Past exam of the laser science and quantum information processing course of the University of Oxford / 2022 (A11045W1)

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 25–30. The cover establishes 2022 despite the attachment filename. The solutions use the original PDF's port labels and distinguish signed carrier-phase conventions.

In Q-switching, hold the optical resonator's quality factor low while pumping builds population inversion, then rapidly reduce the loss. The stored excitation produces a large optical gain, the intracavity light intensity grows, and stimulated emission depletes the population inversion until the optical gain falls below the laser threshold. The resulting optical pulse is a transient discharge of stored excitation.

Its duration depends on the high-quality factor photon lifetime, initial population inversion, saturation dynamics and switching time. For example, a simple laser rate equation model iswhere $S$ is intracavity photon number, $N$ is the available inversion in compatible units, $R_p$ is the pump rate and ${\tau }_c$ is the photon lifetime. During the short discharge, pumping and upper-state decay can often be neglected. The build-up and depletion rates determine the width; there is no universal equality between the width and ${\tau }_c$. The initially stored energy and the output coupling also matter. Typical Q-switching produces nanosecond-scale optical pulses.

In mode locking, fixed relative phases between many longitudinal normal modes produce constructive interference during a short part of each optical resonator round trip. The round-trip time sets the repetition period. The available coherent gain bandwidth, the number of locked normal modes, the spectral phase, and residual group velocity dispersion determine the optical pulse duration.

For $M$ consecutive equally weighted normal modes of angular frequency spacing $\Delta \omega$, their wave amplitude sum is proportional toIts central peak has width of order $1/(M\Delta \omega )$, while its repetition period is $2\pi /\Delta \omega$. A transform-limited pulse therefore becomes shorter as more coherent gain bandwidth is used. This mechanism can give picosecond or femtosecond optical pulses: it does not require depletion of the whole stored population inversion during each peak, as Q-switching does. Mode locking generally accesses a much shorter time-bandwidth product scale than the gain-discharge time of Q-switching.

Use the complex electric field convention $e^{-i\omega t}$ and take the locked wave amplitudes $a({\omega }_p)$ to be real and nonnegative after extracting their common phase. Linear superposition givesThe leading factor has unit modulus, so it does not change the light intensity. The triangle inequality bounds the remaining modulus by ${\sum }_{p}a({\omega }_p)$. Consecutive occupied normal modes attain this bound together atThere is also a peak at $t=0$. The electric field obeys $E(t+T)=e^{-i{\omega }_{\mathrm{ce}}T}E(t)$, although the light intensity is periodic.

The carrier-envelope phase specifies where the carrier oscillation lies relative to the maximum of an optical pulse's envelope. With $s=t-t_m$ and $E=F(s)e^{-i{\omega }_{0}s+i{\psi }_m}$, the signed complex-field phase at the envelope maximum is ${\psi }_m$. Consequently,If instead the real electric field is written $F(s)\cos ({\omega }_{0}s+{\varphi }_{\mathrm{CE},m})$, then ${\varphi }_{\mathrm{CE},m}=-{\psi }_m$ and the conventionally quoted slip is ${\varphi }_{\mathrm{slip}}=+2\pi {\omega }_{\mathrm{ce}}/\Delta \omega$. These are the same physical shift with opposite phase definitions.

The carrier-envelope phase matters for few-cycle optical pulses, when the largest instantaneous electric field affects strong-field dynamics, for high harmonic generation, and for stabilizing an optical frequency comb. An ordinary cycle-averaged light intensity measurement does not determine it.

Put $b=\Delta {\omega }_b>0$ and $s=t-t_m$. At the peak time, $e^{-ip\Delta \omega t_m}=1$, so the exact comb sum becomesFor many occupied normal modes, $b\gg \Delta \omega$, replace the sum by an integral with density $1/\Delta \omega$. Setting $x=\omega -{\omega }_0$ and evaluating the Fourier transform of a Gaussian givesThus the isolated-optical pulse approximation hasThe optical pulse has a Gaussian function envelope and no chirp. Its carrier angular frequency is ${\omega }_0$, and its light intensity full width at half maximum is $2\sqrt{2\ln 2}/b$. Successive envelope maxima have different carrier-envelope phases, consistent with b.

