining here has been shifted towards smaller x by L/2, the ground-state wavefunction has the sine function of Eq. (12.11) replaced by cosine:
ψ(x ≤ 0) =
265
2
πx
cos
L
L
The harmonic oscillator ground-state wavefunction is
ψ(x ≥ 0) =
2
1
π1/4 X 0
e –x 2 /2X 0 =
k 1/4e –k 2 x 2 /2¨ ω .
π¨ ω
For these to join smoothly at x = 0, they must equal each other and have equal
first 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 we
seek:
2 1/2 = k 1/4 = mk 1/4
L
π¨ ω
π¨
or
L=2
π¨
.
mk
12.13 Here, we can exploit the eigenvalue expressions for the creation and annihilation operators given on page 405 in the text:
d
ψ = [2(v + 1)]1/2ψ v + 1 , v = 0, 1, 2, …
dq v
d
q+
ψ = (2v)1/2ψ v – 1 , v = 1, 2, 3, …
dq v
q–
We write the matrix element qnm and expand it as follows:
qnm =
1
2
1
=
2
=
ψn qψm dq =
ψn q –
1
2
d
ψ dq +
dq m
2(m + 1)
d
d
+ q+
ψ dq
dq
dq m
d
ψn q +
ψ dq
dq m
ψn q –
ψn ψm + 1 dq + 2m
ψn ψm – 1 dq .
Since the wavefunctions are orthonormal, the first integral in the last line above
vanishes unless n = m + 1 (in which case it equals 1), and the second integral
vanishes 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 of
these integrals.
266
12.14 The matrix representation of the square if the q operator is the square of the
matrix representation of the q operator alone:
2
qnm
∞
=
∑
qni q im .
i=1
The previous problem gives the pattern of nonzero values for qnm, and armed
with this pattern and the equation above, we can deduce that the pattern of non2
zero values for qnm must be
0 q01 0
q
0 q12
q2 = 10
0 q21 0
2
q01 q10
0
q12 q01
0
0 q12 q21 + q01 q10
0
q23 q12
0
q23 q32 + q12 q21
0
= q21 q10
0
q32 q21
0
q34 q43 + q23 q32
.
2
In particular, we see that q00
is nonzero and equals
2
q00
= q01 q10 =
2
2
2
1
= .
2
2
Problem 11.34 found <x2> = X2/2 where X is the classical turning point. Our
2
result here, q00 = <q2>00 = 1/2, is the dimensionless variant of <x2> for the
ground state for which q = x/X. Thus, <q2>00 = <x2>00/X2 = 1/2.
12.15 Differentiating the generating function gives
2
2
dH v d
2 d v e –q
2 d v + 1 e –q
=
(–1)v e q
= (–1)v 2qe q
= 2qH v – Hv + 1 .
dq
dq
dq v
dq v + 1
Differentiating this gives
267
d 2H v
dH v dH v + 1
d(2qHv – Hv + 1)
= 2Hv + 2q
–
dq
dq
dq
dq 2
= 2H v + 2q(2qH v – Hv + 1) – 2qH v + 1 + Hv + 2
=
= (2 + 4q 2)H v – 4qH v + 1 + Hv + 2 .
If we substitute these expressions for the first and second derivatives of Hv into
Hermite’s differential equation, we find
d 2H v
dq 2
– 2q
dHv
+ 2vH v = (2 + 2v)H v + Hv + 2 – 2qH v + 1 = 0 .
dq
To arrive at the recurrence relation shown in the text, define w = v + 1, so that
v = 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 from
w 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 original
expression for the first derivative of Hv, we find
dH v
= 2qH v – Hv + 1 = 2qH v – 2qH v + 2vH v – 1 = 2vH v – 1 .
dq
12.16 If we expand the Lennard-Jones potential function in a Taylor’s series with the
aim of finding its effective force constant, k we see that we need the second
derivative of V, since k = (d2V/dx2)x = xe. We can write V as
V(x) = D 1 – 2xe6x –6 + xe12x –12
so that the first derivative is
dV(x)
= D 12xe6x –7 – 12xe12x –13
dx
and the second derivative is
268
d 2V(x)
dx 2
= D –12·7xe6x –8 + 12·13xe12x –14 .
Evaluating this at x = xe gives the answer we seek:
k=
d 2V(x)
dx 2
= D –12·7xe6xe–8 + 12·13xe12xe–14 =
x = xe
72D
2
.
xe
SECTION 12.3
12.17 A particle moving freely in 3-D space has the same Hamiltonian whether the
space is confined to a finite region (the 3-D particle-in-a-box potential) or
extends to infinity. The Hamiltonian is given by Eq. (12.21):
2
2
¨
∂2
¨
2
∂2
∂2
∇ .
H=–
+
+
=–
2m ∂x 2 ∂y 2 ∂z 2
2m
Since the three directions in space x, y, and z are equivalent and independent of
each other, the most general 3-D free particle wavefunction has the form
ψ (x, y, z) = Ae i(k xx + kyy + kzz) + Be –i(k xx + kyy + kzz) = Ae ik⋅r + Be –ik⋅r
where kx, ky, and kz are the wavevectors for motion along the x, y, and z directions, respectively, and where k is the vector with components kx, ky, and kz, r
is the vector with components x, y, and z, and k·r represents the vector dot
product of k and r. The three quantum numbers needed to describe the state of
the system (including its energy) are the wavevector components since
2
¨ k x2 + ky2 + kz2
=
, ky =
=
, kz = =
, and E =
kx =
2m
¨
¨
¨
¨
¨
¨
px
mv x
py
mv y
pz
mv z
.
Each wavevector can vary continuously from –∞ to ∞ (although ±∞ implies an
infinite velocity, which relativity does not allow).
