Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 4 / c / Solution

To distinguish the two normalizations, denote the physicists' Hermite polynomial by $h_n(x)=(-1{)}^n{e}^{x^2}(d/dx{)}^n{e}^{-x^2}$. Define the normalized Hermite functions byIntroduce the differential operators $a=(x+d/dx)/\sqrt{2}$, $a^{\ast }=(x-d/dx)/\sqrt{2}$ on the Schwartz space. Direct multiplication gives $[a,a^{\ast }]=1$ andThe Gaussian $H_0={\pi }^{-1/4}{e}^{-x^2/2}$ satisfies $a{H}_0=0$, and repeated differentiation gives $H_n=(a^{\ast }{)}^n{H}_0/\sqrt{n!}$. The commutator identity implies $a{H}_n=\sqrt{n}{H}_{n-1}$ and $a^{\ast }{H}_n=\sqrt{n+1}{H}_{n+1}$. ThereforeFor normalization, integration by parts makes $a^{\ast }$ the adjoint operator of $a$ on these rapidly decreasing functions, so $\Vert (a^{\ast }{)}^n{H}_0{\Vert }^2=n!\Vert H_0{\Vert }^2=n!$. Distinct eigenvalues of the symmetric quantum harmonic oscillator give orthogonality.

The generating identity follows directly from the Rodrigues definition of the Hermite polynomials:The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt{2}$ and the Bargmann-Fock space basis $e_n(z)=z^n/\sqrt{n!}$ to obtainThe convergence is locally uniform in $z$ by the same entire generating series.

For $0<t<1$, $A_x(\sqrt{t}z)$ belongs to the Bargmann-Fock space: after writing $z=u+iv$, its squared modulus times $e^{-|z{|}^2}$ has strictly negative quadratic coefficients $-(1+t)u^2-(1-t)v^2$. Taking the Bargmann-Fock space inner product of the two generating functions and using the orthonormal basis proved in part (b) givesThe coefficient series converges absolutely by the Cauchy-Schwarz inequality. Substitution factors the right side intoUse the Gaussian integral $\int e^{-a{u}^2+bu}du=\sqrt{\pi /a}{e}^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent isThus the Mehler formula for Hermite functions isAs a normalization check, its $t\downarrow 0$ limit is $H_0(x)H_0(y)$. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ yields the harmonic oscillator transition kernel with eigenvalues $2n+1$.