Q1(l). No solution with E = 0 for the particle in a box

Answer in brief

With E = 0 the only solution is ψ = 0 everywhere. It cannot be normalised, so it is not acceptable.

Key points

• Inside the box: −(ℏ²/2m) d²ψ/dx² = Eψ. With E = 0: d²ψ/dx² = 0; integrating twice: ψ = Ax + B.
• Boundary conditions: ψ(0) = 0 ⇒ B = 0; ψ(a) = 0 ⇒ Aa = 0 ⇒ A = 0.
• ψ(x) = 0 for all x ⇒ ∫ψ*ψ dx = 0 ≠ 1: not normalisable, no particle in the box.
• Meaning: n = 0 is not allowed. The lowest level is the zero-point energy E₁ = h²/8ma² (uncertainty principle).