QUANTUM MECHANICS OF

SOME SIMPLE SYSTEMS

The Free Particle

The simplest system in quantum mechanics has the potential energy V
equal to zero everywhere. This is called a free particle since it has no forces
acting on it. We consider the one-dimensional case, with motion only in
the x-direction, giving the Schrödinger equation

h̄2 d2 ψ(x)
− = Eψ(x) (1)
2m dx2

Total derivatives can be used since there is but one independent variable.
The equation simplifies to

ψ 00 (x) + k2 ψ(x) = 0 (2)

with the definition

k 2 ≡ 2mE/h̄2 (3)
Possible solutions of Eq (2) are
(
sin kx
ψ(x) = const cos kx
e±ikx

(4)
There is no restriction on the value of k. Thus a free particle, even in
quantum mechanics, can have any non-negative value of the energy

h̄2 k 2
E= ≥0 (5)
2m

The energy levels in this case are not quantized and correspond to the same
continuum of kinetic energy shown by a classical particle.

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 It is of interest also to consider the x-component of linear momentum
for the free-particle solutions (4). According to Eq (2-32), the eigenvalue
equation for momentum should read
dψ(x)
p̂x ψ(x) = −ih̄ = p ψ(x) (6)
dx
where we have denoted the momentum eigenvalue as p. It is easily shown
that neither of the functions sin kx or cos kx from (4) is an eigenfunction
of p̂x . But e±ikx are both eigenfunctions with eigenvalues p = ±h̄k, respec-
tively. Evidently the momentum p can take on any real value between −∞
and +∞. The kinetic energy, equal to E = p2 /2m, can correspondingly
have any value between 0 and +∞.
The functions sin kx and cos kx, while not eigenfunctions of p̂x , are each
superpositions of the two eigenfunctions e±ikx , by virtue of the trigonometric
identities
1 1
cos kx = (eikx + e−ikx ) and sin kx = (eikx − e−ikx ) (7)
2 2i
The eigenfunction eikx for k > 0 represents the particle moving from left
to right on the x-axis, with momentum p > 0. Correspondingly, e−ikx
represents motion from right to left with p < 0. The functions sin kx
and cos kx represent standing waves, obtained by superposition of opposing
wave motions. Although these latter two are not eigenfunctions of p̂x but
are eigenfunctions of p̂2x , hence of the Hamiltonian Ĥ.

Particle in a Box
This is the simplest non-trivial application of the Schrödinger equation,
but one which illustrates many of the fundamental concepts of quantum
mechanics. For a particle moving in one dimension (again along the x-
axis), the Schrödinger equation can be written
h̄2 00
− ψ (x) + V (x)ψ(x) = E ψ(x) (8)
2m
Assume that the particle can move freely between two endpoints x = 0
and x = a, but cannot penetrate past either end. This is equivalent to a
potential energy dependent on x with
n
0 0≤x≤a
V (x) = (9)
∞ x < 0 and x > a

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 E4=16E1

V(x)

E3=9E1

Figure 1. Potential well

E2=4E1
and lowest energy levels
for particle in a box.
E1=h2/8ma2

0 x a

This potential is represented by the dark lines in Fig. 1. Infinite potential

energy constitute an impenetrable barrier. The particle is thus bound to a
potential well. Since the particle cannot penetrate beyond x = 0 or x = a,

ψ(x) = 0 for x < 0 and x > a (10)

By the requirement that the wavefunction be continuous, it must be true

as well that
ψ(0) = 0 and ψ(a) = 0 (11)
which constitutes a pair of boundary conditions on the wavefunction within
the box. Inside the box, V (x) = 0, so the Schrödinger equation reduces to
the free-particle form (1)

h̄2 00
− ψ (x) = E ψ(x), 0≤x≤a (12)
2m
We again have the differential equation

ψ00 (x) + k 2 ψ(x) = 0 with k 2 = 2mE/h̄2 (13)

The general solution can be written

ψ(x) = A sin kx + B cos kx (14)

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where A and B are constants to be determined by the boundary conditions
(11). By the first condition, we find

ψ(0) = A sin 0 + B cos 0 = B = 0 (15)

The second boundary condition at x = a then implies

ψ(a) = A sin ka = 0 (16)

