PHY3702
Assignment 3
Due 2025
Quantum Mechanics: Concepts
and Applications Zettili (2nd Ed.)
,Exercise 6.2
Problem Statement
A particle of mass m moves in the xy-plane in the potential
1 mω 2 y 2 , 0 < x < a,
V (x, y) = 2
+∞, elsewhere.
(a) Write down the time-independent Schrödinger equation for this particle and reduce
it to a set of familiar one-dimensional equations. (b) Find the normalized eigenfunctions
and the eigenenergies.
Part (a): Time-Independent Schrödinger Equation
Step 1: Write the time-independent Schrödinger equation The
time-independent Schrödinger equation in two dimensions is:
ℏ2 ∂ 2ψ ∂ 2ψ
− + 2 + V (x, y)ψ(x, y) = Eψ(x, y).
2m ∂x2 ∂y
Step 2: Analyze the potential and boundary conditions The potential is
V (x, y) = 21 mω 2 y 2 for 0 < x < a, and V (x, y) = +∞ elsewhere. The infinite potential
implies ψ(x, y) = 0 for x ≤ 0 or x ≥ a. Inside 0 < x < a, the potential is harmonic in y,
suggesting a particle in a box in x and a harmonic oscillator in y.
Step 3: Use separation of variables Assume ψ(x, y) = X(x)Y (y). Substitute into
the Schrödinger equation:
ℏ2 1
− (X ′′ (x)Y (y) + X(x)Y ′′ (y)) + mω 2 y 2 X(x)Y (y) = EX(x)Y (y).
2m 2
Divide by X(x)Y (y):
ℏ2 X ′′ (x) Y ′′ (y)
1
− + + mω 2 y 2 = E.
2m X(x) Y (y) 2
Rearrange:
ℏ2 X ′′ (x) ℏ2 Y ′′ (y) 1
− =E+ − mω 2 y 2 .
2m X(x) 2m Y (y) 2
Set each side equal to a constant Ex :
ℏ2 X ′′ (x) ℏ2 Y ′′ (y) 1
− = Ex , E+ − mω 2 y 2 = Ex .
2m X(x) 2m Y (y) 2
1
, Step 4: Solve the x-equation (particle in a box) The x-equation is:
2mEx
X ′′ (x) + X(x) = 0, X(0) = X(a) = 0.
ℏ2
Solutions are:
r
2 nπx n2 π 2 ℏ2
Xn (x) = sin , Ex = , n = 1, 2, . . . .
a a 2ma2
Step 5: Solve the y-equation (harmonic oscillator) The y-equation is:
ℏ2 ′′ 1
− Y (y) + mω 2 y 2 Y (y) = Ey Y (y), Ey = E − Ex .
2m 2
This is the harmonic oscillator equation with solutions:
mω 1/4 r
1 mω mωy 2
Yny (y) = p Hny y e− 2ℏ ,
πℏ 2ny ny ! ℏ
1
E y = ny + ℏω, ny = 0, 1, 2, . . . .
2
Step 6: Combine the results Total energy:
n2 π 2 ℏ2
1
E = Ex + Ey = + ny + ℏω.
2ma2 2
Wave function:
ψn,ny (x, y) = Xn (x)Yny (y).
Final Answer
q for (a) The Schrödinger equation reduces to: - x: Particle in a box,
2 π 2 ℏ2
Xn (x) = a sin nπx
2
, Ex = n2ma
a 2 , n = 1, 2, . . . . - y: Harmonic oscillator, Yny (y) as
above, Ey = ny + 21 ℏω, ny = 0, 1, . . . .
2
Assignment 3
Due 2025
Quantum Mechanics: Concepts
and Applications Zettili (2nd Ed.)
,Exercise 6.2
Problem Statement
A particle of mass m moves in the xy-plane in the potential
1 mω 2 y 2 , 0 < x < a,
V (x, y) = 2
+∞, elsewhere.
(a) Write down the time-independent Schrödinger equation for this particle and reduce
it to a set of familiar one-dimensional equations. (b) Find the normalized eigenfunctions
and the eigenenergies.
Part (a): Time-Independent Schrödinger Equation
Step 1: Write the time-independent Schrödinger equation The
time-independent Schrödinger equation in two dimensions is:
ℏ2 ∂ 2ψ ∂ 2ψ
− + 2 + V (x, y)ψ(x, y) = Eψ(x, y).
2m ∂x2 ∂y
Step 2: Analyze the potential and boundary conditions The potential is
V (x, y) = 21 mω 2 y 2 for 0 < x < a, and V (x, y) = +∞ elsewhere. The infinite potential
implies ψ(x, y) = 0 for x ≤ 0 or x ≥ a. Inside 0 < x < a, the potential is harmonic in y,
suggesting a particle in a box in x and a harmonic oscillator in y.
Step 3: Use separation of variables Assume ψ(x, y) = X(x)Y (y). Substitute into
the Schrödinger equation:
ℏ2 1
− (X ′′ (x)Y (y) + X(x)Y ′′ (y)) + mω 2 y 2 X(x)Y (y) = EX(x)Y (y).
2m 2
Divide by X(x)Y (y):
ℏ2 X ′′ (x) Y ′′ (y)
1
− + + mω 2 y 2 = E.
2m X(x) Y (y) 2
Rearrange:
ℏ2 X ′′ (x) ℏ2 Y ′′ (y) 1
− =E+ − mω 2 y 2 .
2m X(x) 2m Y (y) 2
Set each side equal to a constant Ex :
ℏ2 X ′′ (x) ℏ2 Y ′′ (y) 1
− = Ex , E+ − mω 2 y 2 = Ex .
2m X(x) 2m Y (y) 2
1
, Step 4: Solve the x-equation (particle in a box) The x-equation is:
2mEx
X ′′ (x) + X(x) = 0, X(0) = X(a) = 0.
ℏ2
Solutions are:
r
2 nπx n2 π 2 ℏ2
Xn (x) = sin , Ex = , n = 1, 2, . . . .
a a 2ma2
Step 5: Solve the y-equation (harmonic oscillator) The y-equation is:
ℏ2 ′′ 1
− Y (y) + mω 2 y 2 Y (y) = Ey Y (y), Ey = E − Ex .
2m 2
This is the harmonic oscillator equation with solutions:
mω 1/4 r
1 mω mωy 2
Yny (y) = p Hny y e− 2ℏ ,
πℏ 2ny ny ! ℏ
1
E y = ny + ℏω, ny = 0, 1, 2, . . . .
2
Step 6: Combine the results Total energy:
n2 π 2 ℏ2
1
E = Ex + Ey = + ny + ℏω.
2ma2 2
Wave function:
ψn,ny (x, y) = Xn (x)Yny (y).
Final Answer
q for (a) The Schrödinger equation reduces to: - x: Particle in a box,
2 π 2 ℏ2
Xn (x) = a sin nπx
2
, Ex = n2ma
a 2 , n = 1, 2, . . . . - y: Harmonic oscillator, Yny (y) as
above, Ey = ny + 21 ℏω, ny = 0, 1, . . . .
2
