Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 4

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

By the two Fourier transform rules in part (b),Consequently transforming the differential equation givesorThe inverse Fourier transform gives the reverse implication under the same regularity hypotheses. This expresses the Fourier symmetry of the quantum harmonic oscillator operator.

Part (b) shows that the transform of $g_n$ is $q_n(k)e^{-k^2/2}$ with $q_n$ of degree $n$. Part (c) shows that it solves the same eigenvalue equation as $g_n$, with the same ${\mu }_n$. The assumed uniqueness of the polynomial solution of each degree therefore forces $q_n=c_n{p}_n$; hence ${\tilde{g}}_n=c_n{g}_n$.

The constants can also be determined. Substituting $g_n=p_n{e}^{-x^2/2}$ into the equation gives the Hermite differential equationComparison of the leading coefficient yields ${\mu }_n=-(2n+1)$. If the leading coefficient of $p_n$ is $a_n$, part (b) gives leading coefficient $\sqrt{2\pi }(-i{)}^n{a}_n$ for $q_n$. ThusThus Hermite functions are Fourier eigenfunctions with the paper's unnormalized Fourier transform; dividing that transform by $\sqrt{2\pi }$ gives the unitary eigenvalues $(-i{)}^n$.