Solving the Schrödinger Equation

Complete analytical solution for infinite square well and methodology for finite square well with numerical approaches

The General Schrödinger Equation

-ħ²/2m · d²ψ/dx² + V(x)ψ(x) = Eψ(x)

This is the time-independent Schrödinger equation in one dimension, where:

Symbol	Meaning	Units
ħ	Reduced Planck constant (ħ = h/2π)	J·s
m	Mass of the particle	kg
ψ(x)	Wavefunction (probability amplitude)	m⁻¹/²
V(x)	Potential energy function	J
E	Total energy of the particle	J

Infinite Square Well

Also called "particle in a box." Perfectly rigid walls with infinite potential outside.

V(x) = 0, for 0 < x < L
V(x) = ∞, for x ≤ 0 or x ≥ L

Boundary Conditions: ψ(0) = 0 and ψ(L) = 0

Solution Type: Exact analytical solution

Difficulty: Undergraduate quantum mechanics

Finite Square Well

More realistic with finite potential barriers. Allows quantum tunneling.

V(x) = 0, for -L/2 < x < L/2
V(x) = V₀, for x ≤ -L/2 or x ≥ L/2
where V₀ > 0 is finite

Boundary Conditions: ψ and dψ/dx must be continuous at x = ±L/2

Solution Type: Transcendental equations requiring numerical solution

Difficulty: Advanced undergraduate/graduate level

Solving Infinite Square Well

1

Set up the differential equation

Inside the well (0 < x < L), V(x) = 0, so the Schrödinger equation simplifies to:

d²ψ/dx² + k²ψ = 0, where k² = 2mE/ħ²

2

General solution

This is a standard second-order differential equation with general solution:

ψ(x) = A sin(kx) + B cos(kx)

where A and B are constants determined by boundary conditions.

3

Apply boundary conditions

At x = 0: ψ(0) = A sin(0) + B cos(0) = B = 0, so B = 0.

At x = L: ψ(L) = A sin(kL) = 0.

Since A cannot be zero (or we get trivial solution ψ(x) = 0), we require sin(kL) = 0.

4

Quantization condition

sin(kL) = 0 implies kL = nπ, where n = 1, 2, 3, ...

Thus: kₙ = nπ/L

5

Energy quantization

Since k² = 2mE/ħ², we have:

Eₙ = (ħ²kₙ²)/(2m) = (n²π²ħ²)/(2mL²)

This shows energy is quantized with n = 1, 2, 3, ...

6

Normalize wavefunction

ψₙ(x) = A sin(nπx/L). Normalization requires ∫₀ᴸ |ψₙ(x)|² dx = 1.

Solving gives A = √(2/L).

Final Solution for Infinite Square Well

ψₙ(x) = √(2/L) · sin(nπx/L)

Eₙ = (n²π²ħ²)/(2mL²), n = 1, 2, 3, ...

Quantum Number (n)	Energy (Relative)	Nodes inside well	Wavefunction symmetry
1 (Ground state)	E₁ = π²ħ²/(2mL²)	0	Even about x = L/2
2 (First excited)	E₂ = 4E₁	1	Odd about x = L/2
3	E₃ = 9E₁	2	Even about x = L/2
n	Eₙ = n²E₁	n-1	Even if n odd, Odd if n even

Solving Finite Square Well

The finite well problem is symmetric about x = 0, so we consider three regions:

Region I: x < -L/2

V(x) = V₀, E < V₀ for bound states

d²ψ/dx² - κ²ψ = 0

where κ² = 2m(V₀ - E)/ħ² > 0

Solution: ψᴵ(x) = A e^{κx} (discard divergent term e^{-κx})

Region II: -L/2 < x < L/2

V(x) = 0

d²ψ/dx² + k²ψ = 0

where k² = 2mE/ħ² > 0

Solution: ψᴵᴵ(x) = B cos(kx) or C sin(kx)

Region III: x > L/2

V(x) = V₀

d²ψ/dx² - κ²ψ = 0

Solution: ψᴵᴵᴵ(x) = D e^{-κx} (discard divergent term e^{κx})

1

Even and odd solutions

Due to symmetry, solutions separate into even and odd parity:

Even solutions: ψᴵᴵ(x) = B cos(kx)

Odd solutions: ψᴵᴵ(x) = C sin(kx)

2

Continuity conditions

At x = L/2, both ψ(x) and dψ/dx must be continuous:

1. ψᴵᴵ(L/2) = ψᴵᴵᴵ(L/2)

2. dψᴵᴵ/dx at x = L/2 = dψᴵᴵᴵ/dx at x = L/2

3

Transcendental equations

For even solutions: k tan(kL/2) = κ

For odd solutions: -k cot(kL/2) = κ

Key Relationship

Since k² = 2mE/ħ² and κ² = 2m(V₀ - E)/ħ², we have:

k² + κ² = 2mV₀/ħ² (constant)

This defines a circle in (k, κ) space. The intersections of this circle with the curves from the transcendental equations give the allowed k and κ values.

