Past exam of the laser science and quantum information processing course of the University of Oxford / 2021 (A11045W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover dates this examination to 11 June 2021; the public bundle's filename is not the examination year. Source bundle pages 28–33. Solutions use the original conventions, with the Ramsey population-sign inconsistency identified explicitly in part 3(c).

Put $s=z-z_0$. In the source's complex beam parameter convention,Substitution separates the transverse wave amplitude and phase:Thus the Gaussian beam is circularly symmetric, has a Gaussian function transverse light intensity, and narrows to a beam waist at $z_0$ before spreading by diffraction. Its wavefront is planar at the beam waist, converging before it and diverging after it. The quadratic transverse phase identifies the signed radius of curvature:Define spot size as the wave amplitude $1/e$ radius, equivalently the light intensity $1/e^2$ radius. ThenThe Rayleigh range is where $w=\sqrt{2}{w}_0$; the far-field beam divergence half-angle is $\lambda /(\pi w_0)$. The factor $1/q$ also produces a longitudinal phase shift relative to a plane wave, changing by $\pi$ across the beam waist; its sign here follows the given $e^{i(kz-\omega t)}$ convention. The on-axis light intensity decreases as $1/(s^2+z_R^2)$ while transverse integration leaves the power constant. These are scalar paraxial approximation results.

A thin lens has ray transfer matrixThe complex beam parameter transformation gives $q_2=q_1/(1-q_1/f)$, henceThe sign is also obtained by multiplying the incident wave amplitude by the thin lens phase $\exp [-ik{\rho }^2/(2f)]$.

Choose the thin lens plane as $z=0$ and put $a=\pi w_0^2/\lambda$, the incident Rayleigh range. Since the incident beam waist lies there, $q_1=-ia$. Immediately after the thin lens,Subsequent free propagation adds $z$, so write $q(z)=z-z_f-i{z}_R^{\prime }$. Comparing real and imaginary parts givesFor a positive focal length, the new beam waist lies downstream but closer than the geometrical focus. At fixed $f,\lambda$, as $w_0\to \infty$,This is the formal unbounded-aperture scalar limit; eventually finite aperture and the paraxial approximation limit the extrapolation.

A chirp is variation of the instantaneous frequency through a signal. Here the group delay relative to the pulse centre at the optical lens isConsequently $\eta >0$ orders lower frequencies before higher frequencies; $\eta <0$ reverses that ordering. A representative strongly chirped optical pulse iswith a smooth localized envelope $A(t)$. The carrier oscillations become closer together towards later times for positive $\eta$ and farther apart for negative $\eta$.

The relation between the spectral phase and a locally assigned instantaneous frequency is a stationary phase method, not an exact reconstruction from spectral phase alone: the spectral wave amplitude also matters. For example, if the spectral envelope is $e^{-(\omega -{\omega }_0{)}^2/(2{\sigma }^2)}$, a Fourier transform of a Gaussian gives the exact real-field phase ${\omega }_{0}t+\eta t^2/(4|a{|}^2)$ up to a constant, where $a=1/(2{\sigma }^2)-i\eta$. The slope retains the sign of $\eta$ and approaches $1/(2\eta )$ for $|\eta |{\sigma }^2\gg 1$.

Place the optical lens at $z=0$. For a sufficiently broad incident Gaussian beam, the beam waist positions approach the focal lengths frequency by frequency:Add the group delay at the optical lens to propagation through vacuum:Here an arbitrary common time origin has been omitted. Differentiating both equations with respect to angular frequency yieldsFor $\beta >0$, define $u=2\eta c/\beta$. The velocity graph in the preceding figure has horizontal asymptote zero and a vertical asymptote at $u=-1$. For $\eta >0$, $0<v_f<c$: the higher-frequency components both enter later and travel farther. At $\eta =0$, $v_f=c$. For $-\beta /(2c)<\eta <0$, $v_f>c$: the farther-focusing components leave earlier, partly compensating their extra travel. At $\eta =-\beta /(2c)$ every component focuses at the same time, so the trajectory is not a finite-velocity function of time. For $\eta <-\beta /(2c)$ the farther positions focus first, giving $v_f<0$; it tends to zero from below as $\eta \to -\infty$.

