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

Normalization implies that $$\int d\omega_{k}|f(\omega_{k})|^{2} = 1$$

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

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

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 $$F(t) = \begin{cases} 1, & \text{if }0<t<T \\ 0, & \text{if }t≥T \end{cases}$$

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
$$\langle \vec{k},\lambda|\vec{k}',\lambda' \rangle=\langle \text{vac}|\hat{a}(\vec{k},\lambda \hat{a}^{\dagger}(\vec{k}',\lambda') \rangle=\delta_{\lambda \lambda'} \delta^{(3)}(\vec{p}-\vec{p}').$$
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?

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

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 :)