PHY3702
ASSIGNMENT 2
Quantum Physics
FULL
SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 27
,SOLUTIONS:
Question 1
(a) The potential energy is
0, 0 < 𝑥 < 𝑎,
𝑉(𝑥) = {
∞, elsewhere.
The time-independent Schrödinger equation inside the box is
ℏ2 𝑑 2 𝜓(𝑥)
− = 𝐸𝜓(𝑥).
2𝑚 𝑑𝑥 2
This can be written as
𝑑2𝜓
2
+ 𝑘 2 𝜓 = 0,
𝑑𝑥
where
2𝑚𝐸
𝑘2 = .
ℏ2
The general solution is
Page 2 of 27
, 𝜓(𝑥) = 𝐴sin(𝑘𝑥) + 𝐵cos(𝑘𝑥).
Since the wavefunction must vanish at the walls,
𝜓(0) = 0.
Therefore,
𝐵 = 0.
Applying the second boundary condition,
𝜓(𝑎) = 𝐴sin(𝑘𝑎) = 0.
For a non-trivial solution,
sin(𝑘𝑎) = 0,
which gives
𝑘𝑎 = 𝑛𝜋, 𝑛 = 1,2,3, …
Hence
𝑛𝜋
𝑘= .
𝑎
The allowed energy levels are
ℏ2 𝑘 2 𝑛 2 𝜋 2 ℏ 2
𝐸𝑛 = = , 𝑛 = 1,2,3, …
2𝑚 2𝑚𝑎2
The corresponding wavefunctions are
𝑛𝜋𝑥
𝜓𝑛 (𝑥) = 𝐴sin ( ).
𝑎
Page 3 of 27
ASSIGNMENT 2
Quantum Physics
FULL
SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 27
,SOLUTIONS:
Question 1
(a) The potential energy is
0, 0 < 𝑥 < 𝑎,
𝑉(𝑥) = {
∞, elsewhere.
The time-independent Schrödinger equation inside the box is
ℏ2 𝑑 2 𝜓(𝑥)
− = 𝐸𝜓(𝑥).
2𝑚 𝑑𝑥 2
This can be written as
𝑑2𝜓
2
+ 𝑘 2 𝜓 = 0,
𝑑𝑥
where
2𝑚𝐸
𝑘2 = .
ℏ2
The general solution is
Page 2 of 27
, 𝜓(𝑥) = 𝐴sin(𝑘𝑥) + 𝐵cos(𝑘𝑥).
Since the wavefunction must vanish at the walls,
𝜓(0) = 0.
Therefore,
𝐵 = 0.
Applying the second boundary condition,
𝜓(𝑎) = 𝐴sin(𝑘𝑎) = 0.
For a non-trivial solution,
sin(𝑘𝑎) = 0,
which gives
𝑘𝑎 = 𝑛𝜋, 𝑛 = 1,2,3, …
Hence
𝑛𝜋
𝑘= .
𝑎
The allowed energy levels are
ℏ2 𝑘 2 𝑛 2 𝜋 2 ℏ 2
𝐸𝑛 = = , 𝑛 = 1,2,3, …
2𝑚 2𝑚𝑎2
The corresponding wavefunctions are
𝑛𝜋𝑥
𝜓𝑛 (𝑥) = 𝐴sin ( ).
𝑎
Page 3 of 27
