 # Temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function psi: psi(t)=1/((2 pi sigma^2))^1/4 e^(-t^2/4 sigma^2) e^(i omega_0 t) Brenton Dixon 2022-07-16 Answered
Temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function $\psi$ :
$\psi \left(t\right)=\frac{1}{\left(2\pi {\sigma }^{2}{\right)}^{1/4}}{\text{e}}^{-\frac{{t}^{2}}{4{\sigma }^{2}}}{\text{e}}^{\text{i}{\omega }_{0}t}\phantom{\rule{thickmathspace}{0ex}}.$
It is normalised
$\int |\psi \left(t\right){|}^{2}\text{d}t=1\phantom{\rule{thickmathspace}{0ex}},$
and the fourier transform is the wave function in the frequency domain,
$\stackrel{~}{\psi }\left(\omega \right)=\frac{1}{\sqrt{2\pi }}\int \psi \left(t\right){\text{e}}^{-\text{i}\omega t}\text{d}t=\left(\frac{2{\sigma }^{2}}{\pi }{\right)}^{\frac{1}{4}}{\text{e}}^{-{\sigma }^{2}\left(\omega -{\omega }_{0}{\right)}^{2}}\phantom{\rule{thickmathspace}{0ex}},$
such that $|\stackrel{~}{\psi }\left(\omega \right){|}^{2}$ represents the frequency distribution of the photon.
However, since a single photon pulse is still a electromagnetic pulse, is there any link between $\psi \left(t\right)$ and the electric field $E\left(t\right)$ of this pulse? Like that
$E\left(t\right)\sim \text{Re}\left[\psi \left(t\right)\right]\sim {\text{e}}^{-\frac{{t}^{2}}{4{\sigma }^{2}}}\mathrm{cos}\left({\omega }_{0}t\right)\phantom{\rule{thickmathspace}{0ex}}?$
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Your single photon pulse wave function is an element of the first Fock layer (the zeroth is the vacuum layer) of the quantised Maxwell field Fock space. The electric field is still an operator but you can obtain its expectation value as $=<\psi |E|\psi >$.

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