Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula

CHAPTER 8

The quantum theory of motion
The three basic modes of motion—translation (motion
through space), vibration, and rotation—all play an important
role in chemistry because they are ways in which molecules
store energy. Gas-phase molecules, for instance, undergo translational motion and their kinetic energy is a contribution to
the total internal energy of a sample. Molecules can also store
energy as rotational kinetic energy and transitions between
their rotational energy states can be observed spectroscopically.
Energy is also stored as molecular vibration, and transitions
between vibrational states also give rise to spectroscopic signatures. In this Chapter we use the principles of quantum theory
to calculate the properties of microscopic particles in motion.

vibrations. We see that the energies of oscillator are quantized.
The acceptable wavefunctions also show that the oscillator may
be found at extensions and compressions that are forbidden by
classical physics.

8C  Rotational motion
The energy of a rotating particle is quantized, but in this Topic
we see that its angular momentum is also restricted to certain values. The quantization of angular momentum is a very
important aspect of the quantum theory of electrons in atoms
and of rotating molecules.

8A  Translation
In this Topic we see that, according to quantum theory, a particle constrained to move in a finite region of space is described
by only certain wavefunctions and their corresponding energies. Hence, quantization emerges as a natural consequence of
solving the Schrödinger equation and the conditions imposed
on it. The solutions also bring to light a number of non-classical
features of particles, especially their ability to tunnel into and

through regions where classical physics would forbid them to
be found.

8B  Vibrational motion
This Topic introduces the ‘harmonic oscillator’, a simple
but very important model for the description of molecular

What is the impact of this material?
‘Nanoscience’ is the study of atomic and molecular assemblies with dimensions ranging from 1 nm to about 100 nm
and ‘nanotechnology’ is concerned with the incorporation of
such assemblies into devices. We encounter several concepts
of nanoscience throughout the text. In Impact I8.1 we explore
quantum mechanical effects that render the properties of a
nanometre-sized assembly dependent on its size.
To read more about the impact of this
material, scan the QR code, or go to
bcs.whfreeman.com/webpub/chemistry/
pchem10e/impact/pchem-8-1.html


8A  Translation
Contents
8A.1 

Free motion in one dimension
Brief illustration 8A.1: The wavefunction
of a freely-moving particle

8A.2 

Confined motion in one dimension
The acceptable solutions
Brief illustration 8A.2: The energy of a particle
in a box
(b) The properties of the wavefunctions
Example 8A.1: Determining the probability
of finding the particle in a finite region
(c) The properties of observables
Example 8A.2: Estimating an absorption
wavelength
(a)

8A.3 

➤➤ What do you need to know already?
317
318
318
318
319
320
320
321
321

Confined motion in two or more dimensions 322
Separation of variables
Brief illustration 8A.3: The distribution of a
particle in a two-dimensional box
(b) Degeneracy

Brief illustration 8A.4: Degeneracies in a
two-dimensional box
(a)

8A.4 Tunnelling

Brief illustration 8A.5: Transmission probabilities
for a rectangular barrier
Checklist of concepts
Checklist of equations

You should know that the wavefunction is the solution of
the Schrödinger equation (Topic 7B), and be familiar with
the techniques of deriving dynamical properties from the
wavefunction by using operators corresponding to the
observables (Topic 7C).

322
323
324
324
324
326
327
328

In this Topic we present the essential features of the solutions of
the Schrödinger equation for translation, one of the basic types
of motion. We see that quantization emerges as a natural consequence of the Schrödinger equation and conditions imposed
on it. The solutions also bring to light a number of non-classical

features of particles, especially their ability to tunnel into and
through regions where classical physics would forbid them to
be found.

8A.1  Free

The Schrödinger equation for a particle of mass m moving
freely in one dimension is (Topic 7B)
−

➤➤ Why do you need to know this material?
The application of quantum theory to translation reveals
the origin of quantization and other non-classical features
of physical and chemical phenomena. This material is
important for the discussion of atoms and molecules that
are free to move within a restricted volume, such as a gas
in a container.

➤➤ What is the key idea?
The translational energy levels of a particle confined to
a finite region of space are quantized, and under certain
conditions particles can pass into and through classically
forbidden regions.

motion in one dimension

2 d 2ψ (x )
= Eψ (x )
2m dx 2

Free motion
in one
dimension 

Schrödinger

equation

(8A.1)

and the solutions are (as in eqn 7B.6)
ψ k = Aeikx + Be − ikx

Ek =

k 2 2
2m

Wave­
Free motion
functions
in one
and
dimension 
energies

(8A.2)

with A and B constants. Note that we are now labelling both

the wavefunctions and the energies with the index k. The wavefunctions in eqn 8A.2 are continuous, have continuous slope
everywhere, are single-valued, and do not go to infinity, and
so—in the absence of any other information—are acceptable
for all values of k. Because the energy of the particle is proportional to k2, all non-negative values, including zero, of the
energy are permitted. It follows that the translational energy of a
free particle is not quantized.


318  8  The quantum theory of motion

• If the particle is shot in the opposite direction,
towards negative x, then its linear momentum is −k
and its wavefunction is proportional to e−ikx. In this
case, A = 0 and B is the normalization factor.

∞

∞

Potential energy, V

• If it is shot towards positive x, then its linear
momentum is +k (Topic 7C), and its wavefunction
is proportional to eikx. In this case B = 0 and A is a
normalization factor.

Physical interpretation

The values of the constants A and B depend on how the state
of motion of the particle is prepared:

0

Location, x

L

moving particle

Figure 8A.1  A particle in a one-dimensional region with
impenetrable walls. Its potential energy is zero between x = 0
and x = L, and rises abruptly to infinity as soon as it touches the
walls.

An electron at rest that is shot out of an accelerator towards
positive x through a potential difference of 1.0 V acquires a
kinetic energy of 1.0 eV or 0.16 aJ (1.6 × 10 −19 J). The wavefunction for such a particle is given by eqn 8A.3 with B = 0 and
k given by rearranging the expression for the energy in eqn
8A.2 into

When the particle is between the walls, the Schrödinger equation is the same as for a free particle (eqn 8A.1), so the general
solutions given in eqn 8A.2 are also the same. However, it will
prove convenient to use e±ikx = cos kx ± i sin kx (Mathematical
background 3) to write

Brief illustration 8A.1  The wavefunction of a freely-

 2m E 
k =  e2 k 
  

1/2

 2 × (9.109 ×10−31 kg ) × (1.6 ×10−19 J) 
=

(1.055 ×10−34 Js)2


1/2

= 5.1 ×109 m −1
or 5.1 nm−1 (with 1 nm = 10−9 m). Therefore the wavefunction is
ψ(x) = Ae5.1ix/nm.
Self-test 8A.1  Write the wavefunction for an electron travel-

ling to the left (negative x) after being accelerated through a
potential difference of 10 kV.

Answer: ψ(x) = Be−510ix/nm

The probability density |ψ|2 is uniform if the particle is in
either of the pure momentum states eikx or e−ikx. According to
the Born interpretation (Topic 7B), nothing further can be said
about the location of the particle. That conclusion is consistent with the uncertainty principle, because if the momentum
is certain, then the position cannot be specified (the operators
corresponding to x and p do not commute and thus correspond
to complementary observables, Topic 8C).

8A.2  Confined

dimension

motion in one

Consider a particle in a box in which a particle of mass m is
confined to a finite region of space between two impenetrable walls. The potential energy is zero inside the box but rises
abruptly to infinity at the walls at x = 0 and x = L (Fig. 8A.1).

ψ k (x ) = Aeikx + Be − ikx = A(cos kx + i sin kx ) + B(cos kx − i sin kx )
= ( A + B) cos kx + ( A − B)i sin kx
If we write C = (A − B)i and D = A + B the general solutions take
the form
ψ k (x ) = C sin kx + D cos kx

General solution for 0 ≤ x ≤ L  (8A.3)

Outside the box the wavefunctions must be zero as the particle will not be found in a region where its potential energy is
infinite:
For x < 0 and x > L, ψ k (x ) = 0



(8A.4)

At this point, there are no restrictions on the value of k and all
solutions appear to be acceptable.

(a)  The acceptable solutions
The requirement of the continuity of the wavefunction (Topic

7B) implies that ψk(x) as given by eqn 8A.3 must be zero at the
walls, for it must match the wavefunction inside the material of
the walls where the functions meet. That is, the wavefunction
must satisfy the following two boundary conditions, or constraints on the function at certain locations:
ψ k (0) = 0 and ψ k (L) = 0



Particle in a onedimensional box 

Boundary
conditions

(8A.5)

As we show in the following Justification, the requirement
that the wavefunction satisfy these boundary conditions
implies that only certain wavefunctions are acceptable and


319

8A  Translation  
that the only permitted wavefunctions and energies of the
particle are

En =

nh
8mL2

2 2

n = 1, 2,

n = 1, 2,

1/2

(8A.6a)



(8A.6b)



where C is an as yet undetermined constant. Note that the
wavefunctions and energy are now labelled with the dimensionless integer n instead of the quantity k.
Justification 8A.1  The energy levels and wavefunctions

of a particle in a one-dimensional box
From the boundary condition ψk(0) = 0 and the fact that, from
eqn 8A.3, ψ k(0) = D (because sin 0 = 0 and cos 0 = 1), we can
conclude that D = 0. It follows that the wavefunction must be
of the form ψk(x) = C sin kx. From the second boundary condition, ψk(L) = 0, we know that ψk(L) = C sin kL = 0. We could
take C = 0, but doing so would give ψ k(x) = 0 for all x, which
would conflict with the Born interpretation (the particle must
be somewhere). The alternative is to require that kL be chosen
so that sin kL = 0. This condition is satisfied if
kL = nπ

n = 1, 2,…

The value n = 0 is ruled out, because it implies k = 0 and
ψ k(x) = 0 everywhere (because sin 0 = 0), which is unacceptable. Negative values of n merely change the sign of sin kL
(because sin(−x) = −sin x) and do not result in new solutions.
The wavefunctions are therefore
ψ n (x ) = C sin (nπx / L)

n =1, 2,…

as in eqn 8A.6a. At this stage we have begun to label the solutions with the index n instead of k. Because k and Ek are related
by eqn 8A.2, and k and n are related by kL = nπ, it follows that
the energy of the particle is limited to En = n2h2/8mL2, as in eqn
8A.6b.