Approximation in the printed assertion. A discrete optical frequency comb gives a train, not one exact isolated Gaussian pulse. Applying the Poisson summation formula to the ideal sum over all integers instead givesNear $t_m$, the $j=m$ term dominates when $bT\gg 1$. The other terms are exponentially small but not identically zero. A single Gaussian function used globally would decay between $0$ and $T$, whereas the exact light intensity has equal maxima there; this supplies a concrete reason the isolated form cannot be an exact global identity. The extension to negative frequencies is the usual narrow-positive-band analytic-field approximation when ${\omega }_0\gg b$.

In second-harmonic generation, the quadratic nonlinear polarization contains products of pairs of comb components. Their angular frequencies areincluding $2{\omega }_p$ when $p=q$. Original comb components have angular frequencies ${\omega }_r={\omega }_{\mathrm{ce}}+r\Delta \omega$, henceWhen the doubled low-frequency region overlaps the original high-frequency region, choosing $r=p+q$ gives the beat at $|{\omega }_{\mathrm{ce}}|$. More generally the nearest positive beat has angular frequencywhich is $|{\omega }_{\mathrm{ce}}|$ for the stipulated small offset. For $0\le {\omega }_{\mathrm{ce}}<\Delta \omega$ it is $\min ({\omega }_{\mathrm{ce}},\Delta \omega -{\omega }_{\mathrm{ce}})$, provided the corresponding teeth are in the overlap.

In comb self-referencing, overlap the doubled red part and the original blue part on a photodiode. The beat measures the carrier-envelope offset frequency. Feedback on an intracavity dispersion control or pump setting can set ${\omega }_{\mathrm{ce}}=0$ modulo $\Delta \omega$, giving $\boxed{{\varphi }_{\mathrm{slip}}=0(\mathrm{mod}2\pi )}$. A phase-sensitive lock or calibrated approach to zero resolves the sign ambiguity of an unsigned beat measurement.^[ref]

The coherent optical spectrum must span an octave, ${\omega }_{\max }\gtrsim 2{\omega }_{\min }$, with sufficient overlapping spectral strength after second-harmonic generation. This calls for very short broadband optical pulses, or coherent external spectral broadening. The two branches must preserve optical coherence, overlap in time and polarization, and have adequate phase matching; bandwidth alone does not guarantee a detectable beat.

Use the complex beam parameter conventionHere $R_b$ is the wavefront curvature, distinguished from the mirror curvature $R$. In an unfolded propagation coordinate, a concave mirror has the ray transfer matrix $\left(\begin{array}{cc}1& 0\\ -2/R& 1\end{array}\right)$: a ray at height $x$ changes direction by $-2x/R$. The return to the original plane is thereforeThe Gaussian beam transformation $q^{\prime }=(Aq+B)/(Cq+D)$ givesFor $R=2L$, $M=\left(\begin{array}{cc}0& L\\ -1/L& 0\end{array}\right)$ and $q^{\prime }=-L^2/q=i{L}^2/z_R$. The returning Gaussian beam consequently has a beam waist at the same plane, withThis alone does not make it the same Gaussian beam; equal Rayleigh ranges require $z_R=L$.

Adding the plane mirror requires a self-consistent optical resonator normal mode, $q^{\prime }=q=i{z}_R$. Since $A=D$, the fixed-point equation reduces to $C{q}^2=B$, givingFor a finite stable Gaussian beam, $0<L<R$. The same formula remains valid at $R=2L$, where self-consistency selects $z_R=L$. At the geometrical hemispherical limit $R=L$ it predicts $z_R=0$; that marginal boundary has no finite-width stable fundamental Gaussian beam. Thus the source's word “hemispherical” is used for the plano-concave family, including the near-hemispherical case below.

