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Let's say I have a creation operator that creates a photon in some spatial mode. It has a spectral distribution given by [itex]f(\omega_{k})[/itex]

So we have [tex]

\mid 1_{p} \rangle=\int d\omega_{k}f(\omega_{k})a^{\dagger}_{k}\mid 0 \rangle[/tex]

Normalization implies that [tex]

\int d\omega_{k}|f(\omega_{k})|^{2} = 1[/tex]

Now, let's see this photon in time. It is given by [tex]

F(t)=\int d\omega_{k}f(\omega_{k})e^{i\omega_{k}t}[/tex]

From a theorem in Fourier transforms, we have[tex]

\int d\omega_{k}|f(\omega_{k})|^{2} = 1 ⇔ \int dt|F(t)|^{2}=1

[/tex]

So my question now is this: Suppose I chose a pulse [itex]F(t)[/itex] but it didn't obey [itex]\int dt|F(t)|^{2}=1[/itex]. It is not a single photon state anymore, so what is this? I can, for instance, consider a rectangular pulse such that [tex]

F(t) =

\begin{cases}

1, & \text{if }0<t<T \\

0, & \text{if }t≥T

\end{cases}

[/tex]

By changing T, I can normalize it to whatever number I want. My question is, what does this correpond to? If I take T very large, it doesn't mean a large number of photons because even a 100 photon state has a specific normalization condition. Classically, this is very easy to see (long rectangular pulse) but I'm not sure how to describe it quantum mechanically.

Thank you!

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# Normalization of photon pulse. I'm confused

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