We conclude that the energy of the particle in a one-dimensional box is quantized and that this quantization arises from
the boundary conditions that ψ must satisfy. This is a general
conclusion: the need to satisfy boundary conditions implies that
only certain wavefunctions are acceptable, and hence restricts
observables to discrete values. So far, only energy has been quantized; shortly we shall see that other physical observables may
also be quantized.
We need to determine the constant C in eqn 8A.6a. To do
so, we normalize the wavefunction to 1 by using a standard
integral from the Resource section. Because the wavefunction is
zero outside the range 0 ≤ x ≤ L, we use
L

∫ ψ dx = C ∫
0

2

2

L

0

 nπx 
sin 
dx
 L 
2

Integral T.2


=

L
 2
C × = 1, so C =  
2
 L
2

1/2

 2

 nπx 
for 0 ≤ x ≤ L
ψ n (x ) =   sin 
L
 
 L 
for x < 0 and x > L
ψ n (x )= 0
En =

n2 h 2
8mL2

n = 1, 2,…





Onedimen­ Wave­
sional functions (8A.7a)
box

One-dimensional
box 

Energy

levels

(8A.7b)

where the energies and wavefunctions are labelled with the
quantum number n. A quantum number is an integer (in some
cases, as we see in Topic 9B, a half-integer) that labels the state
of the system. For a particle in a one-dimensional box there is
an infinite number of acceptable solutions, and the quantum
number n specifies the one of interest (Fig. 8A.2). As well as
acting as a label, a quantum number can often be used to calculate the energy corresponding to the state and to write down
the wavefunction explicitly (in the present example, by using
the relations in eqn 8A.7).
100
Energy, E/E1; E1 = h2/8mL2

 nπx 
ψ n ( x ) = C sin 
 L 

for all n. Therefore, the complete solution for the particle in a
box is

0

n
10
Classically allowed energies

81

9

64

8

49

7

36

6

25
16
9
41

5
4
3
2
1

Figure 8A.2  The allowed energy levels for a particle in a
box. Note that the energy levels increase as n2, and that their
separation increases as the quantum number increases.
Classically, the particle is allowed to have any value of the
energy in the continuum shown as a shaded area.
Brief illustration 8A.2  The energy of a particle in a box

A long carbon nanotube can be modelled as a one-dimensional
structure and its electrons described by particle-in-a-box
wavefunctions. The lowest energy of an electron in a carbon
nanotube of length 100 nm is given by eqn 8A.7b with n = 1:
2

kg m2 s−2 


(1)2 ×  6.626 ×10−34 J s




E1 =
= 6.02 × 10−24 J
8 × (9.109 × 10−31 kg ) × (100 ×10−9 m)2

or 0.00602 zJ and its wavefunction is ψ1(x) = (2/L)1/2sin(πx/L).
Self-test 8A.2  What are the energy and wavefunction for the

next higher energy electron of the system described in this
Brief illustration?

Answer: E2 = 0.0241zJ, ψ 2 (x) = (2/L)1/2 sin(2πx/L)


(b)  The properties of the wavefunctions
Figure 8A.3 shows some of the wavefunctions of a particle in a

one-dimensional box. We see that:

• Shortening the wavelength results in a sharper
average curvature of the wavefunction and therefore
an increase in the kinetic energy of the particle (its
only source of energy because V = 0 inside the box).
• The number of nodes also increases as n increase;
the wavefunction ψn has n − 1 nodes.
• Increasing the number of nodes between walls of a
given separation increases the average curvature of
the wavefunction and hence the kinetic energy of the
particle.

Physical interpretation

• The wavefunctions are all sine functions with the
same amplitude but different wavelengths.

• The probability density for a particle in a onedimensional box is

Wavefunction, ψ

54 3 2



(8A.8)

• and varies with position. The non-uniformity in the
probability density is pronounced when n is small

(Fig. 8A.4). The most probable locations of the particle
correspond to the maxima in the probability density.
Example 8A.1  Determining the probability of finding

the particle in a finite region
The wavefunctions of an electron in a conjugated polyene can
be approximated by particle-in-a-box wavefunctions. What is
the probability, P, of locating the electron between x = 0 (the
left-hand end of a molecule) and x = 0.2 nm in its lowest energy
state in a conjugated molecule of length 1.0 nm?
Method  According to the Born interpretation, ψ (x)2 dx is

the probability of finding the particle in the small region dx
located at x; therefore, the total probability of finding the electron in the specified region is the integral of ψ(x)2dx over that
region, as given in eqn 7B.11. The wavefunction of the electron
is given in eqn 8A.7a with n = 1. The integral you need is in the
Resource section:

1

Answer  The probability of finding the particle in a region
between x = 0 and x = l is
x

0

L

∫

l

P = ψ n2dx =
0

Figure 8A.3  The first five normalized wavefunctions of a
particle in a box. Each wavefunction is a standing wave;
successive functions possess one more half wave and a
correspondingly shorter wavelength.

2
L

l

2

∫ sin 
0

l
1
nπx 
 2πnl 
dx = −
sin
L 2nπ  L 
L 

Now set n = 1, L = 1.0 nm, and l = 0.2 nm, which gives P = 0.05.

The result corresponds to a chance of 1 in 20 of finding the
electron in the region. As n becomes infinite, the sine term,
which is multiplied by 1/n, makes no contribution to P and the
classical result for a uniformly distributed particle, P = l/L, is
obtained.
Self-test 8A.3  Calculate the probability that an electron in the

state with n = 1 will be found between x = 0.25L and x = 0.75L in
a conjugated molecule of length L (with x = 0 at the left-hand
end of the molecule).

n=1
(a)

2
 nπx 
ψ n2 (x)= sin2 
L
 L 

Physical interpretation

320  8  The quantum theory of motion

Answer: P = 0.82

n=2
n=2
n=1

(b)
n=2
n=1
(c)

Figure 8A.4  (a) The first two wavefunctions, (b) the
corresponding probability densities, and (c) a representation of
the probability density in terms of the darkness of shading.

The probability density ψ n2 (x ) becomes more uniform as n
increases provided we ignore the fine detail of the increasingly
rapid oscillations (Fig. 8A.5). The probability density at high
quantum numbers reflects the classical result that a particle
bouncing between the walls spends, on the average, equal times
at all points. That the quantum result corresponds to the classical prediction at high quantum numbers is an illustration of the
correspondence principle, which states that classical mechanics emerges from quantum mechanics as high quantum numbers are reached.


implies that 〈p2〉 ≠ 0, which implies that the particle
must always have nonzero kinetic energy.
|ψ (x)|2

• If the wavefunction is to be zero at the walls, but
smooth, continuous, and not zero everywhere, then
it must be curved, and curvature in a wavefunction
implies the possession of kinetic energy.

Physical interpretation

321

8A  Translation  

The separation between adjacent energy levels with quantum
numbers n and n + 1 is
0

1

x/L

ψ 2(x)

Figure 8A.5  The probability density
for large quantum
number (here n = 50, blue, compared with n = 1, red). Notice
that for high n the probability density is nearly uniform,
provided we ignore the fine detail of the increasingly rapid
oscillations.

(c)  The properties of observables
The linear momentum of a particle in a box is not well defined
because the wavefunction sin kx is not an eigenfunction of the
linear momentum operator. However, each wavefunction is a
linear combination of the linear momentum eigenfunctions eikx
and e−ikx. Then, because sin x = (eix − e−ix)/2i, we can write
1/2

1/2

 2
 nπx  1  2 
ψ n ( x ) =   sin 
(eikx − e − ikx )
=
L
 
 L  2i  L 

k=

nπ
L



(8A.9)

It follows from the discussion in Topic 7C that half the measurements of the linear momentum will give the value +k︀ and
−k for the other half. This detection of opposite directions
of travel with equal probability is the quantum mechanical
version of the classical picture that a particle in a one-dimensional box rattles from wall to wall and in any given period
spends half its time travelling to the left and half travelling to
the right.
Because n cannot be zero, the lowest energy that the particle may possess is not zero (as would be allowed by classical
mechanics, corresponding to a stationary particle) but
E1 =

h2
8mL2

Particle in a box 

Zero-point energy  (8A.10)

• The Heisenberg uncertainty principle requires a
particle to possess kinetic energy if it is confined to a
finite region: the location of the particle is not
completely indefinite (Δx ≠ ∞), so the uncertainty in
its momentum cannot be precisely zero (Δp ≠ 0).
Because Δp = (〈p2〉 − 〈p〉2)1/2 = 〈p2〉1/2 in this case, Δp ≠ 0

Physical interpretation

This lowest, irremovable energy is called the zero-point energy.
The physical origin of the zero-point energy can be explained
in two ways:

En+1 − En =

(n + 1)2 h2 n2 h2
h2
−
= (2n + 1)
2
2
8mL
8mL
8mL2

(8A.11)

This separation decreases as the length of the container
increases, and is very small when the container has macroscopic dimensions. The separation of adjacent levels becomes
zero when the walls are infinitely far apart. Atoms and molecules free to move in normal laboratory-sized vessels may
therefore be treated as though their translational energy is not
quantized.
Example 8A.2  Estimating an absorption wavelength

β-Carotene (1) is a linear polyene in which 10 single and 11
double bonds alternate along a chain of 22 carbon atoms. If
we take each Ce C bond length to be about 140 pm, then the
length L of the molecular box in β-carotene is L = 2.94 nm.
Estimate the wavelength of the light absorbed by this molecule
from its ground state to the next higher excited state.

1 β-Carotene

Method  For reasons that will be familiar from introductory
chemistry, each C atom contributes one p electron to the
π-orbitals. Use eqn 8A.11 to calculate the energy separation
between the highest occupied and the lowest unoccupied
­levels, and convert that energy to a wavelength by using the
Bohr frequency relation (eqn 7A.12).
Answer  There are 22 C atoms in the conjugated chain; each

contributes one p electron to the levels, so each level up to
n = 11 is occupied by two electrons. The separation in energy
between the ground state and the state in which one electron is
promoted from n = 11 to n = 12 is

∆E = E12 − E11
= (2 ×11 + 1)

(6.626 ×10−34 Js)2
8 × (9.109 ×10−31 kg ) × (2.94 ×10−9 m)2

= 1.60 × 10−19 J


322  8  The quantum theory of motion
or 0.160 aJ. It follows from the Bohr frequency condition
(ΔE = hν) that the frequency of radiation required to cause this
transition is
=

1.60 × 10−19 J
∆E
=
= 2.41 × 1014 s −1
h 6.626 × 10−34 Js

or 241 THz (1 THz = 1012 Hz), corresponding to a wavelength
λ = 1240 nm. The experimental value is 603 THz (λ = 497 nm),
corresponding to radiation in the visible range of the electromagnetic spectrum. Considering the crudeness of the model
we have adopted here, we should be encouraged that the computed and observed frequencies agree to within a factor of 2.5.
Self-test 8A.4  Estimate a typical nuclear excitation energy
in electronvolts (1 eV = 1.602 × 10 −19 J; 1 GeV = 109 eV) by calculating the first excitation energy of a proton confined to a
one-dimensional box with a length equal to the diameter of a
nucleus (approximately 1 × 10−15 m, or 1 fm).
Answer: 0.6 GeV

8A.3  Confined

dimensions

motion in two or more

Now consider a rectangular two-dimensional region of a surface with length L1 in the x-direction and L2 in the y-direction;
the potential energy is zero everywhere except at the walls,
where it is infinite (Fig. 8A.6). As a result, the particle is never
found at the walls and its wavefunction is zero there and everywhere outside the two-dimensional region. Between the walls,
because the particle has contributions to its kinetic energy from
its motion in both the x and y directions, the Schrödinger equation has two kinetic energy terms, one for each axis. For a particle of mass m the equation is

∞

V

∞

Particle
confined
to surface

0

∞
∞

−

2  ∂2ψ ∂2ψ 
+
= Eψ
2m  ∂x 2 ∂y 2 

(8A.12)



This is a partial differential equation (Mathematical background
4), and the resulting wavefunctions are functions of both x and
y, denoted ψ(x,y). This dependence means that the wavefunction and the corresponding probability density depend on the
location in the plane, with each position specified by the values
of the coordinates x and y.