It is assumed that A 6= 0, for otherwise ψ(x) would be zero everywhere and

the particle would disappear. The condition that sin kx = 0 implies that

ka = nπ (17)

where n is a integer, positive, negative or zero. The case n = 0 must

be excluded, for then k = 0 and again ψ(x) would vanish everywhere.
Eliminating k between (13) and (17), we obtain

h̄2 π2 2 h2 2
En = n = n n = 1, 2, 3 . . . (18)
2ma2 8ma2
These are the only values of the energy which allow solution of the Schrö-
dinger equation (12) consistent with the boundary conditions (11). The
integer n, called a quantum number, is appended as a subscript on E to
label the allowed energy levels. Negative values of n add nothing new
because the energies in Eq (18) depends on n2 . Fig. 1 shows part of the
energy-level diagram for the particle in a box. The occurrence of discrete
or quantized energy levels is characteristic of a bound system, that is, one
confined to a finite region in space. For the free particle, the absence of
confinement allowed an energy continuum. Note that, in both cases, the
number of energy levels is infinite—denumerably infinite for the particle in
a box but nondenumerably infinite for the free particle.
The particle in a box assumes its lowest possible energy when n = 1,
namely
h2
E1 = (19)
8ma2
The state of lowest energy for a quantum system is termed its ground state.
An interesting point is that E1 > 0, whereas the corresponding classical

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system would have a minimum energy of zero. This is a recurrent phe-
nomenon in quantum mechanics. The residual energy of the ground state,
that is, the energy in excess of the classical minimum, is known as zero point
energy. In effect, the kinetic energy, hence the momentum, of a bound par-
ticle cannot be reduced to zero. The minimum value of momentum is found
by equating E1 to p2 /2m, giving pmin = ±h/2a. This can be expressed as
an uncertainty in momentum given by ∆p ≈ h/a. Coupling this with the
uncertainty in position, ∆x ≈ a, from the size of the box, we can write

∆x ∆p ≈ h (20)

This is in accord with the Heisenberg uncertainty principle, which we will

discuss in greater detail later.
The particle-in-a-box eigenfunctions are given by Eq (14), with B = 0
and k = nπ/a, in accordance with (17):
nπx
ψn (x) = A sin , n = 1, 2, 3 . . . (21)
a
These, like the energies, can be labelled by the quantum number n. The
constant A, thus far arbitrary, can be adjusted so that ψn (x) is normalized.
The normalization condition (2-39) is, in this case,
Z a
[ψn (x)]2 dx = 1 (22)
0

the integration running over the domain of the particle, 0 ≤ x ≤ a. Substi-

tuting (21) into (22),
Z a Z nπ
nπx a a
A2 sin2 dx = A2 sin2 θ dθ = A2 = 1 (23)
0 a nπ 0 2

We have made the substitution θ = nπx/a and used the fact that the
average value of sin2 θ over an integral number of half wavelenths equals
1/2. (Alternatively, one could refer to standard integral tables.) From (23),
we can identify the normalization constant A = (2/a)1/2 , for all values of
n. Finally we can write the normalized eigenfunctions:
µ ¶1/2
2 nπx
ψn (x) = sin , n = 1, 2, 3 . . . (24)
a a

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The first few eigenfunctions and the corresponding probability distributions
are plotted in Fig. 2. There is a close analogy between the states of this
quantum system and the modes of vibration of a violin string. The patterns
of standing waves on the string are, in fact, identical in form with the
wavefunctions (24).

4(x) 4(x)

3(x) 3(x)

2(x) 2(x)

1(x) 1(x)
0 x a 0 x a

Figure 2. Eigenfunctions and probability

densities for particle in a box.

A significant feature of the particle-in-a-box quantum states is the oc-

currence of nodes. These are points, other than the two end points (which
are fixed by the boundary conditions), at which the wavefunction vanishes.
At a node there is exactly zero probability of finding the particle. The
nth quantum state has, in fact, n − 1 nodes. It is generally true that the
number of nodes increases with the energy of a quantum state, which can

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A useful constant in this computation is the Compton wavelength h/mc =
2.426 × 10−12 m. For n = 3, hexatriene, the predicted wavelength is 332
nm, while experiment gives λmax ≈ 250 nm. For n = 4, octatetraene, FEM
predicts 460 nm, while λmax ≈ 300 nm. Clearly the model has been pushed
beyond it range of quantitative validity, although the trend of increasing
absorption band wavelength with increasing n is correctly predicted. Inci-
dentally, a compound should be colored if its absorption includes any part
of the visible range 400–700 nm. Retinol (vitamin A), which contains a
polyene chain with n = 5, has a pale yellow color. This is its structure:

Particle in a Three-Dimensional Box

A real box has three dimensions. Consider a particle which can move freely
with in rectangular box of dimensions a × b × c with impenetrable walls.
In terms of potential energy, we can write
n
0 inside box
V (x, y, z) =
∞ outside box
(32)
Again, the wavefunction must vanish everywhere outside the box. By the
continuity requirement, the wavefunction must also valish in the six surfaces
of the box. Orienting the box so its edges are parallel to the cartesian
axes, with one corner at (0,0,0), the following boundary conditions must be
satisfied:

ψ(x, y, z) = 0 when x = 0, x = a, y = 0, y = b, z = 0 or z = c (33)

