Normalization of photon pulse. I'm confused

[McLaren Rulez]

Hi,

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

So we have <br /> \mid 1_{p} \rangle=\int d\omega_{k}f(\omega_{k})a^{\dagger}_{k}\mid 0 \rangle

Normalization implies that <br /> \int d\omega_{k}|f(\omega_{k})|^{2} = 1

Now, let's see this photon in time. It is given by <br /> F(t)=\int d\omega_{k}f(\omega_{k})e^{i\omega_{k}t}

From a theorem in Fourier transforms, we have<br /> \int d\omega_{k}|f(\omega_{k})|^{2} = 1 ⇔ \int dt|F(t)|^{2}=1<br />

So my question now is this: Suppose I chose a pulse F(t) but it didn't obey \int dt|F(t)|^{2}=1. It is not a single photon state anymore, so what is this? I can, for instance, consider a rectangular pulse such that <br /> F(t) =<br /> \begin{cases}<br /> 1, &amp; \text{if }0&lt;t&lt;T \\<br /> 0, &amp; \text{if }t≥T<br /> \end{cases}<br />

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!

 

[vanhees71]

Your description of a photon doesn't make sense, at least not to me. The standard Fock space is defined via the single-particle momentum-helicity eigenbasis. The photon field is given in terms of the corresponding annihilation and creation operators
A_{\mu}(x)=\sum_{\lambda=\pm 1} \int \frac{\mathrm{d}^3 \vec{p}}{[(2 \pi)^3 2 \omega_{\vec{p}}]^{1/2}} [\epsilon_{\lambda}^{\mu} \hat{a}(\vec{p},\lambda) \exp(-\mathrm{i} x_{\mu} p^{\mu})+ \text{h.c.} ]_{p^0=\omega_{\vec{p}}=|\vec{p}|}.
Here, the creation and annihilation operators are normalized such that
\langle \vec{k},\lambda|\vec{k}&#039;,\lambda&#039; \rangle=\langle \text{vac}|\hat{a}(\vec{k},\lambda \hat{a}^{\dagger}(\vec{k}&#039;,\lambda&#039;) \rangle=\delta_{\lambda \lambda&#039;} \delta^{(3)}(\vec{p}-\vec{p}&#039;).
A true one-photon state is then given by
|\psi,\lambda \rangle=\sum_{\lambda=\pm 1} \int \mathrm{d}^3 \vec{p} \psi(\vec{p},\lambda) \hat{a}^{\dagger}(\vec{p},\lambda) |\text{vac} \rangle.
Here, \psi is a normalized momentum-space wave function, i.e.,
\int \mathrm{d}^3 \vec{p} |\psi(\vec{p},\lambda)|^2=1.

 

[McLaren Rulez]

Aren't these the same thing, apart from me taking it in one polarization while you have two?

<br /> \mid 1_{p} \rangle=\int d\omega_{k}f(\omega_{k})a^{\dagger}_{k}\mid 0 \rangle

A true one-photon state is then given by
|\psi,\lambda \rangle=\sum_{\lambda=\pm 1} \int \mathrm{d}^3 \vec{p} \psi(\vec{p},\lambda) \hat{a}^{\dagger}(\vec{p},\lambda) |\text{vac} \rangle.

I'm not sure what is different between our formulations, could you perhaps explain that bit again? The problem I have is this: I create a photon with a certain spectral distribution. Because it is only one photon, there is a normalization condition. The temporal shape of the photon is determined by the Fourier transform of the spectral distribution. It is normalized automatically.

Now, what I am doing is taking a temporal shape which isn't normalized. Is there a physical meaning to this now? I am not sure of this.

Thank you for you help :)