(a)  Separation of variables
A partial differential equation of the form of eqn 8A.12
can be simplified by the separation of variables technique
(Mathematical background 4), which divides the equation into
two or more ordinary differential equations, one for each variable. We show in the Justification below using this technique
that the wavefunction can be written as a product of functions,
one depending only on x and the other only on y:
ψ (x , y ) = X (x )Y ( y )
and that the total energy is given by
E = E X + EY

Justification 8A.2  The separation of variables technique

applied to the particle in a two-dimensional box

We follow the procedure in Mathematical background 4
and apply it to eqn 8A.12. The first step to confirm that the
Schrödinger equation can be separated and the wavefunction
can be factored into the product of two functions X and Y is to
note that, because X is independent of y and Y is independent
of x, we can write

x

Figure 8A.6  A two-dimensional square well. The particle is
confined to the plane bounded by impenetrable walls. As soon
as it touches the walls, its potential energy rises to infinity.

d 2Y
∂2ψ ∂2 XY
=
=X 2
2
2
dy
∂y
∂y

Note the replacement of the partial derivatives by ordinary
derivatives in each case. Then eqn 8A.12 becomes

y

L1

(8A.13b)

where EX is the energy associated with the motion of the particle parallel to the x-axis, and likewise for EY and motion parallel
to the y-axis.

d2 X
∂2ψ ∂2 XY
=
=Y 2
2
2
dx
∂x
∂x
L2

(8A.13a)

−

d 2Y 
2  d 2 X
Y
X
+
= EXY
2
2m  dx
dy 2 

Next, we divide both sides by XY, and rearrange the resulting
equation into
1 d 2 X 1 d 2Y
2mE
+
=− 2
X dx 2 Y dy 2



8A  Translation  
The first term on the left, (1/X)(d2X/dx2), is independent of y,
so if y is varied only the second term on the left, (1/Y)(d2Y/dy2),
can change. But the sum of these two terms is a constant,
2mE/2, given by the right-hand side of the equation. Therefore,
if the second term did change, then the right-hand side could
not be constant. Consequently, even the second term cannot
change when y is changed. In other words, the second term,
(1/Y)(d2Y/dy2), is a constant, which we write −2mEY/2. By a
similar argument, the first term, (1/X)(d2X/dx2), is a constant
when x changes, and we write it −2mEX/2, with E = EX + EY.
Therefore, we can write
1 d2 X
2mE
=− 2 X
X dx 2


+

–

–

–

+

+

+

–

+

(a)

1 d 2Y
2mE
=− 2 Y
Y dy 2


323

(b)

These expressions rearrange into the two ordinary (that is,
single-variable) differential equations

2 d 2 X
−
= EX X
2m dx 2

2 d 2Y
−
= EY Y
2m dy 2

(8A.14)



(c)

Each of the two ordinary differential equations in eqn
8A.14 is the same as the one-dimensional particle-in-a-box
Schrödinger equation (Section 8A.2). The boundary conditions are also the same, apart from the detail of requiring
X(x) to be zero at x = 0 and L1, and Y(y) to be zero at y = 0
and L2. We can therefore adapt eqn 8A.7a without further
calculation:
1/2

 2
 n πx 
Xn (x ) =   sin  1  for 0 ≤ x ≤ L1
 L1 
 L1 
1

1/2

 2
 n πy 
Yn ( y ) =   sin  2  for 0 ≤ y ≤ L2
 L2 
 L2 

(d)

Figure 8A.7  The wavefunctions for a particle confined to a
rectangular surface depicted as contours of equal amplitude.
(a) n1 = 1, n2 = 1, the state of lowest energy; (b) n1 = 1, n2 = 2;
(c) n1 = 2, n2 = 1; (d) n1 = 2, n2 = 2.

Some of the wavefunctions are plotted as contours in Fig.
8A.7. They are the two-dimensional versions of the wavefunctions shown in Fig. 8A.3. Whereas in one dimension the wavefunctions resemble states of a vibrating string with ends fixed,
in two dimensions the wavefunctions correspond to vibrations
of a rectangular plate with fixed edges.

2

Brief illustration 8A.3  The distribution of a particle in a

Then, because ψ = XY,

two-dimensional box

2

ψ n ,n (x , y ) =
×
(L1L2 )1/2
 n πx   n πy 
sin  1  sin  2 
 L1   L2 
for 0 ≤ x ≤ L1 , 0 ≤ y ≤ L2
ψ n ,n (x , y ) = 0
outside box
1

1

2

2

Twodimensional
box

Wave­
functions

(8A.15a)

ψ 12,2 (x , y ) =



Similarly, because E = EX + EY, the energy of the particle is limited to the values

 n2 n2  h 2
En ,n =  12 + 22 
 L1 L2  8m
1

2



Consider an electron confined to a square cavity of side L, and
in the state with quantum numbers n1 = 1, n 2 = 2. Because the
probability density is

Two-dimensional Energy
levels
box 

(8A.15b)

with the two quantum numbers taking the values n1 = 1, 2, …
and n2 = 1, 2, … independently. The state of lowest energy is
(n1 = 1, n2 = 1) and E1,1 is the zero-point energy.

4
 πx 
 2πy 
sin2   sin2 
L2
 L
 L 

the most probable locations correspond to sin 2(πx/L) = 1 and
sin 2(2πx/L) = 1, or (x,y) = (L/2, L/4) and (L/2, 3L/4). The least
probable locations (the nodes, where the wavefunction passes
through zero) correspond to zeroes in the probability density
within the box, which occur along the line y = L/2.
Self-test 8A.5  Determine the most probable locations of an

electron in a square cavity of side L when it is in the state with
quantum numbers n1 = 2, n2 = 3.

Answer: points (x = L/4 and 3L/4; y = L/6, L/2, and 5L/6)


324  8  The quantum theory of motion
We treat a particle in a three-dimensional box in the same
way. The wavefunctions have another factor (for the z-dependence), and the energy has an additional term in n32 / L23 . Solution
of the Schrödinger equation by the separation of variables technique then gives

–

+

1/2

 8 
 n πx   n πy   n πz 
ψ n ,n ,n (x , y , z ) = 
sin  1  sin  2  sin  3 
 L1L2 L3 

 L1   L2   L3 
for 0 ≤ x ≤ L1 , 0 ≤ y ≤ L2 , 0 ≤ z ≤ L3
1

2

Three-dimensional box 

n n n  h
En ,n ,n =  + + 
 L L L  8m
2

3

2
1
2
1

2
3
2
3

2
2
2
2

2



Wavefunctions  (8A.16a)

Threedimensional
box 

Energy
levels

(8A.16b)

The quantum numbers n1, n2, and n3 are all positive integers 1, 2, … that can be varied independently. The system has a
zero-point energy (E1,1,1 = 3h2/8mL2 for a cubic box).

(b)  Degeneracy
A special feature of the solutions arises when a two-dimensional box is not merely rectangular but square, with L1 = L2 = L.
Then the wavefunctions and their energies are
2  n πx   n πy 
ψ n ,n (x , y ) = sin  1  sin  2 
L  L   L 
for 0 ≤ x ≤ L, 0 ≤ y ≤ L
outside box
ψ n ,n (x , y ) = 0
1

1

2

En ,n = (n12 + n22 )
2

h2
8mL2

(a)

(b)

Figure 8A.8  The wavefunctions for a particle confined to a
square well. Note that one wavefunction can be converted
into the other by rotation of the box by 90°. The two functions
correspond to the same energy. True degeneracy is a
consequence of symmetry.

ψ2,1 are then not degenerate. Similar arguments account for the
degeneracy of states in a cubic box. Other examples of degener­
acy occur in quantum mechanical systems (for instance, in the
hydrogen atom, Topic 9A), and all of them can be traced to the
symmetry properties of the system.
Brief illustration 8A.4  Degeneracies in a two-

dimensional box

2

1

+

3



1

–

Square Wave­
box 
functions

(8.17a)



The energy of a particle in a two-dimensional square box of
side L in the state with n1 = 1, n2 = 7 is
E1,7 = (12 + 72 )

Square box 

Energy levels  (8.17b)

Consider the cases n1 = 1, n2 = 2 and n1 = 2, n2 = 1:
2  πx   2πy 
ψ 1,2 = sin   sin 

L  L   L 

E1,2 =

5h2
8mL2

2  2πx   πy 
ψ 2,1 = sin 
sin
L  L   L 

E2,1 =

5h2
8mL2

Although the wavefunctions are different, they have the same
energy. The technical term for different wavefunctions corres­
ponding to the same energy is degeneracy, and in this case we
say that the state with energy 5h2/8mL2 is ‘doubly degenerate’.
In general, if N wavefunctions correspond to the same energy,
then we say that the state is ‘N-fold degenerate’.
The occurrence of degeneracy is related to the symmetry
of the system. Figure 8A.8 shows contour diagrams of the two
degenerate functions ψ1,2 and ψ2,1. Because the box is square,
one wavefunction can be converted into the other simply by
rotating the plane by 90°. Interconversion by rotation through
90° is not possible when the plane is not square, and ψ1,2 and

h2
50h2
=
2
8mL 8mL2

This state is degenerate with the state with n1 = 7 and n 2 = 1.
Thus, at first sight the energy level 50h2/8mL2 is doubly degener­
ate. However, in certain systems there may be states that are
not apparently related by symmetry but are ‘accidentally’
degenerate. Such is the case here, for the state with n1 = 5 and
n 2 = 5 also has energy 50h 2/8mL 2 . Accidental degeneracy is
also encountered in the hydrogen atom (Topic 9A).
Self-test 8A.6  Find a state (n1, n2) for a particle in a rectangular

box with sides of length L1 = L and L2 = 2L that is accidentally
degenerate with the state (4,4).

