Past exam of the laser science and quantum information processing course of the University of Oxford / 2016 (A11045W1, C2)

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1

Mode locking fixes the relative phases of a laser's longitudinal cavity modes, so they interfere constructively in a regularly repeating short pulse rather than producing an uncontrolled temporal waveform. Active mode locking uses an externally driven modulation of loss, gain or phase at the cavity repetition frequency; an electro-optic modulator acting as a periodic transmission gate is one example. Passive mode locking uses the pulse itself to select a high-intensity waveform: a saturable absorber has lower loss at the pulse peak and can lock the phases without an external clock. A semiconductor saturable-absorber mirror is a practical example. Gain recovery, dispersion and saturation must support a stable circulating pulse.

Using the complex-field convention in the question, add the fields of all the oscillating modes. Set $m=p-q_0$, so that ${\omega }_p={\omega }_{q_0}+m\Delta \omega$. ThenThere are $N+1$ terms with the printed inclusive indexing. The sum is periodic with periodwhich is the separation of adjacent pulses when many modes contribute. The carrier factor need not repeat over this interval.

To specify the carrier-envelope phase sign, write the real field near pulse $j$, with $\tau =t-jT$, as $A(\tau )\cos ({\omega }_{q_0}\tau +{\phi }_j)$. Its cosine phase is ${\phi }_j={\omega }_{q_0}jT-{\varphi }_{q_0}$. Since $q_0\Delta \omega T=2\pi q_0$,Equivalently, the phase ${\varphi }_j$ in the complex convention $E_j=A(\tau )e^{-i{\omega }_{q_0}\tau +i{\varphi }_j}$ advances by $-{\omega }_{\mathrm{ce}}T$. This is the negative of the cosine phase, not a different physical prediction. The carrier-envelope offset frequency fixes this pulse-to-pulse slip. It matters for a few-cycle field's peak electric waveform and for locating an optical frequency comb's lines; a stable intensity envelope alone does not fix the slip.

If $\delta /\Delta \omega \gg 1$ and $N\Delta \omega /2\gg \delta$, many closely spaced modes lie under the spectral Gaussian and the cutoffs have negligible weight. Near one pulse, replace the sum by its continuum integral:The Fourier transform of a Gaussian therefore givesThis is a local single-pulse approximation. The discrete spectrum produces its periodic replicas; extending one Gaussian over all time would lose them. The temporal light intensity is proportional to $e^{-{\delta }^2{t}^2/2}$, while the spectral intensity envelope is proportional to $e^{-2{\Omega }^2/{\delta }^2}$. Their full widths at half maximum areIn ordinary frequency this time-bandwidth product is $\Delta {\nu }_L\tau =2\ln 2/\pi \simeq 0.441$. The spectral phases are flat, so this is a transform-limited pulse; chirp would generally increase its temporal width.

A longitudinal mode makes an integer round-trip phase accumulation. For the same transverse mode and an empty nondispersive cavity, increasing that integer by one changes the optical frequency by the free spectral rangeFrequency-dependent mirror phases or a dispersive intracavity medium would change the round-trip group delay used here.

For a paraxial ray, use the height-slope column $(x,\theta {)}^T$. Free propagation and reflection from a concave mirror are represented, in unfolded coordinates, byThe reflection changes the slope by $-2x/R$. Because the mirrors are identical, the same half-round-trip matrix is applied repeatedly:Its two eigenvalues are reciprocal. They lie on the unit circle without a growing component when $|1-l/R|<1$, whereas $l>2R$ gives a real eigenvalue of modulus greater than one. At $l=2R$, the repeated eigenvalue has a nontrivial Jordan block and generic rays drift out. Thus the nondegenerate stable interval isThe confocal value $l=R$ is included: there $A^2=-I$, so the repeated round trip is bounded. Applying a strict trace inequality to the full-round-trip matrix without checking this exact matrix would incorrectly discard that point. These are the paraxial conditions for a symmetric concave-mirror resonator.

Symmetry places the Gaussian beam waist halfway between the mirrors. Its Rayleigh range is $z_R=\pi w_0^2/\lambda$, and its wavefront radius at distance $z$ from the waist is $R_b(z)=z+z_R^2/z$. Matching that wavefront to the mirror at $z=l/2$ givesHenceA real positive Rayleigh range requires the same stable interval. At its limiting geometries the finite confined Gaussian mode degenerates; the formula is not a way to realize an arbitrarily small physical waist without leaving the paraxial model.

Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the Fock states givesActing with the annihilation operator and shifting the summation index givesThus a coherent state is an annihilation-operator eigenket. It is not also an eigenket of the creation operator: a common eigenket would make $[\hat{a},{\hat{a}}^{†}]$ annihilate it, contradicting $[\hat{a},{\hat{a}}^{†}]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relationWith $\hat{n}={\hat{a}}^{†}\hat{a}$, the photon number expected value is $\mu =|\alpha {|}^2$. Normal ordering ${\hat{n}}^2=({\hat{a}}^{†}{)}^2{\hat{a}}^2+{\hat{a}}^{†}\hat{a}$ gives $⟨n^2⟩={\mu }^2+\mu$. ThereforeAt large mean photon number the relative fluctuations vanish, giving a classical-field limit even though quantum fluctuations remain. For the vacuum $\mu =0$, the relative ratio itself is undefined.