Inside the box, where the potential energy is everywhere zero, the Hamil-
tonian is simply the three-dimensional kinetic energy operator and the
Schrödinger equation reads

h̄2 2
− ∇ ψ(x, y, z) = E ψ(x, y, z) (34)
2m

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subject to the boundary conditions (33). This second-order partial diffential
equation is separable in cartesian coordinates, with a solution of the form

ψ(x, y, z) = X(x) Y (y) Z(z) (35)

subject to the boundary conditions

X(0) = X(a) = 0, Y (0) = Y (b) = 0, Z(0) = Z(c) = 0 (36)

Substituting (35) into (34) and dividing through by (35), we obtain

X 00 (x) Y 00 (y) Z 00 (z) 2mE

+ + + =0 (37)
X(x) Y (y) Z(z) h̄2

Each of the first three terms in (37) depends on one variable only, indepen-
dent of the other two. This is possible only if each term separately equals
a constant, say, −α2 , −β 2 and −γ 2 , respectively. These constants must be
negative in order that E > 0. Eq (37) is thereby transformed into three
ordinary differential equations

X 00 + α2 X = 0, Y 00 + β 2 Y = 0, Z 00 + γ 2 Z = 0 (38)

subject to the boundary conditions (36). The constants are related by

2mE 2 2 2
2 =α +β +γ (39)
h̄

Each of the equations (38), with its associated boundary conditions

in (36) is equivalent to the one-dimensional problem (13) with boundary
conditions (11). The normalized solutions X(x), Y (y), Z(z) can therefore
be written down in complete analogy with (24):
µ ¶1/2
2 n1 πx
Xn1 (x) = sin , n1 = 1, 2 . . .
a a
µ ¶1/2
2 n2 πy
Yn2 (y) = sin , n2 = 1, 2 . . .
b b

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 µ ¶1/2
2 n3 πz
Zn3 (x) = sin , n3 = 1, 2 . . . (40)
c c
The constants in Eq (39) are given by
n1 π n2 π n3 π
α= , β= , γ= (41)
a b c
and the allowed energy levels are therefore
µ ¶
h2 n21 n22 n23
En1 ,n2 ,n3 = + + , n1 , n2 , n3 = 1, 2 . . . (42)
8m a2 b2 c2

Three quantum numbers are required to specify the state of this three-
dimensional system. The corresponding eigenfunctions are
µ ¶1/2
8 n1 πx n2 πy n3 πz
ψn1 ,n2 ,n3 (x, y, z) = sin sin sin (43)
V a b c

where V = abc, the volume of the box. These eigenfunctions form an

orthonormal set [cf. Eq (26)] such that
Z a Z b Z c
ψn01 ,n02 ,n03 (x, y, z) ψn1 ,n2 ,n3 (x, y, z) dx dy dz
0 0 0
= δn01 ,n1 δn02 ,n2 δn03 ,n3 (44)

Note that two eigenfunctions will be orthogonal unless all three quantum
numbers match. The three-dimensonal matter waves represented by (43)
are comparable with the modes of vibration of a solid block. The nodal
surfaces are planes parallel to the sides, as shown here:

Figure 5. Nodal planes for particle

in a box, for n1 = 4, n2 = 2, n3 = 3.

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 When the box has the symmetry of a cube, with a = b = c, the energy
formula (42) simplifies to

h2
En1 ,n2 ,n3 = 2
(n21 + n22 + n23 ), n1 , n2 , n3 = 1, 2 . . . (45)
8ma

Quantum systems with symmetry generally exhibit degeneracy in their en-

ergy levels. This means that there can exist distinct eigenfunctions which
share the same eigenvalue. An eigenvalue which corresponds to a unique
eigenfunction is termed nondegenerate while one which belongs to n different
eigenfunctions is termed n-fold degenerate. As an example, we enumerate
the first few levels for a cubic box, with En1 ,n2 ,n3 expressed in units of
h2 /8ma2 :
E1,1,1 = 3 (nondegenerate)
E1,1,2 = E1,2,1 = E2,1,1 = 6 (3-fold degenerate)
E1,2,2 = E2,1,2 = E2,2,1 = 9 (3-fold degenerate)
E1,1,3 = E1,3,1 = E3,1,1 = 11 (3-fold degenerate)
E2,2,2 = 12 (nondegenerate)
E1,2,3 = E1,3,2 = E2,1,3 = E2,3,1 = E3,1,2 = E3,2,1 = 14 (6-fold degenerate)
The particle in a box is applied in statistical thermodynamics to model
the perfect gas. Each molecule is assumed to move freely within the box
without interacting with the other molecules. The total energy of N mole-
cules, in any distribution among the energy levels (45), is proportional to
1/a2 , thus
E = const V −2/3
From the differential of work dw = −p dV , we can identify

dE 2 E
p=− =
dV 3 V

But the energy of a perfect monatomic gas is known to equal 32 nRT , which
leads to the perfect gas law

pV = nRT

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