Propagate from one beam waist to the other, using the optical lens ray transfer matrix:Optical mode matching requires $q_A=i{z}_A$ to transform to $q_B=i{z}_B$. Multiplying the denominator across givesEquating real and imaginary parts and multiplying by $f$ yieldsSet $\delta =u-f$. The second equation gives $v=f+(z_B/z_A)\delta$. Substituting this into the first cancels the terms $f\delta$:Consequently the two algebraic branches areEquivalently, $u=f+\delta$. The signs in $u$ and $v$ are the same. Real solutions require $\boxed{f^2\ge z_A{z}_B}$; a converging optical lens therefore needs $f\ge \sqrt{z_A{z}_B}$. At equality $u=v=f$. For the stated arrangement, additionally require $u,v\ge 0$ and that the chosen optical lens position is accessible outside both optical resonators. A negative $v$ is an algebraic virtual-waist solution, not a downstream beam waist.

The source optical resonator has Rayleigh rangeThe symmetric target optical resonator has its fundamental Gaussian beam beam waist halfway between the mirrors. At distance $\ell =L_2/2$ from that beam waist, wavefront curvature is $R_2=\ell +z_B^2/\ell$. ThusHere $f^2=2.25{\mathrm{m}}^2>z_A{z}_B=0.866025{\mathrm{m}}^2$, so b gives two real branches:Only the first branch has the target beam waist downstream of the optical lens. It is also geometrically accessible: the optical lens is $u-L_1\simeq 0.62642\mathrm{m}$ beyond the source output mirror, and the target input mirror is $v-L_2/2\simeq 7.44786\mathrm{m}$ beyond the optical lens. The total separation between beam waists is $u+v\simeq 14.07428\mathrm{m}$.

For a symmetric confocal optical resonator, the mirror separation equals either mirror's curvature, $R_2=L_2$, so its symmetric fundamental Gaussian beam hasThe same optical mode matching calculation applies, now with $f_{\min }=\sqrt{z_A{L}_2/2}$. Keeping the example's $L_2=10\mathrm{m}$ gives $z_B=5\mathrm{m}$ and $f_{\min }=0.707107\mathrm{m}$, compared with $0.930605\mathrm{m}$ for the nonconfocal target. With $f=1.5\mathrm{m}$, the two branches becomeAgain only the positive-$v$ branch fits the stated downstream geometry. The confocal target's smaller Rayleigh range means a smaller fundamental beam waist at the same wavelength and a less stringent minimum $f$.

A confocal optical resonator also has transverse normal mode degeneracies: in the ideal paraxial spectrum, increasing transverse order by two shifts the resonance by one longitudinal spacing. Such a cavity can transmit combinations of higher transverse normal modes at a fundamental resonance, so transmitted power alone is a weaker diagnostic of pure fundamental optical mode matching. The degeneracy does not remove the complex beam parameter matching requirement if the objective is specifically the symmetric fundamental Gaussian beam.

For one electromagnetic normal mode, $a$ and $a^{†}$ are the annihilation operator and creation operator, with commutator $[a,a^{†}]=I$. Their action on normalized number states isThey remove or add one photon, whose energy is $\hslash \omega$. The vacuum state $|0⟩$ is normalized and obeys $a|0⟩=0$. It has no photons, but retains zero-point energy $\hslash \omega /2$ for the quantum harmonic oscillator Hamiltonian $H=\hslash \omega (a^{†}a+1/2)$. The coherent state $|\alpha ⟩=D(\alpha )|0⟩$ is a displaced vacuum state.

Set $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$. Their commutator isIt commutes with both operators, so the stated Baker--Campbell--Hausdorff formula terminates:In particular $e^{-{\alpha }^{\ast }a}|0⟩=|0⟩$. Expanding the remaining exponential and using $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$ gives the number state expansionThe orthonormality of the number states givesThis is a Poisson distribution with parameter $\mu =|\alpha {|}^2$. Its probabilities sum to $e^{-\mu }{e}^{\mu }=1$. The phase of $\alpha$ affects the coherent state's relative probability amplitudes but not its photon number probabilities. At $\alpha =0$ the distribution is concentrated at $n=0$.