The Born rule applied to the number-state coefficient gives the Poisson distributionIt depends on the single parameter $\mu =⟨n⟩$, and has equal expected value and variance, consistent with the preceding moments.

For a Fock state $|n⟩$, raising or lowering the number makes an orthogonal state, so the optical quadrature has $⟨\hat{X}⟩=0$. Expanding its square givesThe first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature variance $1/2$.

A crystal needs a nonzero effective second-order nonlinear optical susceptibility for the chosen polarizations. In the bulk electric-dipole approximation, centrosymmetry forbids this response. Transparency at the relevant frequencies and phase matching over the interaction length permit efficient buildup. Nonzero local generation can occur without exact phase matching, but contributions then tend to cancel as they propagate.

Write ${\psi }_j=k_{j}z-{\omega }_{j}t+{\varphi }_j$. Expanding the square and using the product-to-sum identity givesThese are respectively optical rectification, the two second-harmonic generation terms, sum-frequency generation and difference-frequency generation. Their phases and wavevectors are the corresponding sums or differences, not independently adjustable quantities.

For a vector field, $P_i^{(2)}=2{ϵ}_0{\sum }_{jk}{d}_{ijk}{E}_j{E}_k$. In the stated single-frequency amplitude convention, $E_j=(A_j{e}^{i\psi }+A_j^{\ast }{e}^{-i\psi })/2$. If $P_i^{(2\omega )}=[{\mathcal{P}}_i{e}^{2i\psi }+{\mathcal{P}}_i^{\ast }{e}^{-2i\psi }]/2$, comparison givesFor a real instantaneous tensor the zero-frequency contribution is $P_i^{(0)}={ϵ}_0{\sum }_{jk}{d}_{ijk}\mathrm{Re}(A_j{A}_k^{\ast })$. In a dispersive treatment, the tensors must carry their respective input and output frequency arguments.

The identical input-frequency pair is symmetric under $j\leftrightarrow k$, allowing contracted nonlinear susceptibility notation: replace the six pairs $xx,yy,zz,yz,zx,xy$ by $J=1,\dots ,6$. ThusThis contraction of a symmetric pair does not turn the physical susceptibility into a rank-two tensor. The factors of two account for both off-diagonal input orderings, and the prefactor follows from the question's explicit $1/2$ field convention.

For the proposed type-I process, energy conservation requires ${\omega }_1={\omega }_2+{\omega }_3$. HenceThe extraordinary pump and two ordinary output waves can be collinearly phase matched ifPut $n_{\ast }={\lambda }_{1}S$. Substituting the supplied angular extraordinary index and solving for ${\sin }^2\theta$ yieldsThe required index $n_{\ast }\simeq 2.2143$ lies between $2.1782$ and $2.2579$, so a real angle exists. Using the rounded 2922 nm wavelength in the supplied data gives $\theta \simeq 46.94^{\circ }$; retaining the energy-conserving 2922.25 nm value gives about $46.98^{\circ }$. Their common meaningful scale is about 47 degrees.

The spatial walk-off angle is the angle between the extraordinary wave's Poynting vector and its wavevector. For an optic-axis angle $\theta$, the extraordinary dispersion surface givesHere the indices are those of the 780 nm pump. Since this is a negative uniaxial crystal, its energy flow is farther from the optic axis than its wavevector. The ordinary waves' energy flows follow their wavevectors. The pump therefore drifts out of spatial overlap by roughly $L\tan \rho \simeq 0.0357L$, about $36\mu \mathrm{m}$ over a millimetre. This limits the useful nonlinear interaction length even when phase matching is exact.

For $|\alpha {|}^2+|\beta {|}^2=1$, the density operator and its computational-basis matrix areIt is a Hermitian matrix and satisfies ${\rho }^2=\rho$, so $\mathrm{tr}{\rho }^2=1$. Its eigenvalues are one and zero. Thus a pure qubit's density matrix is not unitary: it is singular, and ${\rho }^{†}\rho =\rho \ne I$.

Transposing gives ${\rho }^T=|{\psi }^{\ast }⟩⟨{\psi }^{\ast }|$. It reflects the Bloch vector $(x,y,z)$ to $(x,-y,z)$. Choose the positive axis states $|+⟩$, $|+i⟩=(|0⟩+i|1⟩)/\sqrt{2}$ and $|0⟩$; their target kets are respectively $|+⟩$, $|-i⟩$ and $|0⟩$.

Nonunitarity alone does not make a quantum operation unphysical. The qubit transpose map fails complete positivity: applying it to one half of a Bell pair gives a matrix with a negative eigenvalue. A physical approximation must instead be a valid quantum channel.