A flying focus is a moving pattern assembled from different portions of an optical pulse. Each portion propagates causally at $c$ in this model. The superluminal or backward motion of the focus therefore does not describe a material object or information overtaking light. The calculation assumes negligible additional frequency-dependent group delay from the optical lens; any such delay must be included in $\psi$.

In a direct band gap semiconductor, the small photon momentum makes optical transitions approximately vertical in wave vector. Their energy isThe joint density of states per unit volume per transition energy, including the supplied spin counting, isFor an upper electron state of occupation $f_c$ and a lower electron state of occupation $f_v$, stimulated emission has occupation factor $f_c(1-f_v)$; absorption has factor $f_v(1-f_c)$. Their difference is exactly $f_c-f_v$. Multiplication by the joint density of states therefore givesThe proportionality suppresses the optical transition matrix element and other smoothly varying optical prefactors. It assumes parabolic isotropic energy bands, an allowed direct transition, and negligible linewidth and many-body corrections. Here $f_v$ counts electrons, not electron holes.

A quasi-Fermi level is the chemical potential parameter of an approximately thermalized carrier population within one energy band. Rapid intraband scattering can establish a Fermi-Dirac distribution in each energy band before slower interband recombination equilibrates their populations. Thus separate $E_{\mathrm{Fc}}$ and $E_{\mathrm{Fv}}$ describe a pumped population inversion even though the entire semiconductor is far from thermal equilibrium.

At a common positive temperature, the Fermi-Dirac distribution is strictly decreasing in $E-E_F$. ConsequentlyThe joint density of states is nonzero only above the band gap, soIn particular the quasi-Fermi-level splitting must exceed $E_g$. The upper boundary is transparency, where stimulated emission balances absorption; the lower boundary has zero joint density of states in this ideal three-dimensional parabolic model. The strict occupation comparison uses $T>0$; at exactly zero temperature occupations become step functions, and their supports should be checked explicitly, as in the next part.

Assume equal excess electron and electron hole number densities $N_t$, with no additional doping imbalance. At zero temperature, both occupied carrier spheres have the same Fermi wavevector $k_F$ because their spin counting is the same:For $k<k_F$, the conduction band is occupied and the valence band has electron holes, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-optical gain interval is the supremumEquivalently $E_{\mathrm{Fc}}=E_g+{\hslash }^2{k}_F^2/(2m_c^{\ast })$ and $E_{\mathrm{Fv}}=-{\hslash }^2{k}_F^2/(2m_v^{\ast })$. With the supplied effective masses,Using the vacuum wavelength $\lambda =hc/(\hslash \omega )$ reverses the endpoints:These boundaries are idealized and open: the lower-energy edge has zero joint density of states, and the upper edge is the zero-temperature occupation discontinuity, not a finite-width gain peak. The calculation gives positive material optical gain, not the stronger requirement of exceeding the losses at laser threshold.

A double heterostructure laser sandwiches a narrow-band gap active layer between wider-band gap cladding layers. In a type-I band alignment, the conduction band edge is lower and the valence band edge higher in the active layer. Electrons therefore see an energy well, and electron holes also see a well because their excitation energy increases when the valence band edge decreases. Both carrier populations remain concentrated where stimulated emission occurs.

The active layer also normally has the higher refractive index, forming an optical waveguide. The optical normal mode overlaps the injected carriers strongly, and the wider-band gap cladding has less interband absorption at the emitted photon energy. Carrier confinement reduces leakage, while optical confinement reduces the required pump current and hence heating.

Writing $\Gamma$ for the fraction of the optical normal mode overlapping the active layer, the threshold condition is $\Gamma g={\alpha }_i+{\alpha }_m$, with internal and mirror losses ${\alpha }_i,{\alpha }_m$. A larger overlap and a higher carrier number density at a given current both lower the threshold current. The figure shows composition-induced offsets; a working junction also has band bending under bias, which is omitted to make the confinement mechanism visible.