Answer: (n1 = 2, n2 = 8)

8A.4  Tunnelling
If the potential energy of a particle does not rise to infinity
when it is in the wall of the container, and E < V, the wavefunction does not decay abruptly to zero. If the walls are thin (so
that the potential energy falls to zero again after a finite distance), then the wavefunction oscillates inside the box, varies


8A  Translation  

Incident
wave

Wavefunction, ψ

E

Figure 8A.9  A particle incident on a barrier from the left
has an oscillating wave function, but inside the barrier there
are no oscillations (for E < V). If the barrier is not too thick,
the wavefunction is nonzero at its opposite face, and so
oscillates begin again there. (Only the real component of the
wavefunction is shown.)

smoothly inside the region representing the wall, and oscillates
again on the other side of the wall outside the box (Fig. 8A.9).
Hence the particle might be found on the outside of a container
even though according to classical mechanics it has insufficient
energy to escape. Such leakage by penetration through a classically forbidden region is called tunnelling.
The Schrödinger equation is used to calculate the probability
of tunnelling of a particle of mass m incident from the left on
a rectangular potential energy barrier that extends from x = 0
to x = L. On the left of the barrier (x < 0) the wavefunctions are
those of a particle with V = 0, so from eqn 8A.2 we can write
k = (2mE )1/2



Particle in a
rectangular
barrier

Wave­

function
left of (8A.18)
barrier

The Schrödinger equation for the region representing the barrier (0 ≤ x ≤ L), where the potential energy is the constant V, is
−

2 d 2ψ (x )
+ Vψ (x ) = Eψ (x )
2m dx 2


(8A.19)

We shall consider particles that have E < V (so, according to
classical physics, the particle has insufficient energy to pass
through the barrier), and therefore for which V − E > 0. The
general solutions of this equation are
ψ = Ce + De
κx

−κx

κ  = {2m(V − E )}

1/2

Wave­
Particle in a function
rectangular inside (8A.20)

barrier
barrier

as can be verified by differentiating ψ twice with respect to x.
The important feature to note is that the two exponentials in
eqn 8A.20 are now real functions, as distinct from the complex,
oscillating functions for the region where V = 0. To the right of
the barrier (x > L), where V = 0 again, the wavefunctions are
ψ = A′eikx

k = (2mE )1/2



Particle in a
rectangular
barrier

Wave­
function
right of (8A.21)
barrier

Transmitted
wave
Reflected
wave
x

Figure 8A.10  When a particle is incident on a barrier from the

left, the wavefunction consists of a wave representing linear
momentum to the right, a reflected component representing
momentum to the left, a varying but not oscillating component
inside the barrier, and a (weak) wave representing motion to
the right on the far side of the barrier.

Note that to the right of the barrier, the particle can only be
moving to the right and therefore terms of the form e−ikx do not
contribute to the wavefunction in eqn 8A.21.
The complete wavefunction for a particle incident from the
left consists of (Fig. 8A.10):
• an incident wave (Aeikx corresponds to positive
momentum);
• a wave reflected from the barrier (Be−ikx corresponds
to negative momentum, motion to the left);
• the exponentially changing amplitudes inside the
barrier (eqn 8A.20);
• an oscillating wave (eqn 8A.21) representing the
propagation of the particle to the right after
tunnelling through the barrier successfully.

Physical interpretation

Wavefunction, ψ

V

ψ = Aeikx + Be − ikx

325

The probability that a particle is travelling towards positive x
(to the right) on the left of the barrier (x < 0) is proportional to
|A|2, and the probability that it is travelling to the right on the
right of the barrier (x > L) is |A′|2. The ratio of these two probabilities, |A′|2/|A|2, which reflects the probability of the particle
tunnelling through the barrier, is called the transmission probability, T.
To determine the relationship between |A′|2 and |A|2, we
need to investigate the relationships between the coefficients
A, B, C, D, and A′. Since the acceptable wavefunctions must
be continuous at the edges of the barrier (at x = 0 and x = L,
remembering that e0 = 1):
A+ B =C + D

CeκL + De −κL = A′ eikL

(8A.22a)

Their slopes (their first derivatives) must also be continuous
there (Fig. 8A.11):
ikA − ikB = κ C −κD

κ CeκL −κDe −κL = ikA′ eikL

(8A.22b)


• T approaches, but is still less than, 1 for E > V: there
is still a probability of the particle being reflected by
the barrier even when classically it can pass over it;

Wavefunction, ψ

V

• T ≈ 1 for E ≫ V, as expected classically.

For high, wide barriers (in the sense that κL ≫ 1), eqn 8A.23a
simplifies to
T ≈ 16ε (1− ε )e −2κL

Figure 8A.11  The wavefunction and its slope must be
continuous at the edges of the barrier. The conditions for
continuity enable us to connect the wavefunctions in the three
zones and hence to obtain relations between the coefficients
that appear in the solutions of the Schrödinger equation.

After straightforward but lengthy algebraic manipulations of
the above set of equations 8A.22 (see Problem 8A.6), we find
−1

Rectangular
potential
barrier 

 (eκ L − e −κ L )2 
T = 1+

16ε (1− ε ) 


Transmission
probability

(8A.23a)

• T ≈ 0 for E ≪ V;
• T increases as E approaches V: the probability of
tunnelling increases;
0.5

1

0.4

0.8

0.3

0.6

Physical
interpretation

where ε = E/V. This function is plotted in Fig. 8A.12. The transmission probability for E > V is shown there too. The transmission probability has the following properties:

Rectangular
potential
barrier; κL ≫1

Transmission
probability

(8A.23b)

The transmission probability decreases exponentially with the
thickness of the barrier and with m1/2. It follows that particles of
low mass are more able to tunnel through barriers than heavy
ones (Fig. 8A.13). Tunnelling is very important for electrons
and muons (mμ ≈ 207 me), and moderately important for protons (mp ≈ 1840me); for heavier particles it is less important.
A number of effects in chemistry (for example, the isotopedependence of some reaction rates) depend on the ability of
the proton to tunnel more readily than the deuteron. The very
rapid equilibration of proton transfer reactions is also a manifestation of the ability of protons to tunnel through barriers and
transfer quickly from an acid to a base. Tunnelling of protons
between acidic and basic groups is also an important feature of
the mechanism of some enzyme-catalysed reactions.

Heavy particle

Wavefunction, ψ

x

Transmission probability, T

Physical
interpretation

326  8  The quantum theory of motion

Light particle

2
x

2
0.4

0.2
0.1
0

0.2

4
10

0

Figure 8A.13  The wavefunction of a heavy particle decays
more rapidly inside a barrier than that of a light particle.
Consequently, a light particle has a greater probability of
tunnelling through the barrier.

10

0.2

0.4

0.6

0.8
Incident energy, E/V

1

0

1

2

3

Incident energy, E/V

Figure 8A.12  The transmission probabilities for passage
through a rectangular potential barrier. The horizontal axis
is the energy of the incident particle expressed as a multiple
of the barrier height. The curves are labelled with the value
of L(2mV)1/2/. The graph on the left is for E < V and that on
the right for E > V. Note that T > 0 for E < V whereas classically
T would be zero. However, T < 1 for E > V, whereas classically T
would be 1.

4

Brief illustration 8A.5  Transmission probabilities for a

rectangular barrier
Suppose that a proton of an acidic hydrogen atom is confined to an acid that can be represented by a barrier of height
2.000 eV and length 100 pm. The probability that a proton
with energy 1.995 eV (corresponding to 0.3195 aJ) can escape
from the acid is computed using 8A.23a, with ε = E/V = 1.995 
eV/2.000 eV = 0.9975 and V − E = 0.005 eV (corresponding to
8.0 × 10−22 J).


8A  Translation  

{2 × (1.67 × 10−27 kg ) × (8.0 × 10−22 J)}1/2
= 1.55…× 1010 m −1
1.055 × 10−34 Js

Potential energy, V

κ=

We have used 1 J = 1 kg m2 s −2. It follows that
κL = (1.55… × 1010 m −1 ) × (100 × 10−12 m) = 1.55….
Equation 8A.23a then yields
−1



(e1.55… − e −1.55… )2
−3
T = 1 +
 =1.96 ×10

×
×
−
.
(
.
)
16
0
9975
1
0
9975



327

0

Location, x

L

Figure 8A.14  A potential well with a finite depth.

L(2mV)1/2/

The larger the value of
(here, 31) the smaller is the

value of T for energies close to, but below, the barrier height.
Potential energy, V

Self-test 8A.7  Suppose that the junction between two semiconductors can be represented by a barrier of height 2.00 eV
and length 100 pm. Calculate the probability that an electron
of energy 1.95 eV can tunnel through the barrier.
Answer: T = 0.881

A problem related to tunnelling is that of a particle in a
square-well potential of finite depth (Fig. 8A.14). In this kind
of potential, the wavefunction penetrates into the walls, where
it decays exponentially towards zero, and oscillates within the
well. The wavefunctions are found by ensuring, as in the discussion of tunnelling, that they and their slopes are continuous at
the edges of the potential. Some of the lowest energy solutions
are shown in Fig. 8A.15. A further difference from the solutions
for an infinitely deep well is that there is only a finite number
of bound states. Regardless of the depth and length of the well,
however, there is always at least one bound state. Detailed consideration of the Schrödinger equation for the problem shows
that in general the number of levels is equal to N, with

n=2

n=1
0

Location, x

L

Figure 8A.15  The lowest two bound-state wavefunctions for a

particle in the well shown in Fig. 8A.14.

N −1 <

(8mVL)1/2
h


(8A.24)

where V is the depth of the well and L is its length. We see that
the deeper and wider the well, the greater the number of bound
states. As the depth becomes infinite, so the number of bound
states also becomes infinite, as we have already seen.

Checklist of concepts
☐1.The translational energy of a free particle is not
quantized.
☐2.The need to satisfy boundary conditions implies that
only certain wavefunctions are acceptable and therefore restricts observables to discrete values.
☐3.A quantum number is an integer (in certain cases, a
half-integer) that labels the state of the system.
☐4.A particle in a box possesses a zero-point energy, an
irremovable minimum energy.
☐5.The correspondence principle states that classical
mechanics emerges from quantum mechanics as high
quantum numbers are reached.

☐6.The wavefunction for a particle in a two- or threedimensional box is the product of wavefunctions for the

particle in a one-dimensional box.
☐7.The energy of a particle in a two- or three-dimensional
box is the sum of energies for the particle in two or
three one-dimensional boxes.
☐8.The zero-point energy for a particle in a two-dimensional box corresponds to the state with quantum numbers (n1 = 1, n 2 = 1); for three dimensions, (n1 = 1, n 2 = 1,
n3 = 1).
☐9.Degeneracy occurs when different wavefunctions correspond to the same energy.


328  8  The quantum theory of motion
☐10. The occurrence of degeneracy is a consequence of the
symmetry of the system.
☐11. Penetration into or through a classically forbidden
region is called tunnelling.

☐12. The probability of tunnelling decreases with an increase
in the height and width of the potential barrier.
☐13. Light particles are more able to tunnel through barriers
than heavy ones.

