 Photon-number measurement backaction

An explicit original ideal meter model explains how photon number survives while field quantum coherence changes. Let $\hat{N}$ be the number operator and prepare an atomic qubit in $|+⟩=(|e⟩+|g⟩)/\sqrt{2}$. The dispersive unitary operatorcommutes with $\hat{N}\otimes I$. For a field quantum superposition ${\sum }_n{c}_n|n⟩$, it produces ${\sum }_n{c}_n|n⟩|a_n⟩$, where $|a_n⟩=(e^{-in\chi /2}|e⟩+e^{in\chi /2}|g⟩)/\sqrt{2}$. The atomic meter states encode $n$ through relative quantum phase and can become entangled with the field. The second Ramsey interferometry pulse and detector turn those correlations into the outcome probabilities above when $C=1$.

Tracing out the atom multiplies a field density matrix entry ${\rho }_{nm}$ by $⟨a_m|a_n⟩=\cos ((n-m)\chi /2)$. Hence the nonselective quantum measurement preserves every photon number population ${\rho }_{nn}$ while changing off-diagonal quantum coherence. For $\chi =\pi /2$, coherence between $|0⟩$ and $|1⟩$ is multiplied by $1/\sqrt{2}$ per fresh meter, even with no photons absorbed. Field quantum phase information can therefore be disturbed by an energy-preserving probe. No universal self-adjoint phase operator or naive canonical number–phase commutator is needed for this conclusion.

One possible finite-contrast quantum instrument makes the distinction sharper. Define diagonal Kraus operators $K_e=\cos ((\varphi +\chi \hat{N})/2)$ and $K_g=\sin ((\varphi +\chi \hat{N})/2)$. After their ideal readout, flip the reported label with probability $r=(1-C)/2$. The unnormalized selective maps areTheir traces reproduce $L_e,L_g$ on a Fock state, and their sum reproduces the displayed dephasing quantum channel, independently of $r$. At $C=0$ the recorded labels carry no information although the physical meter interaction can still cause measurement backaction. This is an illustrative instrument, not a reconstruction of all laboratory imperfections; equal likelihoods alone do not determine the disturbance.