Use the source's north-pole assignment $|1⟩\leftrightarrow (0,0,1)$. Fix the ordered basis as $(|1⟩,|0⟩)$ and use the standard Pauli matrices in that order, soA pure state has $|\mathbf{r}|=1$: examples are $|1⟩$ at the north pole and $(|1⟩+|0⟩)/\sqrt{2}$ at $(1,0,0)$. An incoherent mixed statehas $\mathbf{r}=(0,0,2p-1)$ inside the Bloch sphere. At $p=1/2$ it is the origin, distinct from an equatorial pure state despite equal populations. The distinction is the off-diagonal quantum coherence of the density matrix.

For the stated Hamiltonian operator, put $\mathbf{\Omega }=(V,0,-\delta )$ and $\Omega =\sqrt{V^2+{\delta }^2}$. The Bloch vector obeysThus it rotates right-handedly about the fixed axis $\mathbf{\Omega }$, preserving its length and its projection on that axis. For example, on resonance $\delta =0$, a north-pole Bloch vector follows $(0,-\sin Vt,\cos Vt)$ and reaches the south pole at $t=\pi /V$.

Take $V>0$, with the field phase chosen accordingly. The Hadamard gate is $H=({\sigma }_x+{\sigma }_z)/\sqrt{2}$. Align the axis with $(1,0,1)$ and rotate through $\pi$:The factor $-i$ is an irrelevant common global phase. This realizes the whole quantum gate, rather than merely its action on one input quantum state.

There is a basis-ordering convention to make explicit: this $H$ is the Hadamard gate matrix in the ordered basis $(|1⟩,|0⟩)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H_{01}|1⟩=(|0⟩-|1⟩)/\sqrt{2}$, its matrix in our order is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$. For that definition choose $\boxed{\delta =+V,\tau =\pi /(\sqrt{2}V)}$. Both conventions use the same pulse duration; the sign of the Rabi detuning selects the desired axis.

At resonance, the Hamiltonian operator becomes $\hslash V{\sigma }_x/2$. ThereforeIt gives the requested positive rotation about the x-axis. In the ordered basis $(|1⟩,|0⟩)$, $|1⟩$ becomes $(|1⟩-i|0⟩)/\sqrt{2}$, whose Bloch vector is $(0,-1,0)$.

A resonant Rabi pi pulse givesIt swaps $|0⟩$ and $|1⟩$ with a common global phase, implementing the Pauli gate $X$, or NOT. Its Bloch sphere action is a half-turn about the $x$-axis.

Keep the Pauli matrix and north-pole conventions of part (a), and neglect Rabi detuning during each short pulse. Write $P=R_x(\pi /2)$ and $\varphi =\delta T$. During the first pulse the Bloch vector is $(0,-\sin \theta ,\cos \theta )$ for $0\le \theta \le \pi /2$. During free evolution,because the Hamiltonian operator generates $R_z(-\delta t)$. During the final pulse it isThese formulas specify every arc in the two Bloch sphere sketches. For $T=\pi /(2\delta )$ the path is north $\to -y\to -x$, and the last pulse leaves $-x$ fixed. For $T=\pi /\delta$ the path is north $\to -y\to +y\to$ north.

To test the complete quantum gate, multiply its unitary operators:It is proportional to the Hadamard gate when $\cos (\varphi /2)=-\sin (\varphi /2)$, hence the shortest positive waiting time isMore generally $\delta T=-\pi /2(\mathrm{mod}2\pi )$. With the alternative conventional ket definition $H_{01}$ specified in part (b)(i), the required matrix is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$, so $\boxed{T_{H_{01}}=\pi /(2\delta )}$ and $U_R=-i{H}_{01}$. The distinction concerns which basis order defines the named Hadamard gate; the evolution and the return population do not depend on that naming choice. Finally $r_z=-\cos \varphi$, so the actual return population isThe printed plus sign is inconsistent with the two identical pulses. At $T=0$, $U_R=P^2=-i{\sigma }_x$, so $|1⟩$ becomes $|0⟩$ and ${\rho }_{11}(0)=0$; the printed expression gives one. This counterexample is independent of Bloch sphere labeling. A plus sign would describe the complementary population ${\rho }_{00}$, or the return population with an opposite-sign final pulse. That changed sequence also changes the gate question and must not be silently substituted.