12.18 The 2-D particle-in-a-box quantum numbers are nx and ny, and each can range
from 1 to ∞ in integral steps. Since the box is a square (all sides of length L),
the total energy expression is
269
2
E = Ex + Ey =
¨ π2
2mL 2
nx2 + ny2 .
To find all the states with energy E ≤ 9¨2π2/mL2, we can define the dimen2
2
sionless energy ε = E/(¨2π2/mL2) = (nx + ny )/2, which we can tabulate:
n x\n y
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 interested
in here. This shows us that the state with ε = 1 (or E = ¨2π2/mL2) is nondegenerate, 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 at
the square of the wavefunction. If we define the x and y coordinates so that the
edges of the box extend over the ranges 0 ≤ x, y ≤ L, the square of the wavefunction is
2 1/2 2 n xπx
2
2
sin
ψ 2 (x, y) = ψ n (x)ψ n (y) =
x
y
L
L
n
πx
n
πy
2
= sin2 x
sin2 y
.
L
L
L
2 1/2 2 n yπy
sin
L
L
Wherever this function is a maximum for any (nx, ny) choice locates the most
probable position or positions. An easy way to see this probability is as density
plot (an analog of a contour plot) that plots ψ2 in two dimensions as a gradation
in shade that is proportional to the value of ψ2. Such plots are shown at the top
of the next page for the states (nx, ny) = (1, 1), (3, 1), (1, 3), and (2, 2), drawn
so that the darkest shade locates the maximum or maxima in ψ2.
270
(nx, ny) = (1, 1)
L
y
0
y
0
x
L
(nx, ny) = (3, 1)
L
0
0
x
L
(nx, ny) = (2, 2)
L
y
0
(nx, ny) = (1, 3)
L
y
0
x
L
0
0
x
L
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 the
picture of one by 90° to get the picture of the other. State (1, 3) peaks at x = L/2
and 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 parameter
needed to characterize the energy of the system. There are three independent
quantum 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 v x + 1 + v y + 1 + v z + 1
2
2
2
3
= ¨ω v x + v y + v z + ,
2
271
and we can use it to tabulate ε = E/¨ω for various combinations of quantum
numbers:
vx vy vz
0
1
0
0
1
0
0
1
0
1
0
0
0
1
0
ε
ε
vx vy vz
3/2
5/2
5/2
5/2
7/2
1
0
2
0
0
0
1
0
2
0
1
1
0
0
2
vx vy
7/2
7/2
7/2
7/2
7/2
1
2
2
0
1
1
0
1
2
2
vz
ε
1
1
0
1
0
9/2
9/2
9/2
9/2
9/2
vx vy vz
0
1
3
0
0
1
0
0
3
0
2
2
0
0
3
ε
9/2
9/2
9/2
9/2
9/2
We see that the lowest energy state is nondegenerate, at E = 3¨ω/2, the next
highest state is three-fold degenerate at E = 5¨ω/2, the next is six-fold degenerate 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 molecule
units:
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 energy
expression in Example 12.5 with m = mass of one He atom = 6.65 × 10–27 kg
and 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 following
expression for n:
2
En = 3.32 ×
10–19
¨ π2 2
J=
n = (4.08 × 10–21 J)n 2 ,
2
2mrm
and find n = 9.02, so that, since n must be an integer, any state with n ≥ 9 or so
gives the He atom sufficient energy to escape the C60 cage. (Of course, the 200
kJ mol–1 bond energy is not a very accurately known number here, and since
our answer n = 9.02 is so close to 9, it may well be that the state with n = 9 is
not really sufficiently energetic. On the other hand, the model itself is quite
crude, and we should expect the motion of a He atom inside C60 to be more
complex 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 momentum about the center of the cage. These should also be considered.)
272
SECTION 12.4
12.21 We start with the Cartesian representation of Lz given by Eq. (12.31):
Lz =
¨ ∂
∂
x –y
i ∂y
∂x
and follow the steps outlined in the problem to transform to plane polar coordinates. Using the relations r = (x2 + y2)1/2 and θ = tan–1(y/x), we can derive
∂r
x
x
=
=
∂x y
1/2 r
x 2 + y2
∂r
y
y
=
=
∂y x
1/2 r
x 2 + y2
∂θ
y
y
y
=–
=–
=–
∂x y
x 2 1 + (y/x)2
x 2 + y2
r2
∂θ
1
x
x
=
=
= .
∂y x x 1 + (y/x)2 x 2 + y 2 r2
We use these expressions to build the operator through the chain rule:
∂θ ∂
∂
∂r ∂
y∂
x ∂
=x
+
=x
+
∂y x ∂θ
∂y
∂y x ∂r
r ∂r r2 ∂θ
∂θ ∂
∂
∂r ∂
x∂ y ∂
+
=y
–
y =y
∂x y ∂θ
∂x
∂x y ∂r
r ∂r r2 ∂θ
x
so that
x
∂
∂
y∂
x∂ y ∂
x ∂
–y =x
+
–
–y
∂y
∂x
r ∂r r2 ∂θ
r ∂r r2 ∂θ
xy ∂ x 2 ∂ xy ∂ y 2 ∂
=
+
–
+
r ∂r r2 ∂θ r ∂r r2 ∂θ
=
x 2 + y2 ∂
r2
∂θ
=
∂
∂θ
.
This proves that Lz in plane polar coordinates equals (¨/i)(∂/∂θ).
273
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
= = cos θ
∂x y r
∂r
y
= = sin θ
∂y x r
∂θ
sin θ
y
=– =–
∂x y
r
r2
∂θ
x cos θ
.
