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Lecture 8
Particle in a box
(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has
been developed and made available online by work supported jointly by University of Illinois, the
National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus
Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of the sponsoring agencies.
The particle in a box


This is the simplest analytically soluble
example of the Schrödinger equation and
holds great importance in chemistry and
physics.
Each of us must be able to set up the
equation and boundary conditions, solve the
equation, and characterize and explain the
solutions.
The particle in a box


A particle of mass m is
confined on a line segment
of length L.
The Schrödinger equation
is generally:
2
æ
ö
2
çè - 2m Ñ +V ÷ø Y = EY

V = 0 (0 ≤ x ≤ L)
V = ∞ (elsewhwere)
The particle in a box

The Schrödinger equation of this problem is:
d2
Y(x) = EY(x)
2
2m dx
2
0£ x£ L
The particle in a box

What functions stay the same form after two
differentiations?
2
d
2
eikx : 2 eikx   ik  eikx  k 2eikx
dx
d2
sin kx : 2 sin kx  k 2 sin kx
dx

2
d
ekx : 2 ekx  k 2ekx
dx
d2
cos kx : 2 cos kx  k 2 cos kx
dx
And any linear combinations of these with the
same k.
Boundary conditions

Remember there are certain
conditions that a function must
fulfill for it to be a wave
function.




continuous
(smooth)
single valued
square integrable
Boundary conditions





Outside and at the boundary of
the box (x = 0 and x = L), Ψ = 0.
Owing to the continuousness
condition, we must demand that
Ψ(x=0) = 0 and Ψ(x=L) = 0.
ekx cannot satisfy these
simultaneously (ek0 = 1).
cos x is not promising (cos 0 =
1), either.
(Other than this, all four
functions are single-valued and
finite).
Boundary conditions

Boundary conditions are the restrictions imposed on
the solutions of differential equations.


They are typically but not necessarily the numerical values that
solutions must have at the boundaries of their domain.
For example, in classical mechanics, they may be the initial
positions and velocities of the constituent masses; in fluid
dynamics, they may be the shape of the container of the fluid.
Differential equation
<
Boundary conditions
Darth Vader
<
Chancellor
The acceptable solutions

Let us use the most promising “sin kx”:
d
k
sin kx =
sin kx
2
2m dx
2m
2
2
2 2
sin k 0  0 (satisfied already)
sin kL  0 (needs to be enforced)
2 2
d 2 ikx -ikx
k ikx -ikx
(e - e ) =
(e - e )
2
2m dx
2m
2
eik 0  eik 0  (cos k 0  i sin k 0)  (cos k 0  i sin k 0)  0 (ok)
eikL  eikL  2i sin kL  0 (needs to be enforced)
These are two representations of the identical functions
The acceptable solutions

The boundary condition requires sin kL = 0.
kL = np (n = 1,2,… )



n is called a quantum number.
We did not include n = 0 because this makes
the wave function zero everywhere (sin 0x =
0, not normalizable or no particle!).
We did not include negative integers for n
because they lead to the same wave
functions (sin(–kx) = – sin kx).
The particle in a box

We have the solution:
d2
Y = EY
2
2m dx
2
np
Y = N sin
x (n = 1,2,… )
L
2
æ np ö
k
hn
En =
=
=
ç
÷
2m 2m è L ø
8mL2
2 2
2
2 2
+
 (0)   ( L)  0
æ i nLp x -i nLp x ö
Y = N¢çe
-e
÷ (n = 1,2,… )
è
ø
Quantized! Note that boundary condition is
responsible for quantization of energy.
The particle in a box


Now the energy is quantized
because of the boundary
conditions.
The wave functions are the
standing waves. The more
rapidly oscillating the wave
function is and the more the
nodes, the higher the energy.
The zero-point energy



The lowest allowed energy is
nonzero because n = 0 is not a
solution.
This lowest, nonzero energy is called
the zero-point energy. This is a
quantum-mechanical effect.
The particle in a box can never be
completely still (zero momentum =
zero energy)! This is also expected
from the uncertainty principle
(consider the limit L→0).
The ground and excited states


The lowest state corresponds to the
ground state and the n = 2 and higherlying states are the excited states.
The excitation energy from n to n+1 state
is
2
2
2 2
2
h (n  1)
hn
h
En1  En 

 (2n  1)
2
2
8mL
8mL
8mL2

which is quantized.
However, the effect of quantization will
become smaller as L → ∞. In a
macroscopic scale (L very large) or for a
free space (L = ∞), energy and energy
differences become continuous
(quantum classical correspondence).
Quantum in nature
The particle
in a box
Why is
carrot
orange?
β-carotene
Quantum in nature
The particle
in a box
vitamin A
Why do I
have to eat
carrot?
retinal
Normalization

There are two ways of doing this:

Using the original sin kx form.
N
   d 
*
 L 2 n x 
   sin
dx 
0
L


1/ 2
1/ 2
L
 
2
1/ 2
1
1
Differentiate both sides to verify
sin
axdx

x

sin
2
ax

C

this. Use cos2x = cos2x – sin2x.
2
4a
2

N
Using the alternative eikx – e–ikx form

*

 d

1/ 2
 1

 2i
sin kx =
2

L
0

 e

1 ikx -ikx
(e - e )
2i
n
2i x
L
11 e
n
2i x
L
 
 dx 
 
ei 2np = 1 (n is an integer)
1/ 2
L
 
2
1/ 2
Exercise


What is the average value of the linear
momentum of a particle in a box with
quantum number n?
Hints:
1
1
2
 sin
axdx 
2
x
4a
sin 2ax  C
2
n x
*ˆ
    dx
 ( x) 
sin
L
L
np
np
æ
1 2 i L x -i L x ö
Y(x) =
e
-e
ç
÷
2i L è
ø
Exercise

Using the sin kx form of the wave function.
æ 2ö L
¶ö
*æ
p = ò Y p̂Y dx = ç ÷ ò sin kx ç -i
sin kx dx
÷
0
L
¶x
è ø
è
ø
*
np
k=
L
2 L
-i k L
= ò sin kx(-i k)cos kx dx =
sin 2kx dx = 0
ò
0
0
L
L

Alternatively,
1 æ 2ö
p = ò Y p̂Y dx = ç ÷
2i è L ø
*
1/2
2
ò
L
0
æ
¶ ö ikx -ikx
(e ikx - e -ikx )* ç -i
(e - e ) dx
÷
¶x ø
è
1 L -ikx ikx
k L -2ikx 2ikx
ikx
-ikx
=
(e - e ) k(e + e ) dx =
(e
- e ) dx = 0
ò
ò
0
0
2L
2L
More on the momentum

The momentum operator and its eigenfunctions are:
¶
p̂x = -i
¶x

p̂x e = ke
ikx
ikx
The wave function has equal weight on eikx and e–ikx.
1/ 2
1 2
n
ikx
 ikx

 
2i  L 
e
e
,
k
L
The measurement of momentum gives ħk or –ħk
with an equal probability. This is consistent with the
picture that a particle bouncing back and forth.
Probability density

The probability density is
2 2 n x
  sin
L
L
2


Unlike the classical “bouncing particle”
picture, there are places with less
probability (even zero probability at
nodes).
The higher n (quantum number), the more
uniform the probability density becomes,
approaching the uniform probability
density of the classical limit (quantum
classical correspondence).
Summary





The particle in a box: set up the equation and
boundary condition, solve it, and characterize
the solution.
Boundary conditions; quantum number;
quantization of energies.
Wave functions as standing waves.
Zero-point energy and the uncertainty
principle.
Quantum classical correspondence.