Interpret the unspecified Gaussian width $\Delta$ as the standard deviation, soThe characteristic function of this Gaussian distribution isAverage the free-evolution Bloch vectors from part (c):It starts at $-y$ and spirals towards the origin in the equatorial plane, with period $2\pi /{\delta }_0$ and shrinking length $e^{-{\Delta }^2{T}^2/2}$. Individual pure states retain unit-length Bloch vectors; only ensemble averaging produces the apparent loss of quantum coherence.

After the final pulse, the corrected return population isThe oscillations lie between $[1\pm e^{-{\Delta }^2{T}^2/2}]/2$, start at zero, and tend to $1/2$. The upper envelope has no implication of exponential energy relaxation: this is static inhomogeneous dephasing. If instead the intended Gaussian convention was $p\propto e^{-(\delta -{\delta }_0{)}^2/{\Delta }^2}$, every envelope above becomes $e^{-{\Delta }^2{T}^2/4}$. The printed-plus-sign convention would give the complementary fringe.

An ideal instantaneous NOT about $x$ reverses each atom's accumulated phase relative to its subsequent free precession. Algebraically, $X{\sigma }_{z}X=-{\sigma }_z$, so for a static Rabi detuning and $F=e^{i\delta T{\sigma }_z/4}$,The entire Ramsey interferometry sequence is consequently $PXP=-iI$, up to the additional common global phase if NOT is implemented as $-iX$. ThusThis spin echo restores the initial population even for the distribution of static Rabi detunings in part (d). It removes reversible inhomogeneous dephasing; fluctuating Rabi detuning, finite pulse errors and irreversible loss of quantum coherence need not refocus. With an opposite final pulse, the return population would instead be zero; this distinguishes that alternative from the sequence actually specified.

A universal quantum gate set can approximate any finite-register unitary operator to arbitrary accuracy, disregarding a common global phase. A finite example is the Hadamard gate, T gate and CNOT gate. In the ordered one-qubit basis $(|g⟩,|e⟩)$,In the ordered two-qubit basis $(|gg⟩,|ge⟩,|eg⟩,|ee⟩)$, with the first qubit the control,The one-qubit gates generate a dense set of one-qubit unitary operators, and CNOT gates supply entangling operations. Approximate universality is the appropriate qualification for a finite discrete set: its finite circuits cannot exactly equal every member of a continuous uncountable unitary group.

Without interaction, the two-qubit Hamiltonian operator is $H_a\otimes I+I\otimes H_b$, so its evolution factors as $U_a\otimes U_b$. It maps each product state to another product state, and cannot implement an entangling two-qubit gate such as a CNOT gate.

In an ion trap, the Coulomb interaction couples the ions' displacements into collective vibrational normal modes. The qubit states are internal atomic states; lasers coupling internal excitation to creation or annihilation of a shared vibrational quantum allow that motion to carry conditional information between ions. Thus the physical mediator is a collective motional normal mode of the Coulomb-coupled chain, rather than a direct resonant exchange between isolated internal qubit transitions. Ground-state cooling and resolved motional sidebands support the ideal pulse model used here.

Track the $-i$ factors only when the corresponding transition is addressed. They are common within the addressed two-state subspace, but become physically relevant relative phases between computational basis states. Using a semicolon to separate the shared motional number state, the complete paths areThe middle arrow comprises both Y pulses. In the last row those individual steps are $-i|ge;1⟩\to -|gf;0⟩\to +i|ge;1⟩$. The two X pulses contribute $(-i{)}^2=-1$ for either initially excited control; the Y pair supplies a second minus sign only when the second ion is also excited.

The shared motion returns to its ground state for all four inputs, and the auxiliary internal state is empty. Quantum superposition therefore gives the two-qubit gateDiscarding the X-pulse factors would incorrectly identify $G$ with a Controlled-Z gate itself.

The extra Pauli gate on the control is removed by $Z_{a}G=C_Z$. Conjugating the target by a Hadamard gate then givesThus, from left to right in the network, apply $H_b$, then $G$, then $Z_a$ and $H_b$; the last two gates commute.

For a direct proof, the control-zero block of $G$ is $I_b$ and its control-one block is $-Z_b$. Multiplication by $Z_a$ changes the latter into $Z_b$. Since $H_b{I}_b{H}_b=I_b$ and $H_b{Z}_b{H}_b=X_b$, the result iswhich is exactly the CNOT gate on all inputs, including quantum superpositions.

