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

Q7(a)(i). e^(α x) as an eigenfunction of dⁿ/dxⁿ

Question

Show that ψ(x)=e^(α x) is an eigenfunction for the operator A=dⁿdxⁿ. What is the eigenvalue?

A e^(α x)=αⁿe^(α x): the operator returns the same function times a constant, so e^(α x) is an eigenfunction with eigenvalue αⁿ.

A function is an eigenfunction of an operator when the operator acting on it gives back the same function multiplied by a constant: Aψ=aψ. The constant a is the eigenvalue.
d/dxe^(α x)=α e^(α x). Second: d²dx²e^(α x)=/dxe^(α x)=α²e^(α x).
Each further differentiation brings out one more factor α, so after n steps: dⁿdxⁿ e^(α x)=αⁿ e^(α x)=αⁿ
αⁿ does not contain x, so the condition holds for every n and for any constant α, real or complex. By contrast d/dxe^α x²=2α x e^α x²: the factor 2α x varies, so e^α x² is not an eigenfunction of d/dx.

The full model answer

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

A function is an eigenfunction of an operator when the operator acting on it gives back the same function multiplied by a constant: Aψ=aψ. The constant a is the eigenvalue.
First derivative (chain rule): d/dxe^(α x)=α e^(α x). Second: d²dx²e^(α x)=/dxe^(α x)=α²e^(α x).
Each further differentiation brings out one more factor α, so after n steps: dⁿdxⁿ e^(α x)=αⁿ e^(α x)=αⁿ
αⁿ does not contain x, so the condition holds for every n and for any constant α, real or complex. By contrast d/dxe^α x²=2α x e^α x²: the factor 2α x varies, so e^α x² is not an eigenfunction of d/dx.
With α=ik, e^(ikx) is an eigenfunction of pₓ=-iħ d/dx (eigenvalue ħ k, the momentum, n=1) and of -ħ²2md²dx² (eigenvalue ħ²k²/2m, the kinetic energy, n=2).

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