Quantum Wells & Tunneling Toolkit (IPhO Quantum)

IPhO quantum toolbox for 1D wells and barriers: region solutions, boundary matching, parity tricks, and fast tunneling estimates.

  • International Physics Olympiad preparation
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Wells and barriers are the standard “solve once, reuse forever” quantum problems. In olympiad settings, you are rarely asked to solve a new differential equation from scratch. You are asked to (1) write the correct region solutions, (2) apply boundary conditions correctly, and (3) estimate tunneling or bound-state scales quickly.

The 20-second toolkit
  1. Sketch V(x) and mark regions where E > V (oscillatory) and E < V (exponential).
  2. Write the region solutions using k or κ.
  3. Enforce boundary conditions: ψ and dψ/dx continuous (for finite steps).
  4. Use symmetry: even/odd parity in symmetric wells halves the work.
  5. For tunneling, the exponent is the main result: T ∼ exp(-2κ a).

1. Definitions (Must Know)

A. Time-independent Schrodinger equation (1D)

-(ħ²/2m)d²ψ/dx² + V(x)ψ = Eψ

In a region of constant potential V, define: k = square root of ((2m(E-V))/ħ²) (E > V) κ = square root of ((2m(V-E))/ħ²) (E < V)

B. Standard region solutions

  • If E > V (classically allowed): oscillatory ψ(x) = Ae^ikx + Be^(-ikx)
  • If E < V (classically forbidden): exponential ψ(x) = Ce^(κ x) + De^(-κ x)

C. Boundary conditions (finite potential steps)

For a finite jump in V(x) at some boundary:

  • ψ is continuous,
  • dψ/dx is continuous.

D. Reflection and transmission

For a 1D scattering setup, define:

  • transmission coefficient T (probability of passing through),
  • reflection coefficient R (probability of reflecting).

For a single barrier with no loss: T + R = 1.

E. Tunneling estimate (rectangular or WKB)

For a rectangular barrier of width a with E < V₀: T ∼ exp(-2κ a), κ = square root of ((2m(V₀-E))/ħ²)

More generally (WKB form), between turning points x₁ and x₂: T ∼ exp(-2∫_x₁^x₂ square root of ((2m(V(x)-E))/ħ²) dx)

2. Key Ideas (What Earns Marks)

  • Classify regions by E-V. The sign decides the functional form.
  • Use parity in symmetric wells. Even/odd solutions reduce unknown constants.
  • Quantization comes from “no blow-up”. Bound states require ψ to decay outside the well.
  • Tunneling is exponential. Most problems only need the exponent correctly.
  • Check limits. As barrier width grows, T must decrease rapidly; as barrier height drops, T must increase.

3. Detailed Explanations

A. Infinite square well (the reference model)

For an infinite well of width L (wavefunction zero at both walls), the allowed energies are: Eₙ = n²π²ħ²/2mL², n = 1,2,3,…

Key scaling: Eₙ ∝ n² and Eₙ ∝ 1/L².

B. Finite square well (what changes)

For a finite well, the wavefunction leaks into the classically forbidden region, so:

  • energy levels shift downward relative to an infinite well of the same width,
  • there are only finitely many bound states (set by depth and width),
  • matching at boundaries gives transcendental equations (often solved graphically or approximated).

Fast estimate for the number of bound states (order-of-magnitude): N ∼ (L/π) square root of (2mV₀/ħ²)

C. Rectangular barrier tunneling (what to remember)

For E < V₀ and a “thick” barrier, the main result is: T ∼ exp(-2κ a)

This one line often earns most of the marks because it captures the dominant dependence on:

  • width a,
  • mass m,
  • barrier excess V₀-E.

D. Quantum reflection even when E > V

Even if E > V everywhere, there can be reflection at a sharp step because the wavelength changes (mismatch of k). This shows up in the boundary-condition algebra and is a classic exam trap.

4. Common Mistakes

  • Writing exponential solutions in the wrong region (mixing up k and κ).
  • Forgetting that bound states must not diverge at infinity (set the growing exponential coefficient to zero).
  • Dropping the derivative continuity condition at finite steps.
  • Treating tunneling like a small linear correction (it is exponential in a).
  • Mixing amplitudes with probabilities: T is a probability, not a wave amplitude.

5. Exam Tips

  • Always start by sketching V(x) and labeling regions with k or κ.
  • If the potential is symmetric, state “use even/odd parity” explicitly.
  • For tunneling estimates, show the exponent clearly: ln T ∼ -2κ a.
  • Use quick constants if allowed: ħ²/2mₑ ≈ 3.81 eV Ų, hc ≈ 1240 eV nm.

6. Worked Examples

A. Infinite well energy scale (electron in a 1 nm box)

An electron is in a 1D infinite square well of width L = 1.0 nm. Estimate E₁ and the transition energy Δ E = E₂-E₁ (in eV).

Click here to show/hide solution

For an infinite well: Eₙ = n²π²ħ²/2mL²

Use ħ²/(2mₑ) ≈ 3.81 eV Ų and L = 1.0 nm = 10 Å: E₁ ≈ (π² × 3.81)/10² eV ≈ (9.87 × 3.81)/100 eV ≈ 0.38 eV

Since Eₙ ∝ n², E₂ = 4E₁ ≈ 1.5 eV.

So: Δ E = E₂-E₁ = 3E₁ ≈ 1.1 eV

B. Tunneling probability through a rectangular barrier (exponent estimate)

An electron with energy E = 2 eV hits a barrier of height V₀ = 5 eV and width a = 0.50 nm. Estimate the transmission probability.

Click here to show/hide solution

Here V₀-E = 3 eV and E < V₀.

Use: T ∼ exp(-2κ a), κ = square root of ((2mₑ(V₀-E))/ħ²)

A useful numeric form for electrons is: κ (nm⁻¹) ≈ 5.12 square root of ((V₀-E) (eV)) So: κ ≈ 5.12 square root of 3 nm⁻¹ ≈ 8.9 nm⁻¹

Then: 2κ a ≈ 2 × 8.9 × 0.50 ≈ 8.9 Hence: T ∼ e^(-8.9) ∼ 1 × 10⁻⁴

Pre-factors can change this by a factor of a few, but the exponential is the main result.

7. Mind Stretchers

A. Reflection at a potential step even when E > V₀

A particle with E > V₀ encounters a sudden step from V = 0 to V = V₀. Show that there is still reflection, and find the reflection coefficient in terms of k₁ and k₂.

Click here to show/hide hint and solution

Region 1: k₁ = square root of (2mE/ħ²), region 2: k₂ = square root of (2m(E-V₀)/ħ²).

Write: ψ₁ = Ae^ik₁x + Be^(-ik₁x), ψ₂ = Ce^ik₂x

Match ψ and dψ/dx at the boundary. Solving the two equations gives the amplitude reflection coefficient: B/A = (k₁-k₂)/(k₁ + k₂)

So the probability reflection coefficient is: R = ((k₁-k₂)/(k₁ + k₂))² This is nonzero whenever k₁ ≠ k₂ (a wavelength mismatch).

  • In a double-barrier structure, transmission can become close to 1 at special energies (resonant tunneling). What phase condition is being satisfied?
  • For a finite well, why do higher bound states leak more strongly into the classically forbidden region?

8. Practice

Practice
  • Do 2 short drills: one infinite well energy scaling, one rectangular barrier tunneling exponent.
  • Then do 1 symmetry drill: write even and odd solutions for a symmetric well and state the boundary conditions.
Syllabus and review details

No official syllabus alignment is listed for this lesson.