Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 16B (Complex Methods)

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

Multiplication by $x$ transforms to $id/d\lambda$. Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$ and use $\hat{g}=cg$ for the Gaussian $g=e^{-x^2/2}$. ThenSuccessive differentiation of the Gaussian gives a polynomial of degree $j$ with leading term $(-1{)}^j{\lambda }^j$ multiplying that same Gaussian. Thus $q_n$ has leading coefficient $c{a}_n(-i{)}^n\ne 0$. This proves $q_n$ is a polynomial of degree exactly $n$, rather than merely at most $n$.

By part a the transformed function solves the same equation with ${\mu }_n$, and by part c its polynomial factor has degree $n$. The uniqueness proved in part b therefore gives $q_n=c_n{p}_n$. Comparing their leading coefficients givesFor the unitary convention used above, the Gaussian integral gives $c=1$, so $c_n=(-i{)}^n$. With the unnormalized phase-$e^{-i\lambda x}$ convention, $c=\sqrt{2\pi }$ instead. This proves that Hermite functions are Fourier eigenfunctions, with the normalization-dependent scalar fully specified.