Checklist of equations
Property

Equation

Comment

Equation number

Free-particle wavefunctions and

energies

ψk = Aeikx + Be−ikx  Ek = k2ħ2/2m

All values of k allowed

8A.2

≤ x ≤ L
ψn(x) = (2/L)1/2 sin(nπx/L), 0 
ψn(x) = 0,  x < 0 and x > L

n = 1, 2, …

8A.7a

Particle in a box
One dimension:
 Wavefunctions
 Energies

En = n2h2/8mL2

8A.7b

  Zero-point energy

E1 = h2/8mL2

8A.10

Two dimensions:
 Wavefunctions

ψ n1 ,n2 (x , y ) = (2 /(L1L2 )1/2 )sin(n1πx /L1 )sin(n2 πy /L2 )

n1, n2 = 1, 2, …

8A.15a

  0 ≤ x ≤ L1, 0 ≤ y ≤ L2
ψ n1,n2 (x , y ) = 0
outside box
 Energies

En1 ,n2 = (n12 /L21 + n22 /L22 )h2 /8m

Three dimensions:
 Wavefunctions

ψ n1n2n3 (x , y , z ) = 8 /(L1L2 L3 )1/2 ×

8A.15b
n1, n2, n3 = 1, 2, …

8A.16a

sin (n1πx / L1 )sin (n2 πy / L2 )sin (n3 πz / L3 ),

  0 ≤ x ≤ L1, 0 ≤ y ≤ L2, 0 ≤ z ≤ L3

ψ n1,n2 ,n3 (x , y , z ) = 0

outside box

 Energies

En1,n2 ,n3 = (n12 /L21 + n22 /L22 + n32 /L23 )h2 /8m

Transmission probability

T = {1+ (eκ L − e −κ L )2 /16ε (1− ε )}−1

Rectangular potential barrier

8A.23a

T = 16ε(1 − ε)e−2κL

High, wide rectangular barrier

8A.23b

8A.16b


8B  Vibrational motion
Contents
8B.1 

The harmonic oscillator

The energy levels
Brief illustration 8B.1: The vibration of a
diatomic molecule
(b) The wavefunctions
Example 8B.1: Confirming that a wavefunction
is a solution of the Schrödinger equation
Example 8B.2: Normalizing a harmonic oscillator
wavefunction
(a)

8B.2 

The properties of oscillators
Mean values
Example 8B.3: Calculating properties of a
harmonic oscillator
(b) Tunnelling
Example 8B.4: Calculating the tunnelling
probability for the harmonic oscillator
(a)

Checklist of concepts
Checklist of equations

329
330
330
331
332
333

333
334
334
335
335
336
336

Atoms in molecules and solids vibrate around their mean positions as bonds stretch, compress, and bend. Here we consider
one particular type of vibrational motion, that of ‘harmonic
motion’ in one dimension.

8B.1  The

harmonic oscillator

A particle undergoes harmonic motion, and is said to be a
harmonic oscillator, if it experiences a restoring force proportional to its displacement:
F = − kf x

➤➤ What is the key idea?
The quantum mechanical treatment of the simplest model
of vibrational motion, the harmonic oscillator, reveals
that the energy is quantized and the wavefunctions are
products of a polynomial and a Gaussian (bell-shaped)
function.

➤➤ What do you need to know already?
You should know how to formulate the Schrödinger
equation given a potential energy function. You should

also be familiar with the concepts of tunnelling (Topic 8A)
and the expectation value of an observable (Topic 7B).



Parabolic potential energy  (8B.2)

when it is displaced through a distance x from its equilibrium
position. This expression, which is the equation of a parabola (Fig. 8B.1), is the origin of the term ‘parabolic potential
energy’ for the potential energy characteristic of a harmonic
oscillator. The Schrödinger equation for the particle of mass m
is therefore

∞

∞
Potential energy, V

The detection and interpretation of vibrational frequencies
is the basis of infrared spectroscopy (Topics 12D and 12E).
Molecular vibration plays a role in the interpretation
of thermodynamic properties, such as heat capacities
(Topics 5E and 15F), and of the rates of chemical reactions
(Topic 21C).

Restoring force  (8B.1)

where kf is the force constant: the stiffer the ‘spring’, the greater
the value of kf. Because force is related to potential energy by
F = −dV/dx (see Foundations B), the force in eqn 8B.1 corresponds to the particle having a potential energy

V ( x ) = 12 kf x 2

➤➤ Why do you need to know this material?

Harmonic motion 

0

Displacement, x

Figure 8B.1  The parabolic potential energy V = 21 k f x 2 of
a harmonic oscillator, where x is the displacement from
equilibrium. The narrowness of the curve depends on the force
constant k f: the larger the value of k f, the narrower the well.


330  8  The quantum theory of motion

−

2 d 2ψ (x ) 1
+ 2 kf x 2ψ (x ) = Eψ (x )
2m dx 2


Harmonic Schrödinger
(8B.3)
oscillator equation

We can anticipate that the energy of an oscillator will be quantized because the wavefunction has to satisfy boundary conditions (as in Topic 8A for a particle in a box): it will not be

found with very large extensions because its potential energy
rises to infinity there. That is, when we impose the boundary conditions ψ = 0 at x = ±∞, we can expect to find that only
certain wavefunctions and their corresponding energies are
possible.

(a)  The energy levels
Equation 8B.3 is a standard equation in the theory of differential equations and its solutions are well known to mathematicians.1 The permitted energy levels are
Ev = (v + 12 ) ω

ω = (kf /m)1/2

v = 0,1, 2,…



Harmonic
oscillator

Energy
levels

(8B.4)

where ω (omega) is the frequency of oscillation of a classical
harmonic oscillator of the same mass and force constant. Note
that ω is large when the force constant is large and the mass
small. It follows that the separation between adjacent levels is
Ev +1 − Ev = ω

(8B.5)

which is the same for all v. Therefore, the energy levels form
a uniform ladder of spacing ω (Fig. 8B.2). The energy separation ω is negligibly small for macroscopic objects (with
large mass) for which classical mechanics is adequate for
8v

Allowed energies, Ev

7
Potential energy

6
5
4

hω

3

describing vibrational motion; however, the energy separation is of great importance for objects with mass similar to
that of atoms.
Because the smallest permitted value of v is 0, it follows from
eqn 8B.4 that a harmonic oscillator has a zero-point energy
E0 = 12 ω



Harmonic oscillator 

Zero-point energy  (8B.6)

The mathematical reason for the zero-point energy is that
v cannot take negative values, for if it did the wavefunction
would not obey the boundary conditions. The physical reason
is the same as for the particle in a box (Topic 8A): the particle is confined, its position is not completely uncertain, and
therefore its momentum, and hence its kinetic energy, cannot
be exactly zero. We can picture this zero-point state as one in
which the particle fluctuates incessantly around its equilibrium position; classical mechanics would allow the particle to
be perfectly still.
Atoms vibrate relative to one another in molecules with
the bond acting like a spring. The question then arises as to
what mass to use to predict the frequency of the vibration.
In general, the relevant mass is a complicated combination
of the masses of all the atoms that move, with each contribution weighted by the amplitude of the atom’s motion. That
amplitude depends on the mode of motion, such as whether
the vibration is a bending motion or a stretching motion, so
each mode of vibration has a characteristic ‘effective mass’.
For a diatomic molecule AB, however, for which there is only
one mode of vibration, corresponding to the stretching and
compression of the bond, the effective mass, μ, has a very
simple form:
μ=

mA mB
mA + mB

Diatomic molecule 

Effective mass  (8B.7)

When A is much heavier than B, mB can be neglected in the
denominator and the effective mass is μ ≈ mB, the mass of the
lighter atom. This result is plausible, for in the limit of the heavy
atom being like a brick wall, only the lighter atom moves and
hence determines the vibrational frequency.

2
1
0
0

Displacement, x

Figure 8B.2  The energy levels of a harmonic oscillator are
evenly spaced with separation ħω, with ω = (k f/m)1/2. Even in its
lowest energy state, an oscillator has an energy greater than
zero.
1  For details, see our Molecular quantum mechanics, Oxford University
Press, Oxford (2011).

Brief illustration 8B.1  The vibration of a diatomic

molecule
The effective mass of 1H35Cl is
μ=

mHmCl
(1.0078mu ) × (34.9688mu )
= 0.9796mu
=

mH + mCl (1.0078mu ) + (34.9688mu )

which is close to the mass of the proton. The force constant of
the bond is k f = 516.3 N m−1. It follows from eqn 8B.4, with μ in
place of m, that


8B  Vibrational motion  



516.3 N m
=
−27

×
×
0
.
9796
(
1
.
66054
10
kg
)

−1

1

1/2

= 5.634 ×1014 s

or 563.4 THz. (We have used 1 N = 1 kg m s −2 .) Therefore the
separation of adjacent levels is (eqn 8B.5)
Ev +1 − Ev = (1.054 57 × 10−34 Js) × (5.634 ×1014 s) = 5.941 ×10−20 J
or 59.41 zJ, about 0.37 eV. This energy separation corresponds
to 36 kJ mol−1, which is chemically significant. The zero-point
energy, eqn 8B.6, of this molecular oscillator is 29.71 zJ, which
corresponds to 0.19 eV, or 18 kJ mol−1.

0.8

exp(–x2)

k 
ω = f 
 µ

1/2

331

0.6
0.4
0.2
0

–2

–1

0
x

1

2

Figure 8B.3  The graph of the Gaussian function, f ( x ) = e − x .
2

Self-test 8B.1  Suppose a hydrogen atom is adsorbed on the

surface of a gold nanoparticle by a bond of force constant
855 N m−1. Calculate its zero-point vibrational energy.
Answer: 37.7 zJ, 22.7 kJ mol−1, 0.24 eV

The result in Brief illustration 8B.1 implies that excitation
requires radiation of frequency ν = ΔE/h = 90 THz and wavelength λ = c/ν = 3.3 µm. It follows that transitions between adjacent vibrational energy levels of molecules are stimulated by or
emit infrared radiation (Topics 12D and 12E).

(b)  The wavefunctions
Like the particle in a box (Topic 8A), a particle undergoing harmonic motion is trapped in a symmetrical well in which the
potential energy rises to large values (and ultimately to infinity)
for sufficiently large displacements (compare Figs. 8A.1 and
8B.1). However, there are two important differences:

• As the kinetic energy of the oscillator depends on
the displacement in a more complex way (on account
of the variation of the potential energy), the
curvature of the wavefunction also varies in a more
complex way.

The detailed solution of eqn 8B.3 confirms these points
and shows that the wavefunctions for a harmonic oscillator
have the form
ψ (x ) = N × (polynomial in x )
× (bell-shaped Gaussian function)
where N is a normalization constant. A Gaussian function is a
bell-shaped function of the form e − x (Fig. 8B.3). The precise
form of the wavefunctions is
2

y=

2

/2

 2 
α =
 mkf 

x
α

1/4

Harmonic
oscillator



Wave­
functions (8B.8)

The factor Hv(y) is a Hermite polynomial; the form of these poly­
nomials and some of their properties are listed in Table 8B.1.
Hermite polynomials, which are members of a class of functions
called ‘orthogonal polynomials’, have a wide range of important
properties which allow a number of quantum mechanical calculations to be done with relative ease. Note that the first few
Hermite polynomials are very simple: for instance, H0(y) = 1 and
H1(y) = 2y.
Because H0(y) = 1, the wavefunction for the ground state (the
lowest energy state, with v = 0) of the harmonic oscillator is
ψ 0 (x ) = N 0 e− y

2

/2

= N 0 e− x

2

/2α 2

Harmonic
oscillator

Ground-state
wavefunction

(8B.9a)

Table 8B.1  The Hermite polynomials, Hv(y)*
Physical interpretation

• Because the potential energy climbs towards infinity
only as x2 and not abruptly, the wavefunction
approaches zero more slowly at large displacements
than for the particle in a box.