The number operator is $\hat{N}=a^{†}a$. The number state expansion gives $a|\alpha ⟩=\alpha |\alpha ⟩$ by shifting the summation index; taking the adjoint gives $⟨\alpha |a^{†}={\alpha }^{\ast }⟨\alpha |$. Thus the expectation value isFor the second moment, use $a{a}^{†}=a^{†}a+I$:Its expectation value in the coherent state is $⟨{\hat{N}}^2⟩=|\alpha {|}^4+|\alpha {|}^2$. ThereforeThe variance equals the mean, exactly as for the Poisson distribution found in b. For $\mu >0$ the fractional standard deviation is $1/\sqrt{\mu }$, the usual photon shot noise scaling. For the vacuum state, both the mean photon number and its variance vanish; this does not imply zero optical quadrature fluctuations. At $\mu =0$ the fractional fluctuation is undefined.

For the normalized coherent state, $⟨a⟩=\alpha$ and $⟨a^{†}⟩={\alpha }^{\ast }$. The dimensionless electric field and magnetic field optical quadratures consequently have expectation valuesThe physical electric field and magnetic field strengths have the corresponding proportionality factors and mode functions. In free evolution replace $\alpha$ by $\alpha e^{-i\omega t}$ to recover the classical oscillation.

Expanding the squared operators cancels the $a^2$ and $(a^{†}{)}^2$ terms:On the other hand, $(\alpha +{\alpha }^{\ast }{)}^2+[-i(\alpha -{\alpha }^{\ast }){]}^2=4|\alpha {|}^2$. HenceThe difference is the sum of the variances, not an extra mean field. Indeed $\mathrm{Var}(\mathcal{E})=\mathrm{Var}(\mathcal{B})=1$, independent of $\alpha$. The commutator $[\mathcal{E},\mathcal{B}]=2iI$ requires ${\sigma }_{\mathcal{E}}{\sigma }_{\mathcal{B}}\ge 1$, and the coherent state attains equality. The residual two units are the vacuum fluctuations of the two optical quadratures.

Zero-point fields are the irreducible electromagnetic field fluctuations in the vacuum state. Each normal mode has ground energy $\hslash \omega /2$. Although $⟨\mathcal{E}⟩=⟨\mathcal{B}⟩=0$ in the vacuum state, d gives $⟨{\mathcal{E}}^2⟩=⟨{\mathcal{B}}^2⟩=1$. These are quantum fluctuations, not a classical random coherent field with a known instantaneous value.

Two consequences are the Lamb shift, where coupling to the quantized electromagnetic field changes atomic energy levels, and spontaneous emission, where an excited chemical atom can emit a photon into an initially empty normal mode. In the latter process the creation operator has a nonzero matrix element $⟨1|a^{†}|0⟩=1$; stimulated-plus-spontaneous factors have the form $n+1$, which does not vanish at $n=0$. Descriptions splitting spontaneous emission into vacuum fluctuations and radiation reaction depend on operator-ordering conventions; the measurable total rate does not. Zero-point fields do not mean that the vacuum state contains a nonzero mean photon number.

The five DiVincenzo criteria are: a scalable collection of well-characterized qubits; preparation in a known fiducial quantum state; relevant coherence times long compared with quantum gate times; a universal quantum gate set; and qubit-specific quantum measurement. The criterion is useful control of the whole computation, including errors, rather than merely observing two resolved levels.^[ref]

For trapped ions, the implementation is:

For trapped neutral atoms, the corresponding implementation is:

Thus neither platform is forbidden from meeting a criterion in principle. Weak neutral-ground-state interactions make entangling control harder, but claiming that trapped neutral atoms cannot implement a universal quantum gate set would be false. Gate-model quantum algorithms using Rydberg blockade had already been demonstrated in a programmable neutral-atom processor by 2022.^[ref]

Let $x\in \{00,01,10,11\}$ and let $f(x)$ be one only at the marked input. A three-qubit implementation of Grover's algorithm starts with $|00⟩|1⟩$, applies a Hadamard gate to each qubit, queries the reversible quantum oracle $U_f:|x,y⟩\mapsto |x,y\oplus f(x)⟩$, applies the Grover diffusion operator $D=2|s⟩⟨s|-I$ to the first two qubits, and measures those two qubits in the computational basis. Here $|s⟩=\frac{1}{2}{\sum }_x|x⟩$. The full three-wire quantum circuit is the composition

