I consider a one dimensional waveguide. I study free field propagation on it. I apply periodic boundary condition with periodization length $L$ to describe the physics with Fourier series and not fourier transform.

Within this volume $L$, the total Hamiltonian of the system is:

$$H=\sum_{l} \hbar \omega_k a^{\dagger}_k a_k$$

I want to know what is the number of photons per unit time which cross a section $x$ of my waveguide at a time $t$. The photons travels at velocity $c$. What is done on page 68 of Introduction to Quantum Noise, Measurement and Amplification, is to say (without explanations), that this number is basically:

$$\dot{N}(x)=\langle b^{\dagger}_{\rightarrow}(x,t) b_{\rightarrow}(x,t)\rangle $$ With: $b_{\rightarrow}(x,t) \equiv \sqrt{\frac{c}{L}} \sum_{l>0} a_k e^{i(kx-\omega_k t)}$

My main question here is: Why this result ? Why would this represent the average number of photon which cross a section $x$ of my waveguide per unit time ?

Indeed, I would define things differently. Basically to have access to a photon rate, I must know the total number of photon in my system, their velocity, and their distribution along my waveguide. This is represented by the following operator.

$$N_{\rightarrow} = \sum_l a^{\dagger}_l a_l$$

Dividing by my waveguide length $L$, I have access to the number of right moving photons per unit length (I need to assume something about how photons are distributed, else I cannot do anything). Then, the number of photons that will cross a section $x$ of my waveguide during a time $dt$ will correspond to the number of photons contained in a length $cdt$. If I divide by $dt$ I have access to the number of photons that will cross the section $x$ of my waveguide per unit time: the photon rate. It gives:

$$\dot{N}(x)=\frac{c}{L}\sum_{l>0} \langle a^{\dagger}_l a_l\rangle \neq \langle b^{\dagger}_{\rightarrow}(x,t) b_{\rightarrow}(x,t)\rangle $$

Why is my way of defining things incorrect ? Why is the first way of defining things used ? I don't understand.

[edit] I remember this other question I asked some time ago, with the associated answer by @Wolpertinger Quantum input-output theory : Why do we multiply by density of mode to have a number of photon **per unit of time**

I have to think about it more but I think it motivates well the first definition of photon rate. In the end I can also ask the question: does the photon rate definition is partially subjective ? Photon is a concept in itself, we have to define at some point what we call a photon rate. We then have two possible definitions that each have advantages and drewback. Which is why the two definition given here do not always match. The community prefered for some reasons the first definition. Would that make sense ?

  • $\begingroup$ It looks to me like the definitions you've given in fact agree and the does not equals in your last equation should actually be an equals.. what the issue? $\endgroup$ – Jagerber48 Jun 17 '20 at 14:43
  • $\begingroup$ @jgerber it would save my life if that was the case ! Maybe I missed something but I will have crossed term in the first definition, not in the second: terms like $a_l^{\dagger}a_p$ with $p \neq l$. Do you agree ? $\endgroup$ – StarBucK Jun 17 '20 at 14:44
  • $\begingroup$ Yes, you will have the cross terms, but those will not contribute when you take the expectation value. $\endgroup$ – Jagerber48 Jun 17 '20 at 14:46
  • $\begingroup$ @jgerber for a fixed photon number they won't, but for a coherent state (for instance, or any other state) they will. The photon number operator should work for an arbitrary state. $\endgroup$ – StarBucK Jun 17 '20 at 14:47
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    $\begingroup$ Yes, you're right. I would need to think longer about it than I have time for right now to give a satisfying answer. $\endgroup$ – Jagerber48 Jun 17 '20 at 14:53

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