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I'm reading some materials(before eq(129)) about beam splitters while I met the following formula $$ \hat a_{\mathrm{in}}^{\dagger}\hat a_{\mathrm{in}}+\hat b_{\mathrm{in}}^{\dagger}\hat b_{\mathrm{in}}=\hat a_{\mathrm{out}}^{\dagger}\hat a_{\mathrm{out}}+\hat b_{\mathrm{out}}^{\dagger}\hat b_{\mathrm{out}} \tag 1 $$ where $\hat a$ and $\hat b$ stand for annihilation operators for different ports of a beam splitter shown in the following figure enter image description here

My question is, why the eq(1) stands for photon number conservation?

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  • $\begingroup$ Because each pair first annihilates and then creates back the photon, and so must be conserving photon numbers. As long as you do not have an imbalance in the number of creation and annihilation operators in any term, then photon numbers are conserved. $\endgroup$ Jul 1, 2023 at 2:11
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    $\begingroup$ en.wikipedia.org/wiki/Particle_number_operator $\endgroup$
    – Ghoster
    Jul 1, 2023 at 2:14
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    $\begingroup$ @naturallyInconsistent, this looks like an answer. It should not be a comment. $\endgroup$ Jul 1, 2023 at 5:24
  • $\begingroup$ @Ghoster Thank you for your comment. I know $a^\dagger a$ is a photon number operator, but I just don't get why eq(1) means photon number conservation even though it looks like since if we just think $a^\dagger a$ is the number of photons, then we have eq(1) to be(but obviously not a rigorous deduction, just handwaving here) $N_{\mathrm{in}}^{a}+N_{\mathrm{in}}^{b}=N_{\mathrm{out}}^{a}+N_{\mathrm{out}}^{b}$. $\endgroup$
    – narip
    Jul 1, 2023 at 10:23
  • $\begingroup$ I mean the equation you have just written down say "number of photons in = number of photons out" i.e. they are conserved. And I don't think there is anything non-rigorous in how you got there from the equation in your question $\endgroup$ Jul 1, 2023 at 10:52

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For a harmonic oscillator system, the number operator is given by $\hat{N} := \hat{a}^\dagger \hat{a}$ if $\hat{a}^\dagger$ and $\hat{a}$ are creation and annihilation operators. The easiest way to see this is perhaps by noting that the Hamiltonian operator reads $\hat{H} = \hbar\omega (\hat{a}^\dagger\hat{a}+1/2)$ in terms of $\hat{a}$ and $\hat{a}^\dagger$ and that the energy levels are given by $E_n = (n+1/2) \hbar\omega$. (Cf. https://en.m.wikipedia.org/wiki/Quantum_harmonic_oscillator)

Then why is $n$ of "$n$th energy eigenstate" "number?" It's due to an interpretation in second quantization. (Cf. https://en.m.wikipedia.org/wiki/Second_quantization) It does not neccesarily have to be field theory, and the general idea goes like this: particles are excitations of a harmonic oscillator system. Hence $n$th energy eigenstate is interpreted as a state with $n$ quantum particles (bosons). For instance, in our context, the spacing $\hbar\omega$ of the "ladder" becomes the energy of each quantum of light. Hence it is nothing very sophisticated but simply stating Einstein and Planck's concept of photons as energy quanta of light, represented by the formula "$\varepsilon = nhf$."

Finally, your equation translates to $$\hat{N}_{a,\text{in}}+\hat{N}_{b,\text{in}}=\hat{N}_{a,\text{out}}+\hat{N}_{b,\text{out}} \,,$$ as already mentioned in one of the comments. Therefore, according to the photon number interpretation, we conclude that the equation will be interpreted as the conservation of photon number, stating that the sum of numbers should be the same before and after entering the beam splitter.

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