The Particle in a Box Revisited

This is a one-page recap of the infinite square well (“particle in a box”) model. It’s a favourite exam model because it shows how boundary conditions lead to quantised energy levels.

1. Model and boundary conditions

The potential energy is taken to be:

• V(x) = 0 for 0 < x < L
• V(x) → ∞ for x ≤ 0 and x ≥ L

This forces:

• ψ(x) = 0 outside the well
• boundary conditions ψ(0) = 0 and ψ(L) = 0

The figure summarises the exam geometry: infinite walls at x = 0 and x = L, with standing-wave eigenstates that acquire additional interior nodes as n increases.

2. Standing-wave eigenfunctions

Inside the box, the time-independent Schrödinger equation becomes:

-(ħ²/2m)d²ψ/dx² = Eψ

Define k² = 2mE/ħ², giving:

d²ψ/dx² = -k²ψ

General solution:

ψ(x) = A sin(kx) + B cos(kx)

Apply the boundary conditions:

• ψ(0) = 0, so B = 0.
• ψ(L) = 0, so sin(kL) = 0 and therefore kL = nπ.

So

k = nπ/L (n = 1,2,3,…)

and the allowed stationary states are:

ψₙ(x) = square root of (2/L) sin((nπ x)/L) (0 < x < L)

3. Allowed energies and wavelengths

The energy is quantised:

Eₙ = n²π²ħ²/2mL² = n²h²/8mL²

Useful associated results:

λₙ = 2L/n

|pₙ| = h/λₙ = nπħ/L

4. Interpretation and classical limit

• |ψₙ(x)|² gives the probability density for position.
• Higher n gives more oscillations and more interior nodes.
• Increasing L lowers all energies (since Eₙ ∝ 1/L²).
• For large n, averaging |ψₙ|² over several rapid oscillations gives the classical uniform density 1/L.

5. Common traps

1. Allowing n = 0 (ground state is n = 1; n = 0 means ψ = 0 everywhere).
2. Forgetting the boundary conditions ψ(0) = ψ(L) = 0 for an infinite well.
3. Mixing up ψ and |ψ|².
4. Using h instead of ħ inside the Schrödinger equation.

6. Quick check

A. What is λ₃ in a box of length L?

B. If the box length doubles, what happens to E₁?