= =
r
∂y x r2
Using these expressions in the chain rule lets us write
sin θ ∂F
∂F
∂F
= cos θ
–
r ∂θ
∂x y
∂r θ
r
cos θ ∂F
∂F
∂F
= sin θ
+
.
r
∂y x
∂r θ
∂θ
r
We substitute from these six expressions as called for into the general expressions for the second derivatives:
∂θ
∂r ∂ ∂F
∂ ∂F
+
∂x y ∂θ ∂x y
∂x 2 ∂x y ∂r ∂x y
sin θ ∂
∂
∂ sin θ ∂
∂ sin θ ∂
= cos θ
cos θ –
–
cos θ –
r ∂θ
r ∂θ
r ∂θ
∂r
∂r
∂r
∂2F
=
∂θ
∂r ∂ ∂F
∂ ∂F
+
∂y x ∂θ ∂y x
∂y 2 ∂y x ∂r ∂y x
cos θ ∂
∂
∂ cos θ ∂
∂ cos θ ∂
= sin θ
sin θ +
+
sin θ +
r ∂θ
r
r ∂θ
∂r
∂r
∂r
∂θ
∂2F
=
and expand each expression:
274
∂2F
∂x 2
∂y 2
= sin2 θ
+
+
sin2 θ ∂F 2 cos θ sin θ ∂2F
–
r ∂r
r
∂r∂θ
∂r 2
sin2 θ ∂2F
+
∂2F
∂2F
= cos2 θ
r2
∂θ2
∂2F
+
2 cos θ sin θ ∂F
∂θ
r2
cos2 θ ∂F 2 cos θ sin θ ∂2F
+
r
∂r
r
∂r∂θ
∂r 2
cos2 θ ∂2F
r2
+
∂θ
2
2 cos θ sin θ ∂
.
∂θ
r2
–
Adding these and recalling that sin2 θ + cos2 θ = 1 gives us
∇2 =
∂2F
∂x 2
∂2F
+
∂2F
∂y 2
1 ∂F 1 ∂2F
=
+
+
∂r 2 r ∂r r2 ∂θ2
1 ∂ 1 ∂F
1 ∂2F
=
+
.
r ∂r r ∂r r2 ∂θ2
12.23 We start with the expressions for the components of the angular momentum
operator written in spherical polar coordinates as given in Eq. (12.32):
Lx =
¨
∂
∂
–sin φ
– cot θ cos φ
i
∂θ
∂φ
Ly =
¨
∂
∂
cos φ
– cot θ sin φ
i
∂θ
∂φ
Lz =
¨ ∂
,
i ∂φ
then we square each operator:
L x Lx = –¨
2
–sin φ
∂
∂θ
– cot θ cos φ
275
∂
∂φ
–sin φ
∂
∂θ
– cot θ cos φ
∂
∂φ
L y Ly = –¨
L zLz = –¨
2
2
∂
cos φ
– cot θ sin φ
∂θ
∂
∂
∂
cos φ
∂φ
∂
∂θ
– cot θ sin φ
∂
∂φ
,
∂φ ∂φ
and expand the products:
L x Lx = –¨
2
+ 2 cos φ sin φ cot θ
L y Ly = –¨
2
2
∂2
∂θ ∂φ
∂2
2
∂2
∂θ ∂φ
∂φ
+ sin2 φ
cos φ sin φ cot2 θ + csc2 θ
– 2 cos φ sin φ cot θ
L zLz = –¨
∂
–cos φ sin φ cot2 θ + csc2 θ
+ cos2 φ
+ cos2 φ cot2 θ
∂2
∂θ
∂
∂φ
+ sin2 φ cot2 θ
∂2
∂θ
2
+ cos2 φ cot θ
2
+ sin2 φ cot θ
∂2
2
∂φ
∂
∂θ
∂2
2
∂φ
∂
∂θ
.
∂φ
2
We add these to form the operator we seek, L = L xLx + L yLy + LzL z . Fortunately, several terms cancel, and the identity sin2 φ + cos2 φ = 1 simplifies
several others:
2
L = –¨
= –¨
2
2
cot2 θ + 1
∂2
∂θ
= –¨
2
2
+ cot θ
∂2
2
+
∂φ
∂
∂θ
+
∂2
∂θ
2
+ cot θ
∂
∂θ
∂2
1
sin2 θ ∂φ2
∂2
1 ∂ sin θ ∂ + 1
.
sin θ ∂θ
∂θ sin2 θ ∂φ2
2
To express L in Cartesian coordinates, we use Eq. (12.31) for each component operator:
276
Lx =
¨
∂
∂
y
–z
i
∂z
∂y
Ly =
¨
∂
∂
z
–x
i
∂x
∂z
Lz =
¨
∂
∂
x
–y
,
i
∂y
∂x
then we square and expand each component. Since all three operators are so
similar, once we square and expand one in detail, we can immediately write the
other two by analogy:
∂
∂
∂
∂
–z
y
–z
∂z
∂y
∂z
∂y
2
∂
∂
∂ ∂
∂
∂
∂
∂
= –¨ y
y
–y
z
–z
y
+z
z
∂z ∂z
∂z ∂y
∂y ∂z
∂y ∂y
2
2
2
∂
∂
∂
∂
∂2
∂2
–z
+ z2
–y
– yz
– zy
= –¨ y 2
∂z∂y
∂y∂z
∂y
∂z
∂z 2
∂y 2
2
∂2
∂2
∂2
∂
∂
+ z2
= –¨ y 2
– 2yz
–y
–z
.
∂z∂y
∂y
∂z
∂z 2
∂y 2
L x Lx = –¨
2
y
Thus, the squares of the other two components are
L y Ly = –¨
2
z2
∂2
∂2
∂2
∂
∂
+ x2
– 2zx
–z
–x
∂x∂z
∂z
∂x
∂x 2
∂z 2
and
L zLz = –¨
2
x2
∂2
∂2
∂2
∂
∂
+ y2
– 2xy
–x
–y
.
