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Calibrate the two-Hadamard Mach-Zehnder interferometer so that the passive box leaves no relative wave phase or path disturbance. Then $H^2|0⟩=|0⟩$, and port 1 is dark. A photon at port 1 is therefore impossible for the passive box under these assumptions.

For an active ideal detector in arm 1, the first Hadamard gate prepares $(|0⟩+|1⟩)/\sqrt{2}$. Absorption, which triggers the detector, has probability $1/2$. On a no-absorption outcome, the surviving unnormalized state is $|0⟩/\sqrt{2}$. Applying the second Hadamard givesIts squared norm is $1/2$, the survival probability; normalizing it earlier and forgetting that probability would overcount successful trials. The unconditional outcomes for an active detector areThus a port-1 click identifies an active detector without triggering it, with single-pass success probability $1/4$, conditional on testing an active device. A port-0 click is inconclusive. Among trials in which the photon survives, half are successful, but that conditional fraction is not the single-pass success probability.

This is the balanced Hadamard interaction-free measurement version of the Elitzur-Vaidman bomb tester. Conditioning on no absorption changes the quantum coherence available for recombination; a calibrated dark-port click reveals that change. An uncontrolled phase shift in the passive arm could also brighten the port, so the inference depends on the stated optical calibration. It does not assign an observed classical trajectory to the photon before detection. If the same active device is retested after inconclusive clicks, the repeated weak-rotation bomb-test efficiency gives eventual success $(1/4)/(1-1/4)=1/3$; the requested single pass remains $1/4$.

Choose a real unitary matrix and undo it at the second stage:After the first rotation the photon qubit is $\sqrt{q}|0⟩+\sqrt{p}|1⟩$. A dud leaves this quantum superposition unchanged, so the inverse rotation restores $|0⟩$. Its final projective measurement can never return $1$.

For a live bomb the absorbing component explodes with probability $p$. If the photon survives, the normalized state is $|0⟩$, and the second rotation makes it $\sqrt{q}|0⟩-\sqrt{p}|1⟩$. The unconditional probabilities of the three possible live-bomb outcomes are consequentlyThey sum to one. The nonexplosive result $1$ certifies a live bomb because its dud probability is exactly zero. The inconclusive result $0$ resets the qubit to its original state, allowing another identical trial.

The repeated weak-rotation bomb-test efficiency follows by summing the geometric series over the first successful trial:This is conditional on testing a live bomb, not an average over an unspecified prior fraction of duds. To make the finite-run success probability exceed $1/2-\epsilon$, choose positive $p$ so that $p/[2(2-p)]<\epsilon /2$, then choose $N$ so that $q^{2N}<\epsilon /2$. The first condition controls the limiting deficit and the second controls the remaining tail. Exactly $p=0$ yields no explosion but also no information, so the limit must be approached with nonzero weak rotations.

The half-efficiency limit is consistent with the structure of this Elitzur-Vaidman bomb tester. Suppose generally $U|0⟩=a|0⟩+b|1⟩$. A final outcome that certifies a live bomb must be orthogonal to the dud output $U^{\prime }U|0⟩$. Pulling this outcome backwards through $U^{\prime }$ gives the normalized vector orthogonal to $(a,b)$. Its squared overlap with $|0⟩$ is $|b{|}^2$. Thus the safe-certification probability is $|a{|}^2|b{|}^2$, whereas the explosion probability is $|b{|}^2$. The former never exceeds the latter. Summing these mutually exclusive terminal events over repeated, possibly adaptive trials gives $P_{\mathrm{safe}}\le P_{\mathrm{explosion}}$ and hence $P_{\mathrm{safe}}\le 1/2$. This concerns the specified rotate–interact–rotate–measure protocol.