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

The product rule gives and . Substituting in the Schrödinger equation cancels the and constant terms, leaving the Hermite differential equation
For , matching coefficients gives
The power series solutions of the Hermite differential equation split into independent even and odd series. With initial data and , respectively, their first three terms are
Each 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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