PHY3702 Assignment 2 Solutions 2026
UNISA
DUE DATE: 15 JULY 2026
,PHY3702 Assignment 2 – Complete Solutions BY iQ Level
Question 1
The potential well is
0, 0 < 𝑥 < 𝑎
𝑉(𝑥) = {
∞, elsewhere
(Inside the well the potential must be zero, otherwise bound states do not exist.)
Question 1(a)
: Schrödinger equation
Inside the well:
ℏ2 𝑑 2 𝜓
− = 𝐸𝜓
2𝑚 𝑑𝑥 2
Rearranging:
𝑑2 𝜓
+ 𝑘2𝜓 = 0
𝑑𝑥 2
where
2𝑚𝐸
𝑘2 =
ℏ2
The general solution is
𝜓(𝑥) = 𝐴sin(𝑘𝑥) + 𝐵cos(𝑘𝑥)
Apply boundary conditions
At 𝑥 = 0,
𝜓(0) = 0
therefore
𝐵=0
, Hence
𝜓(𝑥) = 𝐴sin(𝑘𝑥)
At 𝑥 = 𝑎,
𝜓(𝑎) = 0
thus
𝐴sin(𝑘𝑎) = 0
For non-trivial solutions,
sin(𝑘𝑎) = 0
which gives
𝑘𝑎 = 𝑛𝜋
where
𝑛 = 1,2,3, …
Therefore
𝑛𝜋
𝑘𝑛 =
𝑎
Energy levels
Using
ℏ2 𝑘 2
𝐸=
2𝑚
ℏ2 𝑛𝜋 2
𝐸𝑛 = ( )
2𝑚 𝑎
Hence
𝑛 2 𝜋 2 ℏ2
𝐸𝑛 =
2𝑚𝑎2
UNISA
DUE DATE: 15 JULY 2026
,PHY3702 Assignment 2 – Complete Solutions BY iQ Level
Question 1
The potential well is
0, 0 < 𝑥 < 𝑎
𝑉(𝑥) = {
∞, elsewhere
(Inside the well the potential must be zero, otherwise bound states do not exist.)
Question 1(a)
: Schrödinger equation
Inside the well:
ℏ2 𝑑 2 𝜓
− = 𝐸𝜓
2𝑚 𝑑𝑥 2
Rearranging:
𝑑2 𝜓
+ 𝑘2𝜓 = 0
𝑑𝑥 2
where
2𝑚𝐸
𝑘2 =
ℏ2
The general solution is
𝜓(𝑥) = 𝐴sin(𝑘𝑥) + 𝐵cos(𝑘𝑥)
Apply boundary conditions
At 𝑥 = 0,
𝜓(0) = 0
therefore
𝐵=0
, Hence
𝜓(𝑥) = 𝐴sin(𝑘𝑥)
At 𝑥 = 𝑎,
𝜓(𝑎) = 0
thus
𝐴sin(𝑘𝑎) = 0
For non-trivial solutions,
sin(𝑘𝑎) = 0
which gives
𝑘𝑎 = 𝑛𝜋
where
𝑛 = 1,2,3, …
Therefore
𝑛𝜋
𝑘𝑛 =
𝑎
Energy levels
Using
ℏ2 𝑘 2
𝐸=
2𝑚
ℏ2 𝑛𝜋 2
𝐸𝑛 = ( )
2𝑚 𝑎
Hence
𝑛 2 𝜋 2 ℏ2
𝐸𝑛 =
2𝑚𝑎2