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

Brenton Dixon

Answered question

2022-07-16

Temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function ψ :
ψ ( t ) = 1 ( 2 π σ 2 ) 1 / 4 e t 2 4 σ 2 e i ω 0 t .
It is normalised
| ψ ( t ) | 2 d t = 1 ,
and the fourier transform is the wave function in the frequency domain,
ψ ~ ( ω ) = 1 2 π ψ ( t ) e i ω t d t = ( 2 σ 2 π ) 1 4 e σ 2 ( ω ω 0 ) 2 ,
such that | ψ ~ ( ω ) | 2 represents the frequency distribution of the photon.
However, since a single photon pulse is still a electromagnetic pulse, is there any link between ψ ( t ) and the electric field E ( t ) of this pulse? Like that
E ( t ) Re [ ψ ( t ) ] e t 2 4 σ 2 cos ( ω 0 t ) ?

Answer & Explanation

Monica Dennis

Monica Dennis

Beginner2022-07-17Added 13 answers

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 < E >=< ψ | E | ψ >.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Electromagnetism

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?