A bipartite pure state is entangled when it cannot be written $|\alpha {⟩}_a\otimes |\beta {⟩}_b$. For mixed states, the corresponding definition excludes every convex mixture of product states, not just a single product state.

Apply a Hadamard gate to ion $a$, followed by the CNOT gate of part (d):The lower network in the figure shows this Bell pair preparation. Either ion alone has reduced density matrix $I/2$, whereas the joint quantum state is pure; hence the joint quantum state cannot be a product state. The entangling operation can be expanded into $H_b,G,Z_a,H_b$ exactly as above.

An AND gate maps four possible pairs to one bit and loses information, so it is not reversible. A reversible embedding retains both inputs and uses a target bit:This is the Toffoli gate; initializing $c=0$ computes $ab$ while preserving enough information to invert the operation.

Classical reversible gates on one or two bits alone cannot synthesize every such operation. Every permutation of two bits is an invertible affine map over the finite field ${\mathbb{F}}_2$: there are $4\cdot |\mathrm{GL}(2,2)|=4\cdot 6=24$ of them, exactly the $4!$ permutations. Their extensions to a larger register and their compositions remain affine. The nonlinear term $ab$ in the Toffoli gate therefore cannot be generated. Three-bit reversible gates remove this obstruction, with ancillary bits initialized appropriately.

There is no conflict with quantum universality: one- and two-qubit gates may pass through quantum superpositions and use relative phases at intermediate steps. They are not restricted to permutations of computational basis states at every step. For example the Hadamard gate, T gate and CNOT gate can decompose a Toffoli gate exactly. Thus the classical affine obstruction does not apply to their intermediate unitary operators.

Use the ordered polarization basis $(|H⟩,|V⟩)$ and define the Bell statesRegrouping the four-photon quantum state by channel pairs $(1,4)$ and $(2,3)$ givesFor instance, projecting channels 2 and 3 on $⟨{\Psi }^{-}|$ retains the two original terms with inner polarizations $HV$ and $VH$, with opposite signs, leaving $(|HV⟩-|VH⟩{)}_{14}/(2\sqrt{2})$. Its squared norm is $1/4$.

By the same orthonormality argument, each Bell-basis measurement outcome prepares the identically named Bell state in channels 1 and 4, withBefore communication of the outcome, Alice and Bob's density matrix is $I_4/4$; the usable conditional entanglement is established by the announced Bell-basis measurement result. This assumes an ideal complete Bell-basis measurement; ordinary passive linear optics alone does not deterministically distinguish all four polarization Bell states.

On Bob's polarization qubit, takeSuitable conditional unitary operators areIndeed $(I\otimes XZ)|{\Phi }^{+}⟩=|{\Psi }^{-}⟩$, $(I\otimes X)|{\Phi }^{-}⟩=|{\Psi }^{-}⟩$, and $(I\otimes Z)|{\Psi }^{+}⟩=-|{\Psi }^{-}⟩$. The last minus sign is a harmless global phase.

For an ideal half-wave plate whose fast axis makes angle $\theta$ with horizontal polarization, discard the common phase and use the Jones matrixIt implements $Z$ at $0^{\circ }$ and $X$ at $45^{\circ }$. Thus for outcome ${\Phi }^{+}$ use two half-wave plates, first $0^{\circ }$ then $45^{\circ }$ in the propagation direction; for ${\Phi }^{-}$ use one at $45^{\circ }$; for ${\Psi }^{+}$ one at $0^{\circ }$; for ${\Psi }^{-}$ use no plate. Ideal retardance is assumed at the photon wavelength, and the setting is chosen after receiving the detector's classical result.

Each source has density matrixAssume the two sources and the detector's random choices are independent. In the $p^2$ branch both resources are ideal. If either resource is $I_4/4$, the remote pair remains $I_4/4$ after any specified Bell-basis measurement outcome. An ideal detector and the correction of part (b) therefore yieldWhen the detector instead reports an independent uniformly random label, the corresponding four Pauli gates average the conditional pair to $I_4/4$. ThusHere $q$ is the probability of the reliable-detector branch, not its total chance of being correct: that chance is $q+(1-q)/4$.

