Bargmann transform

Remarks

For fixed $z$ the kernel is a Gaussian in $x$, so it lies in $L^2(\R)$, and the integral converges by the Cauchy–Schwarz inequality. The function $Bf$ is entire, and $B$ is a unitary operator from $L^2(\R)$ onto $\mathcal{F}^2$. The monomials $z^n / \sqrt{n!}$ form an orthonormal basis of $\mathcal{F}^2$, and $B$ maps the normalized Hermite function $\psi_n$ to $z^n / \sqrt{n!}$. Conventions differ: some authors use the weight $e^{-\pi \abs{z}^2}$ or $e^{-\abs{z}^2/\hbar}$ and rescale the kernel accordingly.

Example: $f = \psi_0 = \pi^{-1/4} e^{-x^2/2}$. Then $Bf(z) = \pi^{-1/2} e^{-z^2/2} \int_\R e^{-x^2 + \sqrt{2} z x}\, dx = \pi^{-1/2} e^{-z^2/2} \cdot \sqrt{\pi}\, e^{z^2/2} = 1$. The constant $1$ has norm $\frac{1}{\pi} \int_\C e^{-\abs{z}^2}\, dA = 1$, equal to $\norm{\psi_0}_{L^2} = 1$.

Non-example: the function $F(z) = e^{z^2}$ is entire but not in $\mathcal{F}^2$. On the real axis $\abs{F(x)}^2 e^{-x^2} = e^{x^2}$, and the integral of $\abs{F}^2 e^{-\abs{z}^2}$ over a neighbourhood of the real axis diverges, so $F$ is not a Bargmann transform.