Energy of particle in infinite well

A particle of mass m moves freely inside a one-dimensional box between two impenetrable walls at x = 0 and x = a.

 V=inf |                    | V=inf
       |   V = 0 inside     |
       |____________________|
       0                    a

Potential: V = 0 for 0 < x < a, V = ∞ for x ≤ 0 and x ≥ a.

Step 1: wave function outside

The particle cannot be outside, so ψ = 0 for x ≤ 0 and x ≥ a.

Step 2: Schrödinger equation inside

Time-independent equation with V = 0:

−(ħ²/2m) d²ψ/dx² = Eψ → d²ψ/dx² + k²ψ = 0, where k² = 2mE/ħ²

Step 3: solution

ψ(x) = A sin kx + B cos kx.

Step 4: boundary conditions

• ψ(0) = 0: B = 0. So ψ = A sin kx.
• ψ(a) = 0: A sin ka = 0. A ≠ 0 (else ψ = 0 everywhere), so ka = nπ, k = nπ/a with n = 1, 2, 3, …

(n = 0 gives ψ = 0, no particle.)

Step 5: energy eigenvalues

Put k = nπ/a into k² = 2mE/ħ²:

Eₙ = n²π²ħ²/(2ma²) = n²h²/(8ma²), n = 1, 2, 3, …

Only these discrete values are allowed, so the energy is quantized. E ∝ n², and the lowest energy (zero-point energy) is E₁ = π²ħ²/(2ma²), not zero.

Step 6: normalisation

∫₀ᵃ |ψ|² dx = A² (a/2) = 1, so A = √(2/a).

ψₙ(x) = √(2/a) sin(nπx/a)

 E4 ---------- n=4   (16 E1)
 E3 ---------- n=3   (9 E1)
 E2 ---------- n=2   (4 E1)
 E1 ---------- n=1