2
∂y∂x
2
∂x
∂y
∂y
∂x
We add these three expressions to form the operator for the square of the total
2
angular momentum, L = L xLx + L yLy + LzLz, shown at the top of the next
page. Note that this version of the operator is considerably more complicated
than the version written in spherical polar coordinates earlier in this problem.
Consequently, the Cartesian representation is rarely used.
277
2
L = –¨
= –¨
∂2
∂2
∂2
∂
∂
+ z2
– 2yz
–y
–z
∂z∂y
∂y
∂z
∂z 2
∂y 2
∂2
∂2
∂2
∂
∂
+ x2
– 2zx
–z
–x
+ z2
∂x∂z
∂z
∂x
∂x 2
∂z 2
∂2
∂2
∂2
∂
∂
+ y2
+ x2
– 2xy
–x
–y
∂y∂x
∂x
∂y
∂y 2
∂x 2
2
y2
∂2
∂2
∂2
∂2
∂2
∂2
+
+ y2
+
+ z2
+
∂y 2 ∂z 2
∂x 2 ∂z 2
∂x 2 ∂y 2
∂2
∂2
∂2
∂
∂
∂
+ zx
+ xy
–2 x
– 2 yz
+y
+z
.
∂z∂y
∂x∂z
∂y∂x
∂x
∂y
∂z
2
x2
12.24 The expression in part (a), [L x , Ly] = i¨L z , is easiest to prove using Cartesian
coordinates:
[L x , Ly] = L x Ly – L y L x
¨
¨
∂
∂ ¨
∂
∂
∂
∂ ¨
∂
∂
y
–z
z
–x
–
z
–x
y
–z
i
i
∂z
∂y i
∂x
∂z
∂x
∂z i
∂z
∂y
2
∂
∂2
∂2
∂2
∂2
– yx
+ zx
+ yz
– z2
=–¨ y
∂z∂x
∂y∂x
∂z∂y
∂x
∂z 2
=
+¨
=¨
2
2
zy
x
∂2
∂2
∂2
∂
∂2
– z2
– xy
+x
+ xz
∂x∂z
∂x∂y
∂z∂y
∂y
∂z 2
¨
∂
∂
∂
∂
–y
= i¨
x
–y
= i¨ L z .
i
∂y
∂x
∂y
∂x
2
The expression in part (b), [L , Lz] = 0, is easiest to prove in spherical polar
coordinates, and the calculus is perhaps a bit easier and more transparent if we
2
write the operator L in the following expanded way, which we first encountered in the previous problem:
2
∂2
1 ∂ sin θ ∂ + 1
sin θ ∂θ
∂θ sin2 θ ∂φ2
2 ∂2
∂
∂2
= –¨
+ cot θ
+ 1
.
2
2
2θ
∂θ
sin
∂θ
∂φ
L = –¨
2
278
2
This lets us write L Lz as
2
L Lz = –¨
2
∂2
∂θ
¨
=–
i
3
2
+ cot θ
∂3
2
∂θ ∂φ
∂
∂θ
+ cot θ
+
∂2 ¨ ∂
1
sin2 θ ∂φ2 i ∂φ
∂2
∂θ∂φ
+
∂3
1
sin2 θ ∂φ3
2
and LzL as
2
L zL =
¨ ∂
i ∂φ
¨
=–
i
3
–¨
2
∂2
∂θ
∂3
2
∂θ ∂φ
2
+ cot θ
+ cot θ
∂2
∂θ∂φ
∂
∂θ
+
∂2
1
sin2 θ ∂φ2
∂3
1
.
sin2 θ ∂φ3
+
Subtracting these to form the commutator clearly gives zero.
For part (c), we substitute the definition L+ = Lx + iL y into the expanded commutator of interest:
[L z , L+] = Lz Lx + iL y – Lx + iL y Lz
= L z L x – L x Lz + i L z L y – L y L z
= [L z , Lx] + i[L z , L y]
= i¨Ly + i(–i¨L x)
= ¨ Lx + iL y = ¨L + .
For part (d), we again substitute the definition L+ = Lx + iL y in the commutator
of interest here and write
2
2
2
2
[L + , L ] = [ Lx + iL y , L ] = [L x , L ] + i[L y , L ] .
(Note how commutator algebra is distributive. This lets us jump directly and
2
confidently from the second to the third step above.) Since L commutes with
every angular momentum component operator, both of the final two commuta-
279
2
tors equal zero, as, therefore, does the operator [L + , L ]. Thus, these two
operators commute.
To see if L + and L– commute, we write
[L + , L–] = [ Lx + iL y , Lx – iL y ]
= Lx + iL y Lx – iL y – Lx – iL y Lx + iL y
2
2
2
2
= L x – iL xLy + iL yLx + L y – L x – iL xLy + iL yLx – L y
= 2i L yLx – L xL y = 2i[L y , L x]
= 2i –i¨Lz = 2¨L z .
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 follows:
L z L –ψ m = L zL –ψ m
= L z L– – L – L z + L – L z ψ m
= [L z ,L–] + L–Lz ψm
= –¨ L– + L –L z ψ m
= L –L zψ m – ¨ L – ψ m
= 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 following
equation:
2
2
2
2
2
2
2
L – Lz ψ1,1 = L ψ1,1 – Lz ψ1,1 = 2¨ ψ1,1 – ¨ ψ1,1 = ¨ ψ1,1
so that L x2 +Ly2 = ¨. This, along with Lz = ¨ and |L| = 2 ¨, lets us construct the diagram at the top of the next page for the classical vector L. The L
280
vector lies somewhere on the surface of the shaded cone in this figure; the
Uncertainty Principle does not allow us to say exactly where.
z
45°
¨
L
=
2¨
y
¨
x
For the general case of l = m, we note that the length of L is l (l + 1) ¨ and the
length of its z component is m¨ = l¨. Thus, we can make the following diagram of the general classical vector geometry:
z
θ
l (l + 1) ¨
l¨
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 polar
forms of L x and L y into the definition of L + . Since we will need L– later on in
this problem, we can derive both at once. We also take advantage of the identity eiφ = cos φ + i sin φ and find
281
L± = Lx ± iL y
∂
∂
∂
=
¨
i
=
¨
∂
∂
(±i cos φ – sin φ) – cot θ(cos φ ± i sin φ)
i
∂θ
∂φ
–sin φ
∂θ
– cot θ cos φ
∂φ
± i cos φ
∂θ
– cot θ sin φ
∂
∂φ
¨ ±iφ ∂
∂
– cot θ e ±iφ
ie
i
∂θ
∂φ
∂
∂
+ i cot θ
= ¨ e ±iφ
.