ψ v (x ) = N v H v ( y )e − y

v

Hv(y)

0

1

1

2y

2

4y2 − 2

3

8y3 − 12y

4

16y4 − 48y2 + 12

5

32y5 − 160y3 + 120y

6

64y6 − 480y4 + 720y2 − 120

* The Hermite polynomials are solutions of the differential equation
H v″ − 2yH v′ + 2vH v = 0
where primes denote differentiation. They satisfy the recursion relation
H v+1 − 2yH v + 2vH v−1 = 0
An important integral is

∫

∞

−∞

if v′ ≠ v
2
 0
H v ′ H v e− y dy =  1/2 v
π 2 v ! if v′ = v


332  8  The quantum theory of motion

Wavefunction, ψ,
and probability density, ψ 2

Example 8B.1  Confirming that a wavefunction is a

solution of the Schrödinger equation
Confirm that the ground-state wavefunction (eqn 8B.9a) is a
solution of the Schrödinger equation, eqn 8B.3.

ψ
ψ2

–4

–2

Method   Substitute the wavefunction given in eqn 8B.9a
into eqn 8B.3. Use the definition of α given in eqn 8B.8
to determine the energy on the right-hand side of eqn 8B.3
and confirm that it matches the zero-point energy given in
eqn 8B.6.

0
Displacement, y = x/α

2

4

Figure 8B.4  The normalized wavefunction and probability
density (shown also by shading) for the lowest energy state of a
harmonic oscillator.

2

2

/α 2



Harmonic
oscillator

2

2

 2
= N1   xe − x /2α
α 
2

Harmonic
oscillator

2



=−

Ground-state
probability (8B.9b)
density

The wavefunction and the probability density are shown in Fig.
8B.4. Both curves have their largest values at zero displacement
(at x = 0), so they capture the classical picture of the zero-point
energy as arising from the ceaseless fluctuation of the particle
about its equilibrium position.
The wavefunction for the first excited state of the oscillator,
the state with v = 1, is
ψ 1 (x ) = N1 (2 y )e − y /2

d

 x  2 2
2
2
N e − x /2α = − N 0  2  e − x /2α
dx 0
α 
d2
d 
 x  − x 2 /2α 2 
2
2
N e − x /2α =
−N
e

dx  0  α 2 
dx 2 0


and the corresponding probability density is
ψ 02 (x ) = N 02 e − y = N 02 e − x

Answer  We need to evaluate the second derivative of the
ground-state wavefunction:

First excitedstate
wavefunction

N 0 − x 2 /2α 2
 x 

2
2
+ N 0  2  e − x /2α
e
2
α
α 

= −(1/ α 2 )ψ 0 + (x 2 /α 4 )ψ 0
Substituting ψ 0 into eqn 8B.3 and using the definition of α
(eqn 8B.8), we obtain
1/2

2  mkf 
2  mkf  2
ψ0 −
x ψ 0 + 12 kf x 2ψ 0 = Eψ 0


2
2m   
2m  2 
and therefore
1/2

(8B.10)

This function has a node at zero displacement (x = 0), and the
probability density has maxima at x = ±α (Fig. 8B.5).

  kf 
ψ 0 − 12 kf x 2ψ 0 + 12 kf x 2ψ 0 = Eψ 0
2  m 
The second and third terms on the left-hand side (in blue) cancel and we obtain E = 12 (kf /m)1/2 in accord with eqn 8B.6 for
the zero-point energy.
Self-test 8B.2  Confirm that the wavefunction in eqn 8B.10 is a
solution of eqn 8B.3 and determine its energy.

Wavefunction, ψ,
and probability density, ψ 2

Answer: yes, with E = 23 ω

–4

ψ
ψ2

–2

0
Displacement, y = x/α

2

4

Figure 8B.5  The normalized wavefunction and probability
density (shown also by shading) for the first excited state of a
harmonic oscillator.

The shapes of several wavefunctions are shown in Fig. 8B.6
and the corresponding probability densities are shown in Fig.
8B.7. At high quantum numbers, harmonic oscillator wavefunctions have their largest amplitudes near the turning points
of the classical motion (the locations at which V = E, so the
kinetic energy is zero). We see classical properties emerging in
the correspondence principle limit of high quantum numbers
(Topic 8A), for a classical particle is most likely to be found at
the turning points (where it travels most slowly) and is least
likely to be found at zero displacement (where it travels most
rapidly).


8B  Vibrational motion  

1

2

3

Example 8B.2  Normalizing a harmonic oscillator

4

wavefunction

Wavefunction, ψ

0

333

Find the normalization constant for the harmonic oscillator
wavefunctions.
Method  Normalization is carried out by evaluating the inte-

–4

–2

0
Displacement, y = x/α

2

gral of |ψ |2 over all space and then finding the normalization
factor from eqn 7B.3 (N = 1/(∫ψ *ψ dτ )1/2 ). The normalized
wavefunction is then equal to Nψ. In this one-dimensional
problem, the volume element is dx and the integration is
from −∞ to +∞. The wavefunctions are expressed in terms of
the dimensionless variable y = x/α, so begin by expressing the
integral in terms of y by using dx = αdy. The integrals required
are given in Table 8B.1.

4

Figure 8B.6  The normalized wavefunctions for the first
five states of a harmonic oscillator. Note that the number
of nodes is equal to v and that alternate wavefunctions

are symmetrical or asymmetrical about y = 0 (zero
displacement).

Answer  The unnormalized wavefunction is

ψ v (x ) = H v ( y )e − y /2
2

It follows from the integrals given in Table 8B.1 that

Note the following features of the wavefunctions:

• The exponent y2 is proportional to x 2 × (mk f )1/2, so
the wavefunctions decay more rapidly for large
masses and stiff springs.
• As v increases, the Hermite polynomials become
larger at large displacements (as xv), so the
wavefunctions grow large before the Gaussian
function damps them down to zero: as a result, the
wavefunctions spread over a wider range as v
increases (Fig. 8B.7).

−∞

ψ v*ψ v dx = α

∫

∞

−∞

ψ v*ψ v dy = α

∫

∞

−∞

H v2 ( y )e − y dy = απ1/2 2v v !
2

where v! = v(v − 1)(v − 2)…1. Therefore,
Physical interpretation

• The Gaussian function goes very strongly to zero as
the displacement increases (in either direction,
extension or compression), so all the wavefunctions
approach zero at large displacements.

∫

∞

1


N v =  1/2 v 
 απ 2 v ! 

1/2

Note that, unlike the normalization constant for a particle in
a box, for a harmonic oscillator Nv is different for each value
of v.
Self-test 8B.3  Confirm, by explicit evaluation of the integral,

that ψ0 and ψ1 are orthogonal.
∞

Answer: Show that ∫ −∞ ψ 0*ψ 1dx = 0 by using the information in
Table 8B.1

0 v
Probability
density

8B.2  The

1
2

Displacement

3
4
18

The average value of a property is calculated by evaluating the

expectation value of the corresponding operator (eqn 7C.11,
 dτ ). Now that we know the wavefunctions of
〈 Ω 〉 = ∫ψ * Ωψ
the harmonic oscillator, we can start to explore its properties by
evaluating integrals of the type
〈Ω 〉 =

Figure 8B.7  The probability densities for the first five
states of a harmonic oscillator and the state with v = 18.
Note how the regions of highest probability density move
towards the turning points of the classical motion as v
increases.

properties of oscillators

∫

∞

 dx
ψ v* Ωψ
v

−∞



(8B.11)

(Here and henceforth, the wavefunctions are all taken to be

normalized to 1.) When the explicit wavefunctions are substituted, the integrals look fearsome, but the Hermite polynomials
have many simplifying features.


334  8  The quantum theory of motion
(a)  Mean values

Now use the recursion relation (Table 8B.1) to form

We show in the following example that the mean displacement,
〈x〉, and the mean square displacement, 〈x2〉, of the oscillator
when it is in the state with quantum number v are
〈x〉 = 0

Harmonic oscillator 

which turns the integral into

Mean displacement  (8B.12a)

∫

∞

−∞


〈 x 2 〉 = (v + 12 )
(mkf )1/2

Harmonic
oscillator



Mean square
displacement

(8B.12b)

The result for 〈x〉 shows that the oscillator is equally likely
to be found on either side of x = 0 (like a classical oscillator).
The result for 〈x2〉 shows that the mean square displacement
increases with v. This increase is apparent from the probability
densities in Fig. 8B.7, and corresponds to the classical amplitude of swing increasing as the oscillator becomes more highly
excited.
Example 8B.3  Calculating properties of a harmonic

oscillator
Consider the harmonic oscillator motion of the He Cl molecule in Brief illustration 8B.1. Calculate the mean displacement of the oscillator when it is in a state with quantum
number v.
Method  Normalized wavefunctions must be used to calcu-

late the expectation value. The operator for position along x is
multiplication by the value of x (Topic 7C). The resulting integral can be evaluated either
• by inspection (the integrand is the product of an odd
and an even function), or
• by explicit evaluation using the formulas in Table 8B.1.
The former procedure makes use of the definitions that an
even function is one for which f(−x) = f(x) and an odd function is one for which f(−x) = −f(x). Therefore, the product of an

odd and even function is itself odd, and the integral of an odd
function over a symmetrical range about x = 0 is zero. The latter procedure using explicit integration is illustrated here to
give practice in the calculation of expectation values. We shall
need the relation x = αy, which implies that dx = αdy.

