Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / 6

Put $s=z-1$ and seek the leading Frobenius series term $y\sim s^{\rho }$. The coefficients of $y^{\prime \prime }$ and $y^{\prime }$ become $-2s(1+s)$ and $2+s$, respectively. The lowest power, $s^{\rho -1}$, therefore has coefficient $-2\rho (\rho -1)+2\rho$. The indicial equation isThe term $-y$ contributes only at the next order and does not enter this equation.

The product rule gives ${\psi }^{\prime }=e^{-x^2/2}(y^{\prime }-xy)$ and ${\psi }^{\prime \prime }=e^{-x^2/2}[y^{\prime \prime }-2x{y}^{\prime }+(x^2-1)y]$. Substituting in the Schrödinger equation cancels the $x^2$ and constant terms, leaving the Hermite differential equationFor $y={\sum }_{n\ge 0}{a}_n{x}^n$, matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data $(y(0),y^{\prime }(0))=(1,0)$ and $(0,1)$, 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 $2|x{|}^2/n$. 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 $N$ requires $\lambda =N$, by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are $\lambda =0,1,2,\dots$. Up to nonzero scalar multiples, the degree-at-most-three solutions are the Hermite polynomials