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Represent the single-photon state by its two path probability amplitudes. The first Hadamard gate sends $|0⟩$ to $(|0⟩+|1⟩)/\sqrt{2}$. With $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$ for the lower-arm phase shifter, the complete Mach-Zehnder interferometer givesThe global phase has no effect on detection. By the Born rule,They sum to one. At $\varphi =0$, port 1 is a dark port; at $\varphi =\pi$, the roles reverse. The result uses coherent addition of the path amplitudes, rather than addition of separate path probabilities.

The two output ports are orthogonal single-photon states. Conservation of total probability at a lossless beam splitter gives $2c^2=1$. With $c$ real and positive,In the ordered input basis $(0,1^{\prime })$ and output basis $(0^{\prime },1)$, the beam splitter scattering matrix isFor the usual reciprocal lossless beam splitter, reversing propagation uses $S^{-1}=S^{†}=S$. Consequently,The relative phase in the first rule is essential to the repeated-path calculation. Overall port phases may be changed consistently; they cannot be changed on one pass alone while retaining the original scattering rules. In the following, take unit mirror reflection amplitudes by incorporating fixed reflection phases into the propagation-mode convention, and keep any unremoved propagation phase explicitly if needed.

Starting at port 1, the first beam splitter encounter sends amplitude $c$ straight out through port 0 and stores amplitude $-c$ towards M2. A return from M2 enters port $1^{\prime }$; it sends the stored amplitude times $-c$ out through port 1 and times $c$ towards M1. A return from M1 enters port $0^{\prime }$; it sends amplitude times $c$ out through port 0 and times $c$ towards M2. Both internal continuations carry $+c$, so two additional encounters multiply the surviving amplitude by $+c^2$.

With the fixed mirror and propagation phases absorbed as in a, the nonzero coefficients of the outgoing single-photon state areand $c_{0,2j}=c_{1,2j+1}=0$. The first nonzero terms are $c_{0,1}=1/\sqrt{2}$, $c_{1,2}=1/2$, $c_{0,3}=-1/(2\sqrt{2})$ and $c_{1,4}=1/4$. If fixed reflection amplitudes $r_1,r_2$ are retained instead, multiply $c_{0,2j+1}$ by $(r_1{r}_2{)}^j$ and $c_{1,2j}$ by $r_2^j{r}_1^{j-1}$, as well as the appropriate delay propagation phases. These unit-modulus factors do not alter the following probabilities.

Distinct exit events are orthogonal because they differ in output port or arrival-time bin. This requires the incident wavepacket to resolve the delay. Summing the squared moduli as geometric series givesTheir sum is one. After $n$ encounters the probability of still being inside is $2^{-n}$, which tends to zero, so no trapped component is omitted.

The source drawing places the delay line in the M2 arm. A visit to M2 crosses that line in both directions. If $\tau$ is its one-way delay, then an exit after $n$ encounters has crossed it $2\lfloor n/2\rfloor$ times:If the phrase “delay time $\tau$” instead denotes the complete M2-arm round-trip delay, replace $2\tau$ by $\tau$ in the floor formula. The wording does not explicitly distinguish these conventions. Consecutive encounters can produce the same exit time in different ports: for example $n=2,3$ both follow one M2 visit. If the incident wavepacket were much longer than the arm delay, different histories in the same port would overlap and their probability amplitudes would have to be added before squaring; the asserted time-bin expansion would then not be an orthonormal event basis.

A single-photon state lies in the one-excitation sector of the quantized electromagnetic field. It can be a quantum superposition over spatial ports or temporal modes while its total photon number remains one. Squared coefficients give exclusive event probabilities only when the corresponding wavepacket modes are orthonormal; overlapping temporal histories must be added coherently.