Answer  The integral we require is

〈x〉 =

yH v = vH v −1 + 12 H v +1

∫

∞

−∞

ψ v* xψ v dx = N v2

∫
=α N ∫
= α 2 N v2
2

2
v

∞

−∞

∞

−∞

∫

∞

−∞

(H v e − y /2 )x(H v e − y /2 )dx
2

(H v e − y /2 ) y(H v e − y /2 )dy
2

2

H v yH v e − y dy
2

2

H v yH v e − y dy = v
2

∫

∞

−∞

H v H v−1e − y dy + 12
2

∫

∞

−∞

H v H v+1e − y dy
2

Both integrals are zero (See Table 8B.1), so 〈x〉 = 0. The mean
displacement is zero because displacements on either side of
the equilibrium position occur with equal probability.
Self-test 8B.4  Calculate the mean square displacement, 〈x 2〉,

of the H − Cl bond distance from its equilibrium position by
using the recursion relation in Table 8B.1 twice.
Answer: (v + 12 ) ×115 pm2 ; eqn 9.12b, with μ in place of m

The mean potential energy of an oscillator, the expectation
value of V = 12 kf x 2 , can now be calculated very easily:
k 
〈V 〉 = kf 〈 x 〉 = ( v + )   f 
 m
2

1
2

1
2

1/2

1
2



or
〈V 〉 = 12 (v + 12 ) ω



Harmonic
oscillator

Mean potential
(8B.13a)
energy

Because the total energy in the state with quantum number v is
(v + 12 )ω , it follows that
〈V 〉 = 12 Ev

Harmonic oscillator 

Mean potential energy  (8B.13b)

The total energy is the sum of the potential and kinetic energies, so it follows at once that the mean kinetic energy of the
oscillator is (as could also be shown using the kinetic energy
operator)
〈 Ek 〉 = 12 Ev



Harmonic oscillator 

Mean kinetic energy  (8B.13c)

The result that the mean potential and kinetic energies of a
harmonic oscillator are equal (and therefore that both are
equal to half the total energy) is a special case of the virial
theorem:
If the potential energy of a particle has the form
V = axb, then its mean potential and kinetic energies are
related by
2〈 Ek 〉 = b〈V 〉

Virial theorem  (8B.14)


8B  Vibrational motion  
For a harmonic oscillator b = 2, so 〈Ek〉 = 〈V〉, as we have found.

The virial theorem is a short cut to the establishment of a number of useful results, and we use it elsewhere (for example, in
Topic 9A).

(b)  Tunnelling
An oscillator may be found at extensions with V > E, which are
forbidden by classical physics, for they correspond to negative
kinetic energy; this is an example of the phenomenon of tunnelling (Topic 8A). As shown in Example 8B.4, for the lowest
energy state of the harmonic oscillator, there is about an 8 per
cent chance of finding the oscillator stretched beyond its classical limit and an 8 per cent chance of finding it with a classically forbidden compression. These tunnelling probabilities are
independent of the force constant and mass of the oscillator.

Example 8B.4  Calculating the tunnelling probability for

the harmonic oscillator
Calculate the probability that the ground-state harmonic
oscillator will be found in a classically forbidden region.
Method  Find the expression for the classical turning point,

x tp, where the kinetic energy vanishes, by equating the potential energy to the total energy E of the harmonic oscillator. The
probability of finding the oscillator stretched beyond a displacement x tp is the sum of the probabilities ψ 2dx of finding
it in any of the intervals dx lying between x tp and infinity, so
evaluate the integral
P=

∫

∞

x tp

2
π1/2

1/2

x tp  2(v + 12 ) ω 
y tp =
=

α  α 2kf


∫

∞

z

e − y dy
2

and evaluated for some values of z in Table 8B.2 (this function
is commonly available in mathematical software packages).
By symmetry, the probability of being found stretched into a
classically forbidden region is the same as that of being found
compressed into a classically forbidden region.
Answer  According to classical mechanics, the turning point,

x tp, of an oscillator occurs when its kinetic energy is zero,
which is when its potential energy 12 kf x 2 is equal to its total

energy E. This equality occurs when

ω = ( kf / m )1/2

=

(2v + 1)1/2

For the state of lowest energy (v = 0), y tp = 1 and the probability
of being beyond that point is
P=

∫

∞

x tp

ψ 02dx = αN 02

∫

∞

1

e − y dy
2

The normalization constant N0 is calculated from the expression for Nv in Example 8B.2 (N v =1/(α π1/2 2v v !)1/2 ) :



1
N0 = 
1/2 0 
 α π 2 0! 

1/2

 1 
=
1/2
 α π 

1/2

The integral in the expression for P is written in terms of the
error function erf(1) as
erf(1) = 1 −

2
π1/2

∫

∞

1

e − y dy so

2

∫

∞

1

1
2
e − y dy = π1/2 (1 − erf(1))
2

It follows that

P =α ×

The variable of integration is best expressed in terms of y = x/α
and the integral to be evaluated is a special case of the error
function, erf z, defined as

1/2

with E given by eqn 8B.4. The variable of integration in the integral P is best expressed in terms of y = x/α with α = (ħ2/mkf )1/4,
and then the right-hand turning point lies at

N02

ψ v2dx

erf (z ) = 1 −

 2E 
or x tp = ±  
 kf 

2E
kf

2
x tp
=

335

∞

∫1 e

− y 2 dy

0.843 

1
1 1/2
1
×
(
1
−

(
1
))
=
1
−
1) = 0.079
π
erf
erf(
2


απ1/2 2

In 7.9 per cent of a large number of observations, any oscillator in the state with quantum number v = 0 will be found
stretched to a classically forbidden extent. There is the same
probability of finding the oscillator with a classically forbidden compression. The total probability of finding the oscillator tunnelled into a classically forbidden region (stretched or
compressed) is about 16 per cent.
Self-test 8B.5  Calculate the probability that a harmonic oscillator in the state with quantum number v = 1 will be found at
a classically forbidden extension. (Follow the argument given
in Example 8B.4 and use the method of integration by parts
(Mathematical background 1) to obtain an integral that can be
expressed in terms of the error function.)
Answer: P = 0.056


336  8  The quantum theory of motion
Table 8B.2  The error function, erf(z)*
z

erf(z)

0

0

0.01

0.0113

0.05

0.0564

0.10

0.1125

0.50

0.5205

1.00

0.8427

1.50

0.9661

2.00

0.9953

The probability of finding the oscillator in classically forbidden regions decreases quickly with increasing v, and vanishes entirely as v approaches infinity, as we would expect
from the correspondence principle. Macroscopic oscillators
(such as pendulums) are in states with very high quantum
numbers, so the tunnelling probability is wholly negligible
and classical mechanics is reliable. Molecules, however, are
normally in their vibrational ground states, and for them
the probability is very significant and classical mechanics is
misleading.

*More values are available in mathematical software
packages.

Checklist of concepts
☐1.A particle undergoing harmonic motion is called a
harmonic oscillator and experiences a restoring force
proportional to its displacement;
☐2.The potential energy of a harmonic oscillator is a parabolic function of the displacement from equilibrium.
☐3.The energy levels of a harmonic oscillator form an
evenly spaced ladder.
☐4.The wavefunctions of a harmonic oscillator are products of a Hermite polynomial and a Gaussian (bellshaped) function.

☐5.There is a zero-point energy, an irremovable minimum
energy, which is consistent with, and can be interpreted
in terms of, the uncertainty principle.
☐6.The probability of finding the harmonic oscillator

in classically forbidden regions is significant for the
ground vibrational state (v = 0) but decreases quickly
with increasing v.

Checklist of equations
Property

Equation

Comment

Equation number

Energy levels of harmonic oscillator

Ev = (v + 12 )ω , ω = (kf /m)1/2

v = 0, 1, 2,…

8B.4

Zero-point energy of harmonic oscillator

E0 = 12 ω

Wavefunction of harmonic oscillator

ψ v (x ) = N v H v ( y )e − y
y = x /α ,
N v = (1/α

8B.6
2

/2

v = 0, 1, 2,…

8B.8

α = (2 /mkf )1/4
π1/2 2v v !)1/2

Mean displacement of harmonic oscillator

〈x〉 = 0

8B.12a

Mean square displacement of harmonic oscillator

〈 x 2 〉 = (v + 12 ) /(mkf )1/2

8B.12b

Virial theorem

2〈Ek〉 = b〈V〉

V = axb

8B.14


8C  Rotational motion
8C.1  Rotation

Contents
8C.1 

Rotation in two dimensions
(a)
(b)

(c)

8C.2 

The qualitative origin of quantized rotation
The solutions of the Schrödinger equation
Example 8C.1: Using the particle on a ring model
Quantization of angular momentum
Brief illustration 8C.1: Nodes in the wavefunction

Rotation in three dimensions
The wavefunctions
Brief illustration 8C.2: The angular nodes
of the spherical harmonics
(b) The energies
Example 8C.2: Using the rotational energy levels

(c) Angular momentum
Brief illustration 8C.3: The magnitude
of the angular momentum
(d) Space quantization
(e) The vector model
Brief illustration 8C.4: The vector model of the
angular momentum
(a)

Checklist of concepts
Checklist of equations

337
337
338
340
340
341
342
342
344
344
344
345

in two dimensions

We consider a particle of mass m constrained to move in a circular path (a ‘ring’) of radius r in the xy-plane with constant
potential energy, which may be taken to be zero (Fig. 8C.1). The
total energy is equal to the kinetic energy, because V = 0 every­

where. We can therefore write E = p2/2m. According to classical mechanics (Foundations B), the angular momentum, Jz,
around the z-axis (which lies perpendicular to the xy-plane) is
Jz = ±pr, so the energy can be expressed as J z2 /2mr 2 . Because mr2
is the moment of inertia, I, of the mass on its path, it follows
that
E=

J z2
2I

I = mr 2



Particle on a ring,
classical expression 

Energy

(8C.1)

345
345
346

We shall now see that not all the values of the angular momentum are permitted in quantum mechanics, and therefore that
both angular momentum and rotational energy are quantized.

347

(a)  The qualitative origin of quantized

347
347

rotation

Because Jz = ±pr, and since the de Broglie relation gives p = h/λ
(Topic 7A) the angular momentum about the z-axis is

➤➤ Why do you need to know this material?
The investigation of rotational motion introduces the
concept of angular momentum, which is central to the
description of the electronic structure of atoms and
molecules and the interpretation of details observed in
molecular spectra.

➤➤ What is the main idea?
The energy and the angular momentum of a rotating
object are quantized.

Jz = ±

hr
λ

(8C.2)

Opposite signs correspond to opposite directions of travel. This
equation shows that the shorter the wavelength of the particle on a circular path of given radius, the greater the angular

momentum of the particle. It follows that if we can see why the
wavelength is restricted to discrete values, then we shall understand why the angular momentum is quantized.
z

➤➤ What do you need to know already?
You should know the postulates of quantum mechanics
(Topic 7C), and be familiar with the concept of angular
momentum in classical physics (Foundations B).

Jz

J

x

y

r
m

This Topic provides a quantum mechanical description of rotation in two and three dimensions. The concepts developed here
form the basis for discussion of atomic structure (Topics 9A
and 9B) and molecular rotation (Topic 12B).

p

Figure 8C.1  The angular momentum of a particle of mass m
on a circular path of radius r in the xy-plane is represented by a
vector J with the single non-zero component Jz of magnitude
pr perpendicular to the plane.

