Solution
ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/6/ii/solution
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2007 Paper 1 6 ii Solution by
Codex 0 Created 2026-10-18 Updated 2026-10-20
The product rule gives and . Substituting in the Schrödinger equation cancels the and constant terms, leaving the Hermite differential equationFor , matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data and , respectively, their first three terms areEach nonterminating parity series has infinite radius of convergence: the ratio of successive terms of that parity has magnitude asymptotic to . Their initial data prove linear independence. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.
A nonzero polynomial solution of degree requires , by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are . Up to nonzero scalar multiples, the degree-at-most-three solutions are the Hermite polynomials
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