Here $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Since $X|-⟩=-|-⟩$, phase kickback gives $U_f(|x⟩|-⟩)=(-1{)}^{f(x)}|x⟩|-⟩$. The third qubit remains separable and unchanged. A direct marked-state phase oracle $O_f|x⟩=(-1{)}^{f(x)}|x⟩$ therefore replaces the three-qubit query using only two qubits:For the explicit choice $f(x)=1$ only at $x=11$, $O_f$ is a Controlled-Z gate. Immediately after it,The mean of these four probability amplitudes is $\overline{a}=1/4$. The Grover diffusion operator maps each probability amplitude $a_x$ to $2\overline{a}-a_x$. The three unmarked probability amplitudes become zero and the marked one becomes one. A single query consequently finds $11$ with probability one.

A gate-level implementation of $D$ is $H^{\otimes 2}(2|00⟩⟨00|-I)H^{\otimes 2}$. Up to an irrelevant global phase, it is $H^{\otimes 2}{X}^{\otimes 2}\mathrm{CZ}{X}^{\otimes 2}{H}^{\otimes 2}$. For any other chosen marked word, conjugate the Controlled-Z gate by Pauli X gates on precisely the positions where the marked word has a zero.

In liquid-state NMR quantum computing, use two distinguishable spin-$1/2$ atomic nuclei in one molecule as the two qubits. Their different resonance frequencies permit selective radiofrequency pulses for one-qubit rotations. In a suitable rotating frame, refocusing removes unwanted one-spin terms while the NMR scalar coupling suppliesAssuming $J>0$, free evolution for $t=1/(2J)$ implements $U_J=e^{-i\pi Z_1{Z}_2/4}$. Define $R_z(\theta )=e^{-i\theta Z/2}$. ThenThis equality follows directly by evaluating the four diagonal entries: the first three are $+1$ and the $11$ entry is $-1$. Negative $J$ can be handled by refocusing or reversing the effective interaction sign. The global phase can be ignored.

Thus the marked-state phase oracle for $11$ requires scalar-coupling evolution and calibrated one-spin rotations. For a marked word $b_1{b}_2$, conjugate by $X_1^{1-b_1}{X}_2^{1-b_2}$, realized as selective $\pi$ pulses. The Grover diffusion operator uses the same entangling operation with additional one-qubit pulses. Liquid-state NMR quantum computing usually reads an ensemble signal and often prepares an effective pseudo-pure deviation from a large mixed background; that reproduces this small quantum algorithm's signal without providing a scalable source of pure initialized qubits.

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.

Expand the possibly unnormalized second-qubit kets as $|\alpha ⟩=a_0|0⟩+a_1|1⟩$ and $|\beta ⟩=b_0|0⟩+b_1|1⟩$. The joint quantum state isThe Born rule in the orthonormal computational basis gives $P_{00}=|a_0{|}^2$, $P_{01}=|a_1{|}^2$, and similarly for the two outcomes beginning with one. Ignoring the second outcome therefore givesEquivalently these are the squared norms of $(|0⟩⟨0|\otimes I)|\psi ⟩$ and $(|1⟩⟨1|\otimes I)|\psi ⟩$. Orthogonality of the first qubit's basis gives $P_0+P_1=⟨\psi |\psi ⟩=1$, even if $⟨\alpha |\beta ⟩$ is nonzero. For a nonzero-probability outcome, the conditional second-qubit quantum state is the corresponding ket divided by $\sqrt{P_i}$.

Read the source network as a Hadamard test: a Hadamard gate on the first qubit, a controlled unitary gate with the first qubit as control, and another Hadamard gate on that control. Starting from $|0⟩|\varphi ⟩$, the successive quantum states areApply b to the last line. Since $U^{†}U=I$ by unitarity,For the Pauli Z gate observable on the first qubit, the two eigenvalues are $+1$ and $-1$. ThusAs a check, $U=I$ returns outcome zero certainly, while $U=-I$ returns one certainly. The controlled unitary gate makes this relative phase measurable even though multiplying an isolated system's quantum state by $-1$ is only a global phase.

