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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper I

Q7(a)(ii). Why n=0 is not allowed for a particle in a box

  • 5 marks
  • Answered
Question

The time-independent Schrödinger equation for a free particle (of mass m) in a box of length L (inside the box potential energy is zero, but outside the box is infinity) is
[d²ψdx²+2mEħ² ψ(x)=0; 0≤ x≤ L]
The solution of the above equation is ψₙ(x)=√(2/L) sin ((nπ x)/(L)); where n=1,2,3,…. Why is n=0 not allowed? Explain.

Answer in brief

For n=0, ψ₀=√(2/L) 0=0 at every point: the particle would be nowhere, and the function cannot be normalised. E₀=0 would also give an exactly known momentum (zero) for a particle confined to length L, against the uncertainty principle. The lowest state is n=1, with the zero-point energy E₁=h²/8mL².

Key points

  • ψ₀(x)=√(2/L) 0=0 for every x in the box (and ψ=0 outside it). It satisfies the equation only trivially, as 0=0.
  • By the Born interpretation |ψ|²dx is the probability of finding the particle between x and x+dx. For ψ₀ this is zero everywhere, so ∫₀^L|ψ₀|²dx=0, not 1: the box would hold no particle. ψ₀ cannot be normalised and is not an…
  • Eₙ=n²h²8mL² gives E₀=0. With V=0 all the energy is kinetic, so p=0 exactly and Δ p=0. The uncertainty principle would then need Δ x=∞, but the particle is held within Δ x≤ L.
  • sin(-θ)=-θ gives the same |ψ|². The allowed states are n=1,2,3,…, and the lowest energy is not zero: E₁=h²8mL² (zero-point energy)

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Putting n=0: ψ₀(x)=√(2/L) 0=0 for every x in the box (and ψ=0 outside it). It satisfies the equation only trivially, as 0=0.
  2. By the Born interpretation |ψ|²dx is the probability of finding the particle between x and x+dx. For ψ₀ this is zero everywhere, so ∫₀^L|ψ₀|²dx=0, not 1: the box would hold no particle. ψ₀ cannot be normalised and is not an acceptable wavefunction.
  3. Energy: Eₙ=n²h²8mL² gives E₀=0. With V=0 all the energy is kinetic, so p=0 exactly and Δ p=0. The uncertainty principle would then need Δ x=∞, but the particle is held within Δ x≤ L.
  4. Negative n adds nothing new: sin(-θ)=-θ gives the same |ψ|². The allowed states are n=1,2,3,…, and the lowest energy is not zero: E₁=h²8mL² (zero-point energy)

The zero-point energy cannot be removed, even at 0 K. A function that is zero at both walls but not zero everywhere must be curved, and curvature means kinetic energy.

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Model answer to Q7(a)(ii), 2026 Paper I: solution page 42
Solution page 42 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 7