PHY3702
ASSIGNMENT 3
Quantum Physics
FULL
SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 16
,SOLUTIONS
Question 1
(a) Time-Independent Schrödinger Equation and Wave Function
The three-dimensional potential is given as:
1 1 1
𝑉(𝑥, 𝑦, 𝑧) = 𝑚𝜔𝑥2 𝑥 2 + 𝑚𝜔𝑦2 𝑦 2 + 𝑚𝜔𝑧2 𝑧 2
2 2 2
This is a 3D anisotropic harmonic oscillator potential.
The time-independent Schrödinger equation is:
ℏ2 𝜕 2 𝜓 𝜕 2 𝜓 𝜕 2 𝜓 1 1 1
− ( 2 + 2 + 2 ) + 𝑚𝜔𝑥2 𝑥 2 𝜓 + 𝑚𝜔𝑦2 𝑦 2 𝜓 + 𝑚𝜔𝑧2 𝑧 2 𝜓 = 𝐸𝜓
2𝑚 𝜕𝑥 𝜕𝑦 𝜕𝑧 2 2 2
We can solve this using the method of separation of variables. Assume a solution of the
form:
𝜓𝑛𝑥 𝑛𝑦 𝑛𝑧 (𝑥, 𝑦, 𝑧) = 𝑋𝑛𝑥 (𝑥)𝑌𝑛𝑦 (𝑦)𝑍𝑛𝑧 (𝑧)
Page 2 of 16
, Substituting this into the Schrödinger equation and dividing by the product 𝑋𝑌𝑍, we get
three independent one-dimensional equations:
ℏ2 𝑑 2 𝑋 1
− + 𝑚𝜔𝑥2 𝑥 2 𝑋 = 𝐸𝑛𝑥 𝑋
2𝑚 𝑑𝑥 2 2
ℏ2 𝑑 2 𝑌 1
− + 𝑚𝜔𝑦2 𝑦 2 𝑌 = 𝐸𝑛𝑦 𝑌
2𝑚 𝑑𝑦 2 2
ℏ2 𝑑 2 𝑍 1
− + 𝑚𝜔𝑧2 𝑧 2 𝑍 = 𝐸𝑛𝑧 𝑍
2𝑚 𝑑𝑧 2 2
Each of these is the familiar one-dimensional quantum harmonic oscillator equation. The
normalized eigenfunctions are given by:
1 𝑚𝜔𝑥 1/4 −𝑚𝜔𝑥 𝑥 2 𝑚𝜔𝑥
𝑋𝑛𝑥 (𝑥) = ( ) 𝑒 2ℏ 𝐻𝑛𝑥 (√ 𝑥)
√2𝑛𝑥 𝑛𝑥 ! 𝜋ℏ ℏ
where 𝐻𝑛𝑥 are the Hermite polynomials. Similar expressions hold for 𝑌𝑛𝑦 (𝑦) and 𝑍𝑛𝑧 (𝑧).
Therefore, the normalized wave function for the 3D oscillator is the product:
𝜓𝑛𝑥 𝑛𝑦 𝑛𝑧 (𝑥, 𝑦, 𝑧) = 𝑋𝑛𝑥 (𝑥)𝑌𝑛𝑦 (𝑦)𝑍𝑛𝑧 (𝑧)
(b) Allowed Eigenenergies
The energy eigenvalues for a 1D harmonic oscillator are:
1
𝐸𝑛𝑥 = ℏ𝜔𝑥 (𝑛𝑥 + ) , 𝑛𝑥 = 0,1,2, …
2
1
𝐸𝑛𝑦 = ℏ𝜔𝑦 (𝑛𝑦 + ) , 𝑛𝑦 = 0,1,2, …
2
1
𝐸𝑛𝑧 = ℏ𝜔𝑧 (𝑛𝑧 + ) , 𝑛𝑧 = 0,1,2, …
2
The total energy of the 3D system is the sum of the energies from each dimension:
𝐸𝑛𝑥 𝑛𝑦 𝑛𝑧 = 𝐸𝑛𝑥 + 𝐸𝑛𝑦 + 𝐸𝑛𝑧
1 1 1
𝐸𝑛𝑥 𝑛𝑦 𝑛𝑧 = ℏ𝜔𝑥 (𝑛𝑥 + ) + ℏ𝜔𝑦 (𝑛𝑦 + ) + ℏ𝜔𝑧 (𝑛𝑧 + )
2 2 2
(c) First Few Energy Levels and Degeneracies
Page 3 of 16
ASSIGNMENT 3
Quantum Physics
FULL
SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 16
,SOLUTIONS
Question 1
(a) Time-Independent Schrödinger Equation and Wave Function
The three-dimensional potential is given as:
1 1 1
𝑉(𝑥, 𝑦, 𝑧) = 𝑚𝜔𝑥2 𝑥 2 + 𝑚𝜔𝑦2 𝑦 2 + 𝑚𝜔𝑧2 𝑧 2
2 2 2
This is a 3D anisotropic harmonic oscillator potential.
