Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normalization of photon pulse. I'm confused

  1. Jan 3, 2013 #1

    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

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

    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!
  2. jcsd
  3. Jan 3, 2013 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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
    [tex]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}|}.[/tex]
    Here, the creation and annihilation operators are normalized such that
    [tex]\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}').[/tex]
    A true one-photon state is then given by
    [tex]|\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.[/tex]
    Here, [itex]\psi[/itex] is a normalized momentum-space wave function, i.e.,
    [tex]\int \mathrm{d}^3 \vec{p} |\psi(\vec{p},\lambda)|^2=1.[/tex]
  4. Jan 3, 2013 #3
    Aren't these the same thing, apart from me taking it in one polarization while you have two?
    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook