UPSC Combined Geo-Scientist, Mains · 2023 · Mains Paper II
Q1(l). Is Peⁱᵏˣ + Qe⁻ⁱᵏˣ an eigenfunction?
Question
Is the function ψ(x) = Peⁱᵏˣ + Qe⁻ⁱᵏˣ an eigenfunction of d/dx and d²/dx²? P, Q and k are real numbers.
Answer in brief
Not an eigenfunction of d/dx (unless P = 0 or Q = 0). An eigenfunction of d²/dx² with eigenvalue −k².
Key points
- dψ/dx = ik(Peⁱᵏˣ − Qe⁻ⁱᵏˣ) ≠ (constant) × ψ for k, P, Q ≠ 0 ⇒ not an eigenfunction of d/dx.
- d²ψ/dx² = (ik)²Peⁱᵏˣ + (−ik)²Qe⁻ⁱᵏˣ = −k²ψ ⇒ an eigenfunction, eigenvalue −k².
- p̂x = −iħ d/dx, so ψ mixes momenta +ħk and −ħk: momentum not sharp; Ĥ gives one value E = ħ²k²/2m: energy sharp.
- P = Q gives 2P cos kx; P = −Q gives 2iP sin kx, the particle-in-a-box form.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Coverseigenfunctioneigenvalueoperatorsmomentum operatorquantum mechanics
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