The time-independent Schrödinger equation is:
ℏ2 𝜕 2 𝜓 𝜕 2 𝜓 𝜕 2 𝜓 1 1 1
− ( 2 + 2 + 2 ) + 𝑚𝜔𝑥2 𝑥 2 𝜓 + 𝑚𝜔𝑦2 𝑦 2 𝜓 + 𝑚𝜔𝑧2 𝑧 2 𝜓 = 𝐸𝜓
2𝑚 𝜕𝑥 𝜕𝑦 𝜕𝑧 2 2 2
We can solve this using the method of separation of variables. Assume a solution of the
form:
𝜓𝑛𝑥 𝑛𝑦 𝑛𝑧 (𝑥, 𝑦, 𝑧) = 𝑋𝑛𝑥 (𝑥)𝑌𝑛𝑦 (𝑦)𝑍𝑛𝑧 (𝑧)
Page 2 of 16
, Substituting this into the Schrödinger equation and dividing by the product 𝑋𝑌𝑍, we get
three independent one-dimensional equations:
ℏ2 𝑑 2 𝑋 1
− + 𝑚𝜔𝑥2 𝑥 2 𝑋 = 𝐸𝑛𝑥 𝑋
2𝑚 𝑑𝑥 2 2
ℏ2 𝑑 2 𝑌 1
− + 𝑚𝜔𝑦2 𝑦 2 𝑌 = 𝐸𝑛𝑦 𝑌
2𝑚 𝑑𝑦 2 2
ℏ2 𝑑 2 𝑍 1
− + 𝑚𝜔𝑧2 𝑧 2 𝑍 = 𝐸𝑛𝑧 𝑍
2𝑚 𝑑𝑧 2 2
Each of these is the familiar one-dimensional quantum harmonic oscillator equation. The
normalized eigenfunctions are given by:
1 𝑚𝜔𝑥 1/4 −𝑚𝜔𝑥 𝑥 2 𝑚𝜔𝑥
𝑋𝑛𝑥 (𝑥) = ( ) 𝑒 2ℏ 𝐻𝑛𝑥 (√ 𝑥)
√2𝑛𝑥 𝑛𝑥 ! 𝜋ℏ ℏ
where 𝐻𝑛𝑥 are the Hermite polynomials. Similar expressions hold for 𝑌𝑛𝑦 (𝑦) and 𝑍𝑛𝑧 (𝑧).
Therefore, the normalized wave function for the 3D oscillator is the product:
𝜓𝑛𝑥 𝑛𝑦 𝑛𝑧 (𝑥, 𝑦, 𝑧) = 𝑋𝑛𝑥 (𝑥)𝑌𝑛𝑦 (𝑦)𝑍𝑛𝑧 (𝑧)
(b) Allowed Eigenenergies
The energy eigenvalues for a 1D harmonic oscillator are:
1
𝐸𝑛𝑥 = ℏ𝜔𝑥 (𝑛𝑥 + ) , 𝑛𝑥 = 0,1,2, …
2
1
𝐸𝑛𝑦 = ℏ𝜔𝑦 (𝑛𝑦 + ) , 𝑛𝑦 = 0,1,2, …
2
1
𝐸𝑛𝑧 = ℏ𝜔𝑧 (𝑛𝑧 + ) , 𝑛𝑧 = 0,1,2, …
2
The total energy of the 3D system is the sum of the energies from each dimension:
𝐸𝑛𝑥 𝑛𝑦 𝑛𝑧 = 𝐸𝑛𝑥 + 𝐸𝑛𝑦 + 𝐸𝑛𝑧
1 1 1
𝐸𝑛𝑥 𝑛𝑦 𝑛𝑧 = ℏ𝜔𝑥 (𝑛𝑥 + ) + ℏ𝜔𝑦 (𝑛𝑦 + ) + ℏ𝜔𝑧 (𝑛𝑧 + )
2 2 2
(c) First Few Energy Levels and Degeneracies
Page 3 of 16
