# Single photon pulse and its electromagnet field

I describe the temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function $\psi$: $$\psi(t) = \tfrac{1}{(2\pi\sigma^2)^{1/4}} \text e^{-\frac{t^2}{4\sigma^2}} \text e^{\text i \omega_0 t} \;.$$ It is normalised $$\int |\psi(t)|^2\text d t = 1\;,$$ and the fourier transform is the wave function in the frequency domain, $$\tilde \psi(\omega) = \frac 1 {\sqrt{2\pi}}\int \psi(t)\text e ^{-\text i \omega t}\text dt = \big(\tfrac{2\sigma^2}{\pi}\big)^{\frac 1 4} \text e^{-\sigma^2(\omega-\omega_0)^2}\;,$$ such that $|\tilde \psi(\omega)|^2$ represents the frequency distribution of the photon. This is, of course, not my invention but I saw it in many papers, like in ref1 or in ref2. Interestingly they always neglect the phase $\text e^{\text i \omega_0 t}$ in $\psi(t)$, but that's a different matter.

However, since a single photon pulse is still a electromagnetic pulse, is there any link between $\psi(t)$ and the electric field $E(t)$ of this pulse? Like that $$E(t) \sim \text{Re}[\psi(t)] \sim \text e^{-\frac{t^2}{4\sigma^2}} \cos(\omega_0 t) \; ?$$

I know that there actually is a conceptual difference. The function $\psi(t)$ is a probability amplitude in the time domain whereas $E(t)$ is a real electric field. At a 50/50 beam splitter, e.g., the electric field would split into two parts where both parts can be measured, whereas the probability amplitude $\psi(t)$, which also splits into two parts, would lead detectors to click either for the transmitted part or for the reflected part.

So, is there now a link between $\psi(t)$ and the electric field $E(t)$ or not?

PS: Of course I know the concept of second quantisation and the quantisation of the electromagnet field. But I never understood how to use that to describe single photon pulses in vacuum...

## 1 Answer

Your single photon pulse wave function is an element of the first Fock layer (the zeroth is the vacuum layer) of the quantised Maxwell field Fock space. The electric field is still an operator but you can obtain its expectation value as $<E>=<ψ|E|ψ>$.

• Thanks for your answer. But applied to my notation, how can a calculate this expectation value? How does the operator $E$ look like? I know the notation where $E$ is written in terms of the creation and annihilation operators... but I don't really see how my $\psi(t)$ is a fock state created by one of these creation operators... – thyme Nov 8 '14 at 10:35
• Your state ψ is given (it can be obtained by acting with the creation operator $a^(ψ)$ on the field vacuum state) and the operator $E$ acts on it. E is given in terms of creation and annihilation operators (en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field). But since the creation and annihilation operators map into different Fock layers $<ψ}E|ψ>$ vanishes. But $<ψ}E^2|ψ>$ does not. – Urgje Nov 8 '14 at 11:06
• Correction: $a^(ψ)$ should be $a#(ψ)$. – Urgje Nov 8 '14 at 11:20
• I think you might be interested in this. When you have a single-photon state $\mid1\rangle$, you can extract what is called the photon wave function by computing $\langle0\mid\hat{\mathbf{E}}\left(\mathbf{x},t\right)\mid1\rangle$. It corresponds to the analytic signal of the electric field carried'' by that photon. If you're interested I can give you some references on that topic. – Vinsanity Jan 20 '15 at 17:24