338  8  The quantum theory of motion

0

π

Wavefunction, ψ

(a)

(b)

Angle, φ

2π

J z = ml 

2π φ
0

Em =
l

0

π

Angle, φ

2π

First, second,... circuits

Suppose for the moment that λ can take an arbitrary value. In
that case, the wavefunction depends on the azimuthal angle ϕ as
shown in Fig. 8C.2a. When ϕ increases beyond 2π, the wavefunction continues to change, but for an arbitrary wavelength it gives
rise to a different value at a given point after each circuit, which is
unacceptable because a wavefunction must be single-valued. An
acceptable solution is obtained only if the wavefunction reproduces itself on successive circuits, as in Fig. 8C.2b. Because only
some wavefunctions have this property, it follows that only some
angular momenta are acceptable and therefore that only certain
rotational energies are allowed. That is, the energy of the particle
is quantized. Specifically, an integer number of wavelengths must
fit the circumference of the ring (which is 2πr):
nλ = 2πr n = 0, 1, 2,…

Particle on
a ring 

ml = 0, ± 1, ± 2, …

Angular
momenta

(8C.4)

Positive values of ml correspond to rotation in a clockwise

sense around the z-axis (as viewed in the direction of increasing z, Fig. 8C.3) and negative values of ml correspond to counter-clockwise rotation around z. It then follows from eqns 8C.1
and 8C.4 that the energy is limited to the values

Figure 8C.2  Two solutions of the Schrödinger equation for
a particle on a ring. The circumference has been opened out
into a straight line; the points at ϕ = 0 and 2π are identical. The
solution in (a) is unacceptable because it is not single-valued.
Moreover, on successive circuits it interferes destructively with
itself, and does not survive. The solution in (b) is acceptable: it is
single-valued, and on successive circuits it reproduces itself.

ml2 2
2I

ml = 0, ± 1, ± 2, …



Particle on
a ring

Energy
levels

(8C.5)

We explore this result further by noting that:
• The energies, labelled by ml, are quantized because
ml must be an integer.
• The occurrence of ml as its square means that the

energy of rotation is independent of the sense of
rotation (the sign of ml), as we expect physically.
That is, states with a given nonzero value of |ml| are
doubly degenerate.
• The state described by ml = 0 is non-degenerate,
consistent with the interpretation that, when ml is
zero, the particle has an infinite wavelength and is
‘stationary’; the question of the direction of rotation
does not arise.
• There is no zero-point energy in this system: the
lowest possible energy is E0 = 0.

(b)  The solutions of the Schrödinger
equation
To obtain the wavefunctions for the particle on a ring and to
confirm that the energies from eqn 8C.5 are correct, we need
ml > 0

(8C.3)

The value n = 0 corresponds to λ = ∞; a ‘wave’ of infinite wavelength has a constant height at all values of ϕ. It follows from
eqns 8C.2 and 8C.3 that the angular momentum is therefore
limited to the values
Jz = ±

values. At the same time we recognize the presence of h/2π = 
and obtain

Physical interpretation

Wavefunction, ψ

First circuit
Second
circuit

hr
nhr
nh
=±
=±
n = 0, 1, 2, …
λ
2πr
2π


The sign of Jz (which indicated the sense of the rotation) can be
absorbed into the quantum number by replacing n by ml = 0,
±1, ±2, … where we have allowed ml (the conventional notation
for this quantum number) to have positive and negative integer

(a)

(b)

ml < 0

Figure 8C.3  The angular momentum of a particle confined to a
plane can be represented by a vector of length |ml| units along

the z-axis and with an orientation that indicates the direction of
motion of the particle. The direction is given by the right-hand
screw rule, so (a) corresponds to ml > 0, clockwise as seen from
below and (b) corresponds to ml < 0, anticlockwise as seen from
below.


8C  Rotational motion  
to solve the Schrödinger equation explicitly. We show in the
following Justification that the normalized wavefunctions and
corresponding energies are
ψ m (φ ) =
l

Em =
l

eim φ
(2π)1/2
l

ml = 0, ± 1, ± 2, …

ml2 2
2I



Particle
Wave­

on a
functions
ring 

Particle on a ring 

(8C.6a)

Energy levels  (8C.6b)

The wavefunction with ml = 0 is ψ0(ϕ) = 1/(2π)1/2, corresponding
to uniform amplitude around the ring, and its energy is E0 = 0.

Justification 8C.1  The solutions of the Schrödinger

equation for a particle on a ring
The hamiltonian for a particle of mass m travelling on a circle
in the xy-plane (with V = 0) is the same as that for free motion
in a plane (eqn 8A.1 of Topic 8A),
2  ∂ 2
∂2 
+ 2
Hˆ = −
2

2m  ∂x ∂y 

but with the constraint to a path of constant radius r. It is
always a good idea to use coordinates that reflect the full symmetry of the system, so we introduce the coordinates r and ϕ
(The chemist’s toolkit 8C.1). By standard manipulations we can

write
∂2
∂2
∂2 1 ∂ 1 ∂2
+
+
2 =
2 +
∂x ∂y
∂r 2 r ∂r r 2 ∂φ 2

(8C.8)

However, because the radius of the path is fixed, the (blue)
derivatives with respect to r can be discarded. Only the last
term in eqn 8C.8 then survives and the hamiltonian becomes
simply
2 d 2
Hˆ = −
2mr 2 dφ 2

Particle on a ring 

Particle on a ring 

Hamiltonian  (8C.9b)

2 d 2ψ
= Eψ
2I dφ 2

We rewrite this equation as
d 2ψ
2 IE
=− 2 ψ

dφ 2


Particle on
a ring

(8C.10b)



The (unnormalized) general solutions of this equation are
ψ ml (φ )= eiml φ

(8C.11)



as can be verified by substitution.
We now select the acceptable solutions from among these
general solutions by imposing the condition that the wavefunction should be single-valued. That is, the wavefunction ψ
must satisfy a cyclic boundary condition, and match at points
separated by a complete revolution: ψ (ϕ + 2π) = ψ(ϕ). On substituting the general wavefunction into this condition, we find
ψ ml (φ + 2π) = eiml (φ +2 π) = eiml φ e2 πiml = ψ ml (φ )e2 πiml

= ψ ml (φ )(e πi)2ml



As eiπ = −1 (Euler’s formula, Mathematical background 3), this
relation is equivalent to
ψ ml (φ + 2π) = (−1)2ml ψ ml (φ )



Because cyclic boundary conditions require (−1)2ml = 1, 2ml
must be a positive or a negative even integer (including 0), and
therefore ml must be an integer: ml = 0, ±1, ±2, ….
We now normalize the wavefunction by finding the normal∞
ization constant N given by eqn 7B.3 (N = (∫ −∞ ψ *ψ dx)−1/2 ),
which in this case becomes:
N=

1



∫


ψ *ψ dφ 

0
2π

1/2

=

1




∫

2π
0


e − iml φ eiml φ dφ 

1

1/2

=

1
(2π)1/2

(8C.12)

and the normalized wavefunctions for a particle on a ring are
those given by eqn 8C.6a. The expression for the energies of
the states (eqn 8C.6b) is obtained by rearranging the relation
ml2 = 2IE /2 into E = ml2 2 /2I .
The chemist’s toolkit 8C.1  Cylindrical coordinates

For systems with cylindrical symmetry it is best to work in
cylindrical coordinates, r, ϕ, and z (Sketch 1), with

and the Schrödinger equation is
−

d 2ψ
= −ml2ψ
dφ 2

Hamiltonian  (8C.9a)

The partial derivative has been replaced by a complete derivative because ϕ is now the only variable. The moment of inertia,
I = mr2 has appeared automatically so Hˆ may be written
2 d 2
Hˆ = −
2 I dφ 2


For a given energy, 2IE/2 is a constant, which for convenience
(and an eye on the future) we write as ml2 . At this stage ml is
just a dimensionless number with no restrictions. Then the
equation becomes

(8C.7)



339

Schrödinger
(8C.10a)
equation

x = r cosθ

y = r sin φ



and where
r ranges from
0 to ∞

ϕ ranges from
0 to 2π

z ranges from
−∞ to ∞


340  8  The quantum theory of motion
Self-test 8C.1  Use the particle on a ring model to calculate the

minimum energy required for the excitation of a π electron in
coronene, C24H12 (1). Assume that the radius of the ring is three
times the carbon–carbon bond length in benzene and that the
electrons are confined to the periphery of the molecule.

z

r
x

φ

y

Sketch 1  Cylindrical coordinates
The volume element is
d τ = r dr d φ d z



1 Coronene

For motion in a plane we set z = 0 and for the volume element
use
d τ = r dr d φ

(model ring in red)
Answer: For transition from ml = +3 to ml = +4:
ΔE = 0.0147 zJ or 8.83 J mol−1

Example 8C.1  Using the particle-on-a-ring model

(c)  Quantization of angular momentum

The particle-on-a-ring model is a crude but illustrative model
of cyclic, conjugated molecular systems. Treat the π electrons
in benzene as particles freely moving over a circular ring of
carbon atoms and calculate the minimum energy required for
the excitation of a π electron. The carbon–carbon bond length
in benzene is 140 pm.

We have seen that the angular momentum around the z-axis
is quantized and confined to the values given in eqn 8C.4
(Jz = ml). The wavefunction for the particle on a ring is given
by eqn 8C.6a:

Method  Because each carbon atom contributes one π electron,
six electrons in the conjugated system move along the perimeter of the ring. Each state is occupied by two electrons, so
only the ml = 0, +1, and −1 states are occupied (with the last two
being degenerate). The minimum energy required for excitation corresponds to a transition of an electron from the ml = +1
(or −1) state to the ml = +2 (or −2) state. Use eqn 8C.6b, and the
mass of the electron, to calculate the energies of the states.
Answer  From eqn 8C.6b, the energy separation between the

ml = +1 and the ml = +2 states is
∆E = E +2 − E −1 = (4 −1) ×

(1.055 ×10−34 Js)2

2 × (9.109 ×10−31 kg) × (1.40 ×10−10 m)2

= 9.35 × 10−19 J

l

1
eim φ
=
(cos mlφ + i sin mlφ )
(2π)1/2 (2π)1/2

l

Therefore, as |ml| increases, the increasing angular momentum
is associated with:
• an increase in the number of nodes in the real (cos mlϕ)
and imaginary (sin mlϕ) parts of the wavefunction (the
complex function does not have nodes but each of its real
and imaginary components does);
• a decrease in the wavelength and, by the de Broglie
relation, an increase in the linear momentum with which
the particle travels round the ring (Fig. 8C.4).



Therefore the minimum energy required to excite an electron
is 0.935 aJ or 563 kJ mol−1. This energy separation corresponds
to an absorption frequency of 1.41 PHz (1 PHz = 1015 Hz) and
a wavelength of 213 nm; the experimental value for a transition of this kind is 260 nm. That such a primitive model gives

relatively good agreement is encouraging. In addition, even
though the model is primitive, it gives insight into the origin
of the quantized π electron energy levels in cyclic conjugated
systems (Topic 10D).

A note on good practice  Note that, when quoting the
value of ml, it is good practice always to give the sign,
even if ml is positive. Thus, we write ml = +1, not ml = 1.

ψ m (φ ) =

ml = 0

|ml| = 1

|ml| = 2

Figure 8C.4  The real parts of the wavefunctions of a particle on
a ring. As shorter wavelengths are achieved, the magnitude of
the angular momentum around the z-axis grows in steps of .