= Solution
To distinguish the two normalizations, denote the physicists' <Hermite polynomial> by $h_n(x)=(-1)^ne^{x^2}(d/dx)^ne^{-x^2}$. Define the normalized <Hermite functions> by
$$
\boxed{H_n(x)=\frac{h_n(x)e^{-x^2/2}}{\pi^{1/4}\sqrt{2^nn!}}.}
$$
Introduce the differential operators $a=(x+d/dx)/\sqrt2$, $a^*=(x-d/dx)/\sqrt2$ on the <Schwartz space>. Direct multiplication gives $[a,a^*]=1$ and
$$
-d^2/dx^2+x^2=2a^*a+1.
$$
The Gaussian $H_0=\pi^{-1/4}e^{-x^2/2}$ satisfies $aH_0=0$, and repeated differentiation gives $H_n=(a^*)^nH_0/\sqrt{n!}$. The <commutator> identity implies $aH_n=\sqrt nH_{n-1}$ and $a^*H_n=\sqrt{n+1}H_{n+1}$. Therefore
$$
\boxed{(-d^2/dx^2+x^2)H_n=(2n+1)H_n.}
$$
For normalization, <integration by parts> makes $a^*$ the <adjoint operator> of $a$ on these rapidly decreasing functions, so $\|(a^*)^nH_0\|^2=n!\|H_0\|^2=n!$. Distinct <eigenvalues> of the symmetric <quantum harmonic oscillator> give orthogonality.
The generating identity follows directly from the Rodrigues definition of the <Hermite polynomials>:
$$
\sum_{n\geq0}\frac{h_n(x)}{n!}w^n
=e^{x^2}\sum_{n\geq0}\frac{(-w)^n}{n!}(d/dx)^ne^{-x^2}
=e^{x^2}e^{-(x-w)^2}=e^{2xw-w^2}.
$$
The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt2$ and the <Bargmann-Fock space> basis $e_n(z)=z^n/\sqrt{n!}$ to obtain
$$
\boxed{\sum_{n\geq0}H_n(x)e_n(z)=A_x(z):=\pi^{-1/4}e^{-x^2/2+\sqrt2xz-z^2/2}.}
$$
The 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) gives
$$
\sum_{n\geq0}t^nH_n(x)H_n(y)
=\frac1\pi\int_{\mathbb C}A_x(\sqrt t\,z)\overline{A_y(\sqrt t\,z)}e^{-|z|^2}\,du\,dv.
$$
The coefficient series converges absolutely by the <Cauchy-Schwarz inequality>. Substitution factors the right side into
$$
\frac{e^{-(x^2+y^2)/2}}{\pi^{3/2}}
\int_{\mathbb R}e^{-(1+t)u^2+\sqrt{2t}(x+y)u}\,du
\int_{\mathbb R}e^{-(1-t)v^2+i\sqrt{2t}(x-y)v}\,dv.
$$
Use the <Gaussian integral> $\int e^{-au^2+bu}\,du=\sqrt{\pi/a}\,e^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent is
$$
-\frac{x^2+y^2}{2}+\frac{t(x+y)^2}{2(1+t)}-\frac{t(x-y)^2}{2(1-t)}
=\frac{4xyt-(1+t^2)(x^2+y^2)}{2(1-t^2)}.
$$
Thus the <Mehler formula for Hermite functions> is
$$
\boxed{\sum_{n\geq0}t^nH_n(x)H_n(y)=\frac1{\sqrt{\pi(1-t^2)}}\exp\!\left(\frac{4xyt-(1+t^2)(x^2+y^2)}{2(1-t^2)}\right).}
$$
As a normalization check, its $t\downarrow0$ 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> $2n+1$.
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