∂θ
∂φ
=
Next, we take l = m, Θl,l(θ) = sinl θ, and Φl(φ) = eimφ = eilφ and operate on
ΘΦ with L +. Since L + raises the z component one ¨ unit, but for m = l, this
component is already as large as it can be, the operator should yield zero:
L +Θ l, l(θ)Φl(φ) = L + sinl θ e ilφ
∂
∂
+ i cot θ
= ¨ e iφ
sinl θ e ilφ
∂θ
∂φ
= ¨ e iφ le ilφ cos θ sinl – 1 θ – l e ilφ cot θ sinl θ
= ¨ e iφ le ilφ cos θ sinl – 1 θ – l e ilφ cos θ sinl θ
sin θ
iφ
ilφ
l
–
1
ilφ
= ¨ e le cos θ sin
θ – l e 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 e 2iφ
= ¨e –iφ i cot θ
∂
∂φ
–
∂
∂θ
sin2 θ e 2iφ
= ¨e –iφ –2e 2iφ cot θ sin2 θ – 2e 2iφ cos θ sin θ
= ¨e –iφ –2e 2iφ cos θ sin2 θ – 2e 2iφ cos θ sin θ
sin θ
–iφ
2iφ
= ¨e
–2e cos θ sin θ – 2e 2iφ cos θ sin θ
= –4¨e iφ cos θ sin θ .
282
Since we have not included normalization constants in our wavefunctions, this
result is only proportional to ψ2,1, but we can see that it has the correct functional dependence on θ and φ for this wavefunction: for m = 1, we should find
a 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 harmonic function (see Table 12.2), and we do.
12.28 With the help of Table 12.2, which lists the spherical harmonic functions we
need 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 times
its complex conjugate gives simply 1, we write Unsöld’s Theorem for the
particular case of l = 1 as
1
∑
2
Y 1m (θ, φ) = Y1,–1 2 + Y1,0 2 + Y1,1 2
m = –1
1 3 2
3
3
sin θ + cos2 θ + sin2 θ
2π 4
2
4
3
3
sin2 θ + cos2 θ =
= constant.
=
4π
4π
=
For l = 2, we follow the same argument, using two trigonometric identities to
simplify our result, sin2 θ = 1 – cos2 θ and sin4 θ – cos4 θ = sin2 θ – cos2 θ:
2
∑
2
Y 2m (θ, φ) = Y2,–2 2 + Y2,–1 2 + Y2,0 2 + Y2,1 2 + Y2,2 2
m = –2
15 4
15 2
2
1
5
sin θ +2
sin θ cos2 θ + 3cos2 θ – 1
2
4
2π 16
8
5
=
9cos4 θ – 6cos2 θ + 3sin4 θ + 12cos2 θ sin2 θ + 1
16π
5
9cos4 θ – 6cos2 θ + 3sin4 θ + 12cos2 θ 1 – cos2 θ + 1
=
16π
5
6cos2 θ + 3 sin4 θ – cos4 θ + 1
=
16π
5
6cos2 θ + 3 sin2 θ – cos2 θ + 1
=
16π
5
5
3sin2 θ + 3cos2 θ + 1 =
= constant.
=
16π
4π
=
283
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 in
Eq. (12.42) with R = a0, the Bohr radius of Eq. (12.45) with µ = me. We find
E=
l(l + 1) ¨
2mea02
2
=
l(l + 1) ¨
2
2
e 4 me
= l(l + 1)
2
2(4πε0)2 ¨
4πε0¨
2me
me e 2
= l(l + 1) × (2.18 × 10–18 J) = l(l + 1) × (13.6 eV) .
2
Note that the energy constant e4me/2(4πε0)2¨2 = e2/2(4πε0)a0 = –E1, which is
the 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 centrifugal
force of an orbiting electron in the Bohr model gives us
e2
2
4πε0a 2
= meaω .
The Bohr quantization condition, mea2ω = n¨, gives us an expression for the
orbiting frequency ω:
ω=
n¨
,
mea 2
that, when substituted in the first equality, gives
e 2 = m a n¨
e
4πε0a 2
mea 2
2
,
or
a=
4πε0¨
n2
mee 2
2
= n 2a0 .
If E = –T = –mea2ω2/2, we can use the expressions above for ω to write an
expression for a2ω2 that, when substituted in the energy expression, gives
something we can arrange in the form of Eq. (12.52):
284
2
2
n 2¨
m a 2ω2
n 2¨
= –me
E=– e
=–
2
2me2a 2
2mea 2
2
n 2¨
=–
2me
e 4me2
4
=–
n 4¨ (4πε0)2
e 4me
e2
1 =–
1 ,
2
2
2(4πε
)a
0 0 n2
2¨ (4πε0)2 n
where, in the last step, we have collected those constants that define the Bohr
radius, a0 = (4πε0)¨2/mee2.
