Q4(a)(i). Particle in a box: probability between 0 and a/2

1. Normalised wavefunctions. Inside the box (0≤ x≤ a, V=0) ψₙ=√(2/a)sin(nπ x)/(a), n=1,2,3,…; outside, ψ=0. Energies Eₙ=n²h²8ma².
2. Probability of finding the particle between x₁ and x₂ is ∫_(x₁)^(x₂)ψ² dx (Born interpretation). Here P=2/a∫₀^(a/2)²(nπ x)/(a) dx
3. Use ²θ=12(1-2θ): P=1/a∫₀^(a/2)(1-cos(2nπ x)/(a))dx=1/a[x-(a)/(2nπ)sin(2nπ x)/(a)]₀^(a/2)
4. Put in the limits: P=1/a[a/2-(a)/(2nπ)sin nπ]=12-(sin nπ)/(2nπ)
5. sin nπ=0 for every integer n, so P(0→ a/2)=12 for all n
6. Reason: ψₙ² is symmetric about x=a/2 for every state: odd n have a maximum at the centre, even n a node there. The particle is equally likely to be in either half.
7. Contrast with a quarter of the box. For n=1, P(0→ a/4)=14-(1)/(2π)=0.091, not 0.25: in the ground state the particle stays away from the walls. As n grows, P(0→ a/4)→14, the classical value (correspondence principle).
8. The result is independent of the mass of the particle and of the length a.

Application. the particle in a box models the π electrons of conjugated polyenes and dyes: Δ E=(N+1)h²/8mL² predicts why longer conjugated chains absorb at longer wavelength (for example β-carotene is coloured).