I describe the temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function $\psi$: $$ \psi(t) = \tfrac{1}{(2\pi\sigma^2)^{1/4}} \text e^{-\frac{t^2}{4\sigma^2}} \text e^{\text i \omega_0 t} \;. $$ It is normalised $$ \int |\psi(t)|^2\text d t = 1\;, $$ and the fourier transform is the wave function in the frequency domain, $$ \tilde \psi(\omega) = \frac 1 {\sqrt{2\pi}}\int \psi(t)\text e ^{-\text i \omega t}\text dt = \big(\tfrac{2\sigma^2}{\pi}\big)^{\frac 1 4} \text e^{-\sigma^2(\omega-\omega_0)^2}\;, $$ such that $|\tilde \psi(\omega)|^2$ represents the frequency distribution of the photon. This is, of course, not my invention but I saw it in many papers, like in ref1 or in ref2. Interestingly they always neglect the phase $\text e^{\text i \omega_0 t}$ in $\psi(t)$, but that's a different matter.
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 $$ E(t) \sim \text{Re}[\psi(t)] \sim \text e^{-\frac{t^2}{4\sigma^2}} \cos(\omega_0 t) \; ? $$
I know that there actually is a conceptual difference. The function $\psi(t)$ is a probability amplitude in the time domain whereas $E(t)$ is a real electric field. At a 50/50 beam splitter, e.g., the electric field would split into two parts where both parts can be measured, whereas the probability amplitude $\psi(t)$, which also splits into two parts, would lead detectors to click either for the transmitted part or for the reflected part.
So, is there now a link between $\psi(t)$ and the electric field $E(t)$ or not?
PS: Of course I know the concept of second quantisation and the quantisation of the electromagnet field. But I never understood how to use that to describe single photon pulses in vacuum...
