Since I am not fully aware of the creation and annihilation operator formalism for single photons, I want to ask, if the following is correct:

I am considering a photon in the vacuum which travel just in one direction, so no different spatial modes are considered.

I interpret the creation operator $a_\omega^\dagger$ as creating a single photon with angular frequency $\omega$ when in acts on the vacuum $|0\rangle$. The commutator relation is $[a_{\omega'},a_\omega^\dagger]=\delta(\omega'-\omega)$.

If I now want to describe a single photon that does not have a sharp frequency but rather a frequency distribution $f(\omega)$, where $|f(\omega)|^2\text d \omega$ is the probability to measure a frequency in the interval $[\omega,\omega+\text d\omega]$, I write the photon state like $$ |\psi\rangle = \int \text d \omega\; f(\omega) \, a_\omega^\dagger\, |0\rangle \;. $$ I can still say that this is a single photon since it is an eigenstate of the photon number operator $$ N = \int \text d \omega\; a_\omega^\dagger a_\omega $$ with eigenvalue 1: $$ N|\psi\rangle = |\psi\rangle \;. $$ If I now change the basis like $$ a_t^\dagger = \frac{1}{\sqrt{2\pi}} \int \text d \omega \; a_\omega^\dagger\,\text e^{\text i \omega t}\;,\qquad a_\omega^\dagger = \frac{1}{\sqrt{2\pi}}\int \text d t \; a_t^\dagger\,\text e^{-\text i \omega t}\;, $$ I can write my photon state as $$ |\psi\rangle = \int \text d t\; \tilde f(t) \, a_t^\dagger\, |0\rangle \;. $$ where $\tilde f$ is the Fourier transform of $f$: $$ \tilde f(t) = \frac{1}{\sqrt{2\pi}} \int \text d \omega \; f(\omega)\,\text e^{\text i \omega t}\;. $$ Furthermore I interpret $\tilde f(t)$ as the photons distribution in the time domain such that $|\tilde f(t)|^2$ is the shape of the photon.

Is this correct? If so, how can I write the operator of the electric field in terms of $a_\omega^\dagger$ or $a_t^\dagger$?

PS: Actually I could also write $k$ instead of $\omega$, since $k=\omega/c$. And I could also write $x$ instead of $t$, since I just mean $t=x/c$.

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    $\begingroup$ There are no photons in one spatial dimension, since light has no longitudinal polarization states. $\endgroup$ – ACuriousMind Nov 13 '14 at 13:11

Something like what you're proposing is theoretically sound. For a one-photon state, you can define the following probability amplitudes that uniquely specify the one-photon state:

$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\psi\right>\right. \end{array}\tag{1}$$

Here $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$ are the positive frequency parts of the (vector valued) electric and magnetic field observables and, of course, $\left<\left.0\right.\right|$ is the unique ground state of the quantum photon field. By dint of the structure of $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$, these work out to be very like the Fourier transforms that you're speaking of. The vector valued "probability amplitudes" $\vec{\phi}_E(\vec{r},\,t)$ and $\vec{\phi}_B(\vec{r},\,t)$ fulfill the classical Maxwell equations and they have the physical meaning that the probability density to destructively detect the photon, i.e. absorb it with an ideal detector, when the state is properly normalised, is the analogue of the classical energy density (normalisation makes the classical energy density into a probability denstity), i.e.

$$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2\tag{2}$$

Given the interpretation of (2), i.e. as a probability amplitude to destructively detect a photon, you can in this sense indeed think of something like your Fourier transform to be the photon's shape. Note that the probability amplitude for destructive detection is different from the probability amplitude "to find" an electron that is used to interpret nonrelativistic Schrödinger equations and define a "shape" of an orbital. There is no "position" representation of the photon in this latter sense.

Given $\vec{\phi}_E(\vec{r},\,t),\,\vec{\phi}_B(\vec{r},\,t)$ you can uniquely find your state superposition:

$$ |\psi\rangle = \int \text d \omega\; f(\omega) \, a_\omega^\dagger\, |0\rangle \;. $$

or at least a three-dimensional analogue thereof and contrariwise. Every classical solution of Maxwell's equations defines such a one photon state and every one photon state uniquely defines a classical solution to Maxwell's equations.

So, in fact, each classical Maxwell equation solution can stand for one of three quite distinct physical entities:

  1. A one-photon state (number observable eigenstate), as discussed above;
  2. A quantum coherent state, wherein the means of the electric and magnetic field observables in the Heisenberg picture propagate like classical electrical and magnetic fields
  3. A classical light state.

No, unfortunately I believe that the identification with time is wrong.

It is clearer to have a mode index for the creation operator $a_k^\dagger$ each of which refers to some spatial mode. For free space, you could choose plane $k$, but don't have to. Now each mode, if you choose it to be an eigenmode of the Hamiltonian, has a well-defined energy $\omega$ (as long as you leave interactions out of the game). Thus, if the spectrum of $H$ is non-degenerate, you could simply label this mode by $\omega$, because it is clear what you mean. However, if you have multiple modes and a continuous spectrum, you need to include the density of states in constructing the number operator.

So those parts of your statements are not wrong. But performing a Fourier transform, also if you formally do it with $\omega$, does not get you into any meaningful time domain.

Imagine you have some other index notation, which is just a labeling scheme: it is not always inherently clear what you mean by performing a Fourier transform with respect to that labeling index. For momentum $k$ to real-space it is.

Here is an easy hint: the labels are mostly nothing different from what you would write into the ket of a single-particle state. If it's clear what you mean with the Fourier transform there, e.g. $$ \int dk \, e^{-ikx} \, | k\rangle $$, then it is also clear for the corresponding ladder operator.

  • $\begingroup$ Thanks for the answer. I edited my question: I am just considering a one dimensional space and the photons can just travel in one direction. So I have just one spatial mode. Furthermore in vacuum I have k=ω/c, so I see no reason to distinguish between ω and k. What do you say, if I change all my ω to k and all t to x? Is it then all right? I don't want to refer to any time evolution with t... thats what I meant with "shape of the photon"... its more a spatial shape, since $t=x/c$. $\endgroup$ – thyme Nov 13 '14 at 12:26

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