12.31 If we scale a 1.2 × 10–15 m proton radius up to 1 mm (a magnification factor
of 8.33 × 1011), and if we define the proton distance from the center of mass
to be rp and the electron distance re, then we can write two expressions relating
rp 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 definition 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, becomes
about 66 m. Solving the two expressions for rp and re gives
rp =
me
<r>
m p + me
and
re =
mp
<r> .
m p + me
The magnified numerical values are rp = 36 mm and re = 66 m. (All current
theories and experiments on electrons, by the way, indicate that the electron has
no radius; it is a true point.) In terms familiar to American readers, if we place
the center of mass on the goal line of an American football field, the proton is
about 1.4 in. into the end zone and the electron is on the opponent’s 28 yard
line.
12.32 From the general expression ∆E = hc/λ for the emission wavelength λ accompanying an energy change ∆E, we can write
λ=
hc
hc
hc
hcn 2m 2
hc n 2m 2
=
=
=
=
.
∆E En – Em E1 E1 E m 2 – n 2 –E1 n 2 – m 2
1
–
n2 m 2
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 in
Eq. (11.2), the Balmer formula.
285
12.33 The differences in size and ionization energy for the three isotopic forms of
hydrogen are due to the different reduced masses µ of each isotope. The reduced mass enters the Bohr radius expression and the ground-state energy expression (the negative of which is the ionization potential):
4πε0¨
a0 =
µe 2
2
and
IP = –E1 =
e2
.
2(4πε0)a0
Using the full number of significant figures in the mass of the electron along
with the accurate nuclear masses given in the problem and the conversion factor
between atomic mass units and kg, we can calculate reduced masses:
memH
= 9.104 434 × 10–31 kg
m e + mH
memD
= 9.106 909 × 10–31 kg
µD =
m e + mD
memT
= 9.107 733 × 10–31 kg .
µT =
m e + mT
µH =
From these, the following radii and ionization potentials follow:
2
4πε0¨
a0(H) =
= 5.294 653 × 10–11 m
2
e µH
e2
= 2.178 688 × 10–18 J
IPH =
2(4πε0)a0(H)
2
4πε0¨
= 5.293 214 × 10–11 m
e 2µD
e2
= 2.179 281 × 10–18 J
IPD =
2(4πε0)a0(D)
a0(D) =
286
2
4πε0¨
a0(T) =
= 5.292 735 × 10–11 m
2
e µT
e2
= 2.179 478 × 10–18 J .
IPT =
2(4πε0)a0(T)
The size differences are inconsequential and unobservable, but the ionization
energy differences are experimentally observable.
12.34 The general rules that guide us here are: the total number of nodes = n – 1 and
the total number of spherical nodes = n – l – 1. For wavefunction (a), “spherically 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), the
three 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 cannot
uniquely identify m. Wavefunction (c) has one spherical and one planar node
so 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 is
cylindrically 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. This
is the 5dz2 wavefunction. For (e), the symmetry tells us m = 0, and the single
nodal plane (which must be the xy plane) tells us l = 1 so that n = 2. This is the
2pz wavefunction of in Figure 12.14(d).
12.35 From Table 12.4, we see that the angular portion of the 2pz wavefunction is
simply 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.
287
2pz
3dz 2
These plots show the symmetry and rough spatial extent of the electron probability 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 given
implicitly by the equation
π
2π
dφ
0.90 =
0
r*
sin θ dθ
0
0
2
ψ1s r 2 dr
=
4
3
ao 0
If we let x = r/a0, this integral becomes
x*
e –2x x 2 dx ,
0.90 = 4
0
288
r*
e –2r/a0 r2 dr .
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 probability
versus r for this state just to see how rapidly the probability is approaching 1:
Probability
1.0
0.8
0.6
0.4
0.2
0
0
1
3
r/a0
2
4
5
6
At the <r> value, which is 1.5 on the scale of this figure, the probability is only
0.6 or so.
2
12.37 From Example 12.8, the radial distribution function, rdf, is defined as 4πr2R nl
where Rnl is the radial part of the wavefunction for the state with quantum numbers n and l. These radial factors are listed in Table 12.3, and thus we can write
rdf1s = 4πr 2
4
3
a0
rdf2pz = 4πr 2
e –2r/a0 =
16π 2 –2r/a0
r e
3
a0
r2 –r/a0 π 3 –r/a0
e
=
r e
.
3 2
5
24a0 a0
6a0
1
We find the maxima in these functions by differentiation:
drdf1s 16π
2
=
2re –2r/a0 – r 2 e –2r/a0
dr
3
a0
a0
drdf2pz
π
r3
=
3r 2 e –r/a0 – e –r/a0 .
dr
5
a03
6a0
289
When these derivatives equal zero, we are at the maximum. For 1s, this happens when (2r – 2r2/a0) = 0 or when r = a0. For 2pz, we want 3r2 – r3/a0 = 0
or r = 3a0. Finally, we find the 2s radial distribution function:
rdf2s = 4πr 2
1
8a03
2–
r 2 –r/a0 πr 2
r 2 –r/a0
e
=
2–
e
.
a0
a0
2a03
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 in
the 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 of
nuclear 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
a
<r> = 0 3n 2 – l(l + 1) .
2Z
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 exceed
the H atom 1s size, given by <r> = 3a0/2. Thus, we write
3a0 3a0
=
n*2
2
2·92
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 expression
for Z and equate the results. We find <r> = (3a0/2) × (–E1/IP)1/2, which is
graphed below in a log–log plot.
290
10
<r>/a0
H
1.0
0.1
U
0.01
1
10
100
–IP/E1
1000
10 000
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 number l. (The correct quantum expression L = l(l + 1) ¨ reduces to L = l¨ in the
large l limit.) An H atom must have n at least this large in order to have such an
angular momentum. Thus, if we take n = l = 2.6 × 1074 and substitute into
the general expression for <r>, we find
a
a
<r> = 0 3n 2 – l(l + 1) = 0 3n 2 – n 2 – n ≅ a0n 2
2
2
or <r> ≅ 6.6 × 10148 a0 = 3.5 × 10138 m. This huge number, larger than the
radius of the known Universe, indicates that while n can increase without limit
in the simplest theory of atomic hydrogen, states with very large n do not have
physical significance.
12.40 The Lz operator is (¨/i)(∂/∂φ), and the 2px and 2py hybrid functions depend on
φ as shown in the text on page 434: 2px ∝ cos φ and 2py ∝ sin φ. Since we
291
know ∂cos φ/∂φ = sin φ and ∂sin φ/∂φ = –cos φ, these hybrid functions are not
eigenfunctions of Lz. In fact, the Lz operator turns one function into something
at least proportional to the other.
12.41 Table 12.5 lists the general form of the hybrid H-atom wavefunctions, and for
the 3px and 3py functions, we find
3px =
1
Ψ311 + Ψ31–1
2
and
3py =
–i
Ψ311 – Ψ31–1 .
2
Table 12.4 in turn gives the hydrogenic wavefunctions Ψ31±1 as
Ψ31±1 =
1 1 –r/3a0 6r r2
e
–
sin θe ±iφ .
a0 a2
π a3/2
0
0
Substitution and simplification (using 2 cos φ = eiφ + e–iφ and 2i sin φ = eiφ –
e–iφ) gives
3px =
6r r2
–
cos φ sin θ e –r/3a0
3/2 a0
2
π a0
a0
3px =
6r r2
–
sin φ sin θ e –r/3a0 .
3/2 a0
2
π a0
a0
2
2
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 the
y-z plane with x = 0, exactly like the 2px hybrid, and the 3py hybrid has a nodal
plane wherever sin φ = 0, which is the x-z plane with y = 0, exactly like the 2py
hybrid.
12.42 We follow the logic of the previous problem again here, starting with the general forms of the hybrids in Table 12.5:
292
1
Ψ321 + Ψ32–1
2
–i
3dyz =
Ψ321 – Ψ32–1
2
–i
3dxy =
Ψ322 – Ψ32–2 .
2
3dxz =
The hydrogenic wavefunctions we need are in Table 12.4:
1
Ψ32±1 =
π a03/2
81
e –r/3a0
1
Ψ32±2 =
162
π a03/2
r2
a02
e –r/3a0
sin θ cos θ e ±iφ
r2
a02
sin2 θ e ±2iφ .
Substitution and simplification again gives the hybrids:
3dxz =
r2 –r/3a0
e
cos φ cos θ sin θ
81 π a03/2 a02
3dyz =
r2 –r/3a0
e
sin φ cos θ sin θ
3/2 2
81 π a0 a0
3dxy =
r2 –r/3a0
e
cos φ sin φ sin2 θ .
3/2 2
81 π a0 a0
2
2
2
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 3dyz
hybrid 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 the
other two hybrids.
SECTION 12.5
293
12.43 Equation (12.53b) gives us the eigenvalues for the S z operator: S z α = (¨/2)α
and S z β = –(¨/2)β. The eigenvalue expressions for the square of S z are therefore
2
2
Sz α
¨
¨
¨
α = S zα =
α
= S z S zα = S z
2
2
4
2
Sz β
¨
¨
¨
= S z S zβ = S z –
β = – S zβ =
β.
2
2
4
2
Since the expectation value of an operator that gives an exact eigenvalue is just
2
the eigenvalue, we can write, for either the α or β spin state, <S z > = ¨2/4.
2
2
Thus, <S > = 3<S z > = 3¨2/4 = (1/2)(1/2 + 1)¨2, in accord with Eq.
(12.53a).
GENERAL PROBLEMS
V(θ)
12.44 For scenario (a), the H2O bending angle, we note first that the H–O–H bending
angle θ varies between 0° (the two H atoms overlap) and 180° (the molecule is
linear). Moreover, we know that water has an equilibrium (lowest energy)
structure with θ ≅ 105°. As θ → 0°, the potential energy should rise rapidly as
the two H atoms bump into and repel each other; the θ = 0° limit should not be
reachable. As θ → 180°, the potential should reach a maximum, since, by
symmetry, the linear configuration divides configurations that are bent to one
side or the other of linear. This leads to the qualitative sketch shown below.
0°
30°
60°
90°
θ
294
120°
150°
180°
We can graph the potential for scenario (b) exactly: it is the Coulombic repulsive
potential energy between two identical charges, V(r12) = e2/(4πε0)r12, with the
zero of energy taken at r12 = ∞. It looks like this:
V(r12)/10–18 J
4
3
2
1
0
0
1
2
3
4
5
/10–10 m
r12
V(θ)
The methyl cyanide, CH3CN, to methyl isocyanide, CH3NC, isomerization in
part (c) is characterized by an isomerization angle θ that we can take to be 0° in
the cyanide configuration and 180° in the isocyanide configuration. Since the
cyanide configuration is the more stable of the two, V(θ) must have an absolute
minimum at θ = 0°. But since the isocyanide is also stable, there must be a secondary minimum in V(θ) at 180° that is higher in energy than the absolute minimum. For all angles in between, the energy must rise smoothly to a maximum
and then fall again, leading to the following qualitative diagram:
0°
30°
60°
90°
θ
120°
150°
180°
As the O–H group is rotated around the C–O bond, the alcoholic H alternately
eclipses and staggers hydrogens on the methyl group. Thus, every 120°, there
295
is a maximum in V(θ) at each eclipse, and 60° past each maximum, half way
between successive maxima, we encounter a minimum representing each staggered conformation. All the maxima have the same height, all the minima have
the same height, and θ ranges from 0° to 360°. These are the characteristics of a
function we can write as V(θ) = V0 [1 + cos(3θ)] where the constant V0 measures the potential difference between the minima and maxima. This function is
shown below.