For the usual two-setting CHSH inequality, the spin singlet state has correlation $-\mathbf{a}\cdot \mathbf{b}$ and the white-noise part has zero correlation, so $E(\mathbf{a},\mathbf{b})=-v\mathbf{a}\cdot \mathbf{b}$. With Alice's axes $z,x$ and Bob's axes $(z+x)/\sqrt{2},(z-x)/\sqrt{2}$, the absolute CHSH inequality combination is $2\sqrt{2}v$. These settings are optimal for an isotropic correlation tensor, givingThis is the threshold for the standard CHSH inequality with optimized projective polarization measurements. The unqualified phrase “the Bell inequality” needs this choice of test; entanglement alone is a weaker requirement, and no claim about every possible Bell inequality is needed for this calculation.

After the first Bell-basis measurement and the announced correction, the shared Werner state has visibility $v=p^{2}q$. Ideal quantum teleportation through that resource is the depolarizing channelOne way to see this is to separate the singlet branch, which teleports perfectly with the appropriate known correction convention, from $I_4/4$, which has no correlations and yields $I_2/2$.

The additional Bell-basis measurement at Alice's station has its own independent reliable branch of probability $q$. In its random-label branch the correction averages the four Pauli gates, again giving $I_2/2$. Hence the actual teleported density matrix isOn sifted BB84 rounds, Bob measures in Alice's preparation basis. The preserved component gives the correct bit with certainty; the maximally mixed component gives an error with probability $1/2$. ThereforeFor $p=q=1$ it vanishes; if either parameter is zero it is $1/2$. The result presumes successful full Bell-basis measurement outcomes and independent stated imperfections, without extra loss, polarization drift or dark counts.

Quantize a single electromagnetic normal mode of angular frequency ${\omega }_c$ and represent one isolated atomic transition as a two-level system with raising operator ${\sigma }_{+}=|e⟩⟨g|$. The bare Hamiltonian operator isThe first main approximation is the electric-dipole approximation: the field varies negligibly across an atom, $k{a}_{\mathrm{atom}}\ll 1$, and the leading coupling is $H_d=-\mathbf{d}\cdot \mathbf{E}({\mathbf{r}}_0)$. For states of definite opposite parity, the relevant dipole part is off-diagonal. After choosing operator phases, this gives $H_d=\hslash g(a+a^{†})({\sigma }_{+}+{\sigma }_{-})$, with $g$ fixed by the transition dipole and the one-photon field wave amplitude.

In the interaction picture, $a{\sigma }_{+}$ and $a^{†}{\sigma }_{-}$ oscillate at the difference ${\omega }_a-{\omega }_c$, whereas $a{\sigma }_{-}$ and $a^{†}{\sigma }_{+}$ oscillate at the sum ${\omega }_a+{\omega }_c$. The second main approximation is the rotating-wave approximation: near resonance and for $g\sqrt{n+1}\ll {\omega }_a+{\omega }_c$, the latter terms average away on the interaction time scale. The resulting Jaynes-Cummings model hasThe two-level system and single-normal mode truncations are additional assumptions: other atomic transitions and optical normal modes must be sufficiently far off resonance or weakly coupled. Dissipation is neglected over the interaction time. The retained terms exchange one atomic excitation for one photon, conserving $a^{†}a+{\sigma }_{+}{\sigma }_{-}$.

Use the resonant interaction picture and define $g>0$ by the one-atom Jaynes-Cummings model in part (a). Equal coupling to both atoms gives the Tavis-Cummings modelThe initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution toThe antisymmetric atomic spin singlet state is uncoupled and unpopulated. Applying the annihilation operator and creation operator givesThe first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt{2}$ enhancement for creating the second photon. Its eigenvalues are $0,\pm \sqrt{6}g$. Put $\Omega =\sqrt{6}g$ and $K=H_I/\hslash$, so $K^3={\Omega }^{2}K$. Its exponential isApplying this to $|A⟩$ yieldsAll three states have the same bare energy on resonance, so restoring the free Schrödinger-picture evolution adds only a common global phase. The expression starts at $|ee;0⟩$, has initial derivative $-i\sqrt{2}g|B⟩$, and its three squared probability amplitudes sum to one.