For second-qubit inputs $|0⟩$ and $|1⟩$, c gives control expectation values $\mathrm{Re}{U}_{00}$ and $\mathrm{Re}{U}_{11}$. Preparing the maximally mixed state $I/2$ is equivalent to choosing these inputs independently with equal probability and discarding the choice. Linearity of the density operator evolution and quantum measurement probabilities therefore givesHenceThe factor two is essential: the Hadamard test directly estimates the normalized trace. More generally, a target density operator $\rho$ gives $⟨Z⟩=\mathrm{Re}\mathrm{Tr}(\rho U)$. This argument never requires knowledge of the eigenvectors of $U$.

Yes. Replace the target qubit by a register of $k$ qubits and implement the controlled unitary gateThe algebra in c uses only normalization and unitarity, so for an arbitrary pure register quantum state $|\varphi ⟩$ it still gives $⟨Z⟩=\mathrm{Re}⟨\varphi |U|\varphi ⟩$. For a mixed register quantum state $\rho$, linearity gives $\mathrm{Re}\mathrm{Tr}(\rho U)$. With $\rho =I/2^k$,Only the control qubit needs to be pure. A uniformly randomized computational basis preparation supplies the target maximally mixed state. This is the principle of trace estimation by a maximally mixed Hadamard test. The extension is algebraic; efficiently implementing controlled-$U$ still requires a known suitable gate construction. Moreover, estimating the unnormalized trace to fixed additive precision requires increasingly precise estimation of the normalized signal as $k$ grows. There is no automatic efficient reconstruction of all entries of $U$.

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.

Let $S$ denote immediate transmission through the BS1 subsystem between Bob's lower-right port and Alice's upper-left port; let $L$ denote transmission after one visit to each internal mirror. From the same beam splitter rules as in b, the corresponding amplitudes in either direction areup to a common fixed two-mirror propagation phase. The minus sign in $L$ comes from the $1^{\prime }\to 1$ or reverse $1\to 1^{\prime }$ scattering. The two paths selected by D0's time gate are $L$ on the outward journey followed by $S$ on the return, and $S$ outward followed by $L$ on the return. They have the same total delay and fixed propagation phase, so their probability amplitudes interfere.

For the first history, Alice applies $e^{i{\varphi }_A}$ to the late component, while Bob's return-time modulation is not encountered in the return delay arm. For the second, Alice's early component is unmodulated and Bob applies $e^{i{\varphi }_B}$ in the return delay arm. Bob must time his modulation to the return passage, leaving the outward reference preparation unchanged; this is implicit in the stated encoding. Each enabled modulator is traversed twice, so its $\pi /2$ single-pass phase becomes the stipulated $\pi$.

Include the amplitude $c$ to enter BS1 through BS2, and the amplitude of modulus $c$ to reach D0 through BS2 on return. Both histories therefore contain six 50:50 beam splitter factors. Ignoring one common unobservable global phase,Hence the unconditional probability per emitted photon, with no renormalization after discarding paths, isFor the allowed values $0,\pi$, this is $1/16$ when ${\varphi }_A={\varphi }_B$ and zero otherwise. The ignored trajectories exit other ports or other time bins; excluding them from D0's acceptance does not increase the success probability. Equal mirror reflection phases, channel phases and Alice's mirror phase multiply both accepted histories equally and cancel from the interference condition. A residual relative phase would need calibration.

Encode $0$ by phase zero and $1$ by phase $\pi$. Alice and Bob choose their bits independently and keep their choices private. Bob announces the trial numbers with a D0 click over an authenticated public channel. For each retained trial, c guarantees that their bits agree. In the ideal symmetric setup, the two equal-phase cases $00$ and $11$ give the same click probability, so announcing success does not itself reveal which shared bit was used.

The matching-phase cases occur with probability $1/2$, and a matching case succeeds with probability $1/16$. ThereforeFor independent ideal trials, the raw-key length has a binomial distribution with standard deviation $\sqrt{31N}/32$. These are bits before sacrificing test positions or performing privacy amplification.