4

Graphical/numerical solution

Define dimensionless variables:

ξ = kL/2, η = κL/2

Then ξ² + η² = mV₀L²/(2ħ²) ≡ R²

Even solutions: η = ξ tan(ξ)

Odd solutions: η = -ξ cot(ξ)

Solve graphically by finding intersections of these curves with circle ξ² + η² = R².

Finite Well Solution Properties

Aspect	Finite Well Behavior	Difference from Infinite Well
Number of bound states	Finite, depends on V₀ and L	Infinite well has infinite bound states
Ground state energy	E₁ > 0 but less than infinite well E₁	Lower than infinite case due to tunneling
Wavefunction at boundaries	Non-zero, decays exponentially outside	Exactly zero at boundaries in infinite case
Energy spacing	Not exactly n² pattern, spacing decreases	Perfect E ∝ n² in infinite case
Tunneling probability	Non-zero (particle found outside well)	Zero probability outside infinite well

Numerical Solution Approach

For finite wells, we typically solve numerically. Here's a Python approach:

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import fsolve

# Parameters
m = 9.11e-31      # electron mass (kg)
hbar = 1.0546e-34 # reduced Planck constant
V0 = 10.0         # well depth (eV)
L = 1e-9          # well width (m)

# Convert V0 to Joules
V0_J = V0 * 1.602e-19

# Define transcendental equations for even states
def even_eq(xi):
    """Equation for even solutions: eta = xi*tan(xi)"""
    R_sq = m * V0_J * (L**2) / (2 * hbar**2)
    eta_sq = R_sq - xi**2
    if eta_sq < 0:
        return np.inf
    eta = np.sqrt(eta_sq)
    return eta - xi * np.tan(xi)

# Define transcendental equations for odd states
def odd_eq(xi):
    """Equation for odd solutions: eta = -xi*cot(xi)"""
    R_sq = m * V0_J * (L**2) / (2 * hbar**2)
    eta_sq = R_sq - xi**2
    if eta_sq < 0:
        return np.inf
    eta = np.sqrt(eta_sq)
    return eta + xi / np.tan(xi)

# Find solutions graphically or using root-finding
xi_vals = np.linspace(0.01, np.sqrt(m*V0_J*L**2/(2*hbar**2))-0.01, 1000)
even_roots = []
odd_roots = []

# Simple root-finding by sign changes
for i in range(len(xi_vals)-1):
    if even_eq(xi_vals[i]) * even_eq(xi_vals[i+1]) < 0:
        root = fsolve(even_eq, (xi_vals[i] + xi_vals[i+1])/2)[0]
        even_roots.append(root)
    if odd_eq(xi_vals[i]) * odd_eq(xi_vals[i+1]) < 0:
        root = fsolve(odd_eq, (xi_vals[i] + xi_vals[i+1])/2)[0]
        odd_roots.append(root)

# Calculate energies from xi values
even_energies = [(2*xi/L)**2 * hbar**2 / (2*m) for xi in even_roots]
odd_energies = [(2*xi/L)**2 * hbar**2 / (2*m) for xi in odd_roots]

print(f"Found {len(even_roots)} even bound states and {len(odd_roots)} odd bound states")

Key Differences Summary

Feature	Infinite Square Well	Finite Square Well
Mathematical Solution	Exact analytic: ψₙ = √(2/L) sin(nπx/L)	Transcendental equations requiring numerical solution
Energy Formula	Eₙ = n²π²ħ²/(2mL²), exact	No closed form, must solve k tan(kL/2)=κ or -k cot(kL/2)=κ
Number of States	Infinite bound states (n=1,2,3,...)	Finite number of bound states, depends on V₀L²
Ground State Energy	E₁ = π²ħ²/(2mL²)	0 < E₁ < π²ħ²/(2mL²) (lower due to tunneling)
Wavefunction Outside	Exactly zero	Exponential decay: ψ ∝ e^{-κ|x|}
Tunneling	Not possible	Possible, probability ∝ e^{-2κL}
Physical Realism	Idealized model for teaching	More realistic, describes actual quantum systems

Physical Insight: As V₀ → ∞, the finite well solutions approach the infinite well solutions. The exponential decay constant κ → ∞, making the wavefunction at the boundaries approach zero, and the transcendental equations reduce to sin(kL/2)=0 or cos(kL/2)=0, giving kL = nπ.

Quantum Mechanics Solutions | The infinite square well has exact solutions while the finite well requires numerical methods

For deeper exploration: Implement the numerical solution, compare energy levels for different well depths, or extend to 2D/3D wells.