$\psi (t)={\textstyle \frac{1}{(2\pi {\sigma}^{2}{)}^{1/4}}}{\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}{\text{e}}^{\text{i}{\omega}_{0}t}\phantom{\rule{thickmathspace}{0ex}}.$

It is normalised

$\int |\psi (t){|}^{2}\text{d}t=1\phantom{\rule{thickmathspace}{0ex}},$

and the fourier transform is the wave function in the frequency domain,

$\stackrel{~}{\psi}(\omega )=\frac{1}{\sqrt{2\pi}}\int \psi (t){\text{e}}^{-\text{i}\omega t}\text{d}t={\textstyle (}{\textstyle \frac{2{\sigma}^{2}}{\pi}}{{\textstyle )}}^{\frac{1}{4}}{\text{e}}^{-{\sigma}^{2}(\omega -{\omega}_{0}{)}^{2}}\phantom{\rule{thickmathspace}{0ex}},$

such that $|\stackrel{~}{\psi}(\omega ){|}^{2}$ represents the frequency distribution of the photon.

However, since a single photon pulse is still a electromagnetic pulse, is there any link between $\psi (t)$ and the electric field $E(t)$ of this pulse? Like that

$E(t)\sim \text{Re}[\psi (t)]\sim {\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}\mathrm{cos}({\omega}_{0}t)\phantom{\rule{thickmathspace}{0ex}}?$