UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q8. Zero-point energy and the n² dependence in a one-dimensional box
Question
For a particle in a one-dimensional box: (i) Why is the ground state energy not zero, even though the potential inside the box is zero? (ii) Why does the energy increase as n² and not linearly with n?
Answer in brief
(i) n=0 gives ψ≡0 (no particle), so the lowest allowed state is n=1 with E₁=h²/8mL²>0: confinement forces a minimum momentum (zero-point energy). (ii) The boundary conditions fit n half wavelengths, λ=2L/n, so p=nh/2L and E=p²/2m∝ n².
Key points
- Inside, V=0, so E=T=p²/2m. ψ=√(2/L)sin(nπ x/L) must vanish at x=0 and L. n=0 gives ψ=0 everywhere: no particle. So n≥1 and E₁=(h²)/(8mL²)≠0.
- Δ x≤ L so Δ p≥ħ/2L; the kinetic energy cannot be zero. A particle with E=0 would have p=0 exactly and Δ x=∞, outside the box.
- the lowest wavefunction has one half-wave, with non-zero curvature, T∝-ψ”/ψ.
- The standing wave has n half wavelengths in L: (nλ)/(2)=L⇒λ=2L/n.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
(i) Ground-state energy
- Inside, V=0, so E=T=p²/2m. ψ=√(2/L)sin(nπ x/L) must vanish at x=0 and L. n=0 gives ψ=0 everywhere: no particle. So n≥1 and E₁=(h²)/(8mL²)≠0.
- Uncertainty principle: Δ x≤ L so Δ p≥ħ/2L; the kinetic energy cannot be zero. A particle with E=0 would have p=0 exactly and Δ x=∞, outside the box.
- Curvature: the lowest wavefunction has one half-wave, with non-zero curvature, T∝-ψ”/ψ.
(ii) E∝ n²
- The standing wave has n half wavelengths in L: (nλ)/(2)=L⇒λ=2L/n.
- de Broglie: p= hλ=(nh)/(2L); E=(p²)/(2m)=(n²h²)/(8mL²).
- Since E depends on p² (not p), levels are E₁:4E₁:9E₁, spaced Eₙ₊₁-Eₙ=(2n+1)E₁, increasing with n.
Coversparticle in a box
All parts of Question 8
- 8 Zero-point energy and the n² dependence in a one-dimensional box10 marks