V(θ)
V0
0
0°
60°
120°
180°
θ
240°
300°
360°
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). Any
function with the right number of wiggles of uniform height could perhaps do
as well or better, but the cosine is the simplest function with a chance at being
correct, and experiments have shown that it can do a good job of reproducing
some 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 these
equations for the original variables: x = (x′ – y′)/ 2 and y = (x′ + y′)/ 2, and
substitute these expressions into the potential energy function V(x, y):
V(x, y) = k0(5x 2 + 5y 2 + 6xy)
5
5
6
= k0
x′ – y′ 2 + x′ + y′ 2 + x′ – y′ x′ + y′
2
2
2
2
2
= k0 (8x′ + 2y′ )
1
1
= (16k0) x′2 + (4k0) y′2 .
2
2
296
Not only does this variable change remove the cross term that precluded a separation of variables solution, the final form of V shows at a glance that the potential is simply a sum of harmonic oscillator potentials in the two independent
variables x′ and y′. Written this way, we can easily identify the force constants
kx′ 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′ v x′ + 1 + ¨ ωy′ v y′ + 1 , ω x′ = 4
2
2
k0
m and ωy′ = 2
k0
m .
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 the
same mass, is always half way between them. For states of zero orbital angular
momentum, these two particles move along a line. The only difference between
the Ps atom quantum mechanical problem and the ordinary H atom problem is
the 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 negative 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 proportional
to 1/µ (see Eq. (12.45)), the Ps Bohr radius is twice that of H. The energy
level expression for Ps is identical to that for H with the exception of the change
in the energy constant:
En(H) = –
2.179 × 10–18 J
n2
, En(Ps) = –
2.179 × 10–18 J
2n 2
=–
1.090 × 10–18 J
n2
.
For H, the longest wavelength a ground-state atom can absorb is that which
takes the atom from n = 1 to n = 2. It is 121.5 nm, the longest wavelength of
the Lyman series mentioned on page 351 and shown in Figure 11.1 in the text.
The same calculation holds for Ps:
λ=
=
hc
hc
=
∆E E2(Ps) – E1(Ps)
hc
–
1.090 × 10–18 J
22
– –
1.090 × 10–18 J
297
12
= 243 nm ,
exactly twice the H atom n = 2 → 1 wavelength. In a classical picture of the Ps
ground state, the positron–electron motion is along a line connecting both particles through the center of mass. The classical turning point of this motion for
Ps is 4a0, found by writing E1(Ps) = the total energy = V(r = classical turning
point) = –e2/(4πε0)r and solving for r. The time τ it takes the electron (say) to
go from its turning point, in to r = 0, out to the opposite turning point, then
back to r = 0 and on to the original turning point must be four times the time to
move from the first turning point to r = 0. (This time is the same for the positron to make its mirror-image motion.) It can be shown that this total time is
four times as long as the time for the electron in the Bohr model of H to make
one 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 make
about 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 20.1
on page 754 in the text. We can follow the logic shown there to derive the
density of states for a particle-on-a-ring, introduced in Example 12.6 on page
419 in the text. The energy expression is E = m2¨2/2MR2 for a particle of
mass M with quantum number m = 0, ±1, ±2, …. One difference between this
system and the particle-in-a-box is that every state here except m = 0 is doubly
degenerate. It introduces little error to derive the general expression assuming
all states are doubly degenerate, and thus we write the density of states ρ(E) as
ρ(E) = 2 dm = 2 dE
dm
dE
–1
=2
m¨
2 –1
MR 2
2
= 2 2MR
2
¨
1/2
¨
2 –1
MR 2
=R
¨
2M .
E
12.48 If V(r) = (constant)rn, the general form of the Virial Theorem becomes
2<T> = <r
∂V
> = <(constant)rnr n – 1> = n<V> .
∂r
Since E = <T> + <V>, we can write
E = <T> + <V> =
n
n+2
2
n+2
<V> + <V> =
<V> = <T> + <T> =
<T> .
2
2
n
n
If n = 2 (harmonic), we see that <T> = <V>. For n = –1 (Coulombic), <T> =
–<V>/2.
298
12.49 If we equate the total energy of a state of the H atom, En, to the effective potential energy (the sum of the centrifugal potential and the Coulombic potential),
we can write
2
2
e2
1 = l(l + 1)¨ – e 2 , a = (4πε0)¨ .
En = –
2(4πε0)a0 n 2
(4πε0)r 0
2µr 2
µe 2
With the definition suggested in the problem, ρ = r/a0, this can be written
2
2
e2
1 – l(l + 1)¨
–
0= e
(4πε0)r 2(4πε0)a0 n 2
2µr 2
2
2
e2
1 – l(l + 1)a0e
–
= e
(4πε0)r 2(4πε0)a0 n 2 2(4πε0)r 2
l(l + 1)a0 1
l(l + 1)
= 1r – 1 –
= – 1 –
ρ 2n 2
2r 2
2n 2
2ρ2
or 2n2ρ – ρ2 – l(l + 1)n2 = 0. A quadratic solution for ρ gives
ρ = n 2 ± n n 2 – 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 simply
means that r = 0 can be reached in s states. In units of a0, the outer classical
turning points (the values for ρ with the + sign chosen) are 32 for 4s, 30.967
for 4p, 28.649 for 4d, and 24 for 4f. These can be compared to the <r> values
for 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/meR 3 ,
we can write the force balance equation
Fout + Fin =
¨
2
m eR 3
–
e2
=0.
(4πε0)R 2
Solving this expression for R gives R = ¨2(4πε0)/mee2 = a0, the usual Bohr
radius. The Virial Theorem says E = –T = –¨2/2mea0 = E1, the exact H atom
ground-state energy. Thus, this simple model gives the same energy and size
parameter as the Bohr theory, which, in turn, gives the same energy and size as
the complete quantum mechanical theory.
299