Use the squared quantum fidelity specified in the question, comparing each actual output to the transposed target. For a pure input Bloch vector $\mathbf{r}$, the identity approximation has $F_I=1-y^2$. Conjugation by the Pauli X gate produces $(x,-y,-z)$, giving $F_X=1-z^2$. Measuring in the computational basis and preparing the corresponding basis state is the dephasing channel $(x,y,z)\mapsto (0,0,z)$, giving $F_Z=(1+z^2)/2$. ThereforeThe stipulated three-axis averages also agree with the uniform Bloch-sphere averages, since each squared coordinate averages to $1/3$.

The complete perfect-fidelity input sets, up to a global phase, are:

Each condition follows by setting the corresponding fidelity formula to one, so these lists include all perfect cases rather than just the three test inputs.

The Hadamard gate H and T gate $T=\mathrm{diag}(1,e^{i\pi /4})$ form a universal quantum gate set for single-qubit operations: finite sequences approximate any target single-qubit unitary arbitrarily well, up to a physically irrelevant global phase. Universality is not a claim that every arbitrary rotation has a finite exact H,T word. With products acting right to left, exact Pauli constructions areIndeed $T^2=S$, $T^6=S^{†}$, and $SX{S}^{†}=Y$.

Initialize the search register in $|00⟩$ and the output qubit in $|1⟩$, then apply H to all three. The search state becomes $|s⟩=(|00⟩+|01⟩+|10⟩+|11⟩)/2$, and the output becomes $|-⟩$. The oracle obeysThis phase kickback leaves the output qubit factored out, so subsequent analysis can use the phase oracle $O_f$ on the two search qubits alone. Apply one Grover diffusion operator $D=2|s⟩⟨s|-I$, then measure the search register.

A promise restricts which Boolean functions may be supplied. For the exact one-iteration four-entry search here, promise exactly one marked input. The four functions are $f_{jk}(x)=1$ precisely when $x=jk$, for $jk\in \{00,01,10,11\}$. Other known numbers of marked entries require a correspondingly chosen iteration count; uniqueness is the promise for this particular exact circuit.

In basis order $00,01,10,11$, the requested explicit calculation isThus the promised marked entry is found with probability one.

For two weakly coupled spin-half nuclei in nuclear magnetic resonance, a useful rotating-frame Hamiltonian iswhere $J$ is in hertz and $I_z=Z_1/2$, $S_z=Z_2/2$. Resonant pulses and refocusing separate local rotations from the scalar coupling. For positive $J$, evolution for $t=1/(2J)$ supplies $e^{-i\pi Z_1{Z}_2/4}$. Combining it with local Z phases gives the phase oracleThe scalar global phase is immaterial. Local $\pi$ pulses interchange zero and one, soThis implements the desired change of marked input without an additional output qubit.

Encode $|0⟩=|H⟩$ and $|1⟩=|V⟩$ in two orthogonal linear polarizations. Then $|\pm ⟩=(|H⟩\pm |V⟩)/\sqrt{2}$ are linear polarizations at opposite forty-five-degree angles, while $(|H⟩\pm i|V⟩)/\sqrt{2}$ are the two circular polarizations. Their right/left names depend on the viewing convention. A half-wave plate with its axis at $22.5^{\circ }$ to H implements H up to a common phase, since its Jones matrix has entries $\cos 2\theta ,\sin 2\theta$ and $-\cos 2\theta$.

In principle, a nonlinear interaction producing $e^{i\pi n_{1V}{n}_{2V}}$ supplies a deterministic Controlled-Z gate: only the joint VV component acquires a minus sign. A strongly coupled matter mediator can also provide the required conditional interaction. At optical single-photon energies the direct nonlinear interaction is weak; loss, unwanted noise and maintaining indistinguishable spatial and temporal modes make a high-fidelity deterministic gate difficult. Passive linear optics by itself does not supply this deterministic entangling interaction; measurement-based optical gates introduce additional resources and success conditions.

For the printed circuit, the first three H gates create $|+{⟩}^{\otimes 3}$. The two controlled-Z operations from the first qubit to the others giveThe final H gates on the last two qubits therefore produce the GHZ stateDefine $|s_X⟩=(|0⟩+s|1⟩)/\sqrt{2}$ and $|s_Y⟩=(|0⟩+is|1⟩)/\sqrt{2}$ for $s=\pm 1$. In any of the XYY, YXY and YYX bases, direct overlaps givewith the same coefficient formula when the X position is permuted. Hence the three explicit expansions can be written compactly asIn each sum the four sign triples are $(+,+,-),(+,-,+),(-,+,+),(-,-,-)$. They have equal probabilities $1/4$. Thus two Y outcomes determine the X outcome with certainty: their product times the X outcome is $-1$.

In the XXX basis,The four allowed triples are $(+,+,+),(+,-,-),(-,+,-),(-,-,+)$, so the XXX outcome product is $+1$.

The GHZ theorem follows without a statistical inequality. If local predetermined values $X_i,Y_i\in \{\pm 1\}$ reproduced the three mixed-basis correlations, multiplying them would givebecause every Y value is squared. Quantum mechanics instead predicts the XXX product $+1$. This is an all-versus-nothing contradiction with local realism using ideal perfect correlations. Standard two-qubit Bell-state tests compare statistical correlations through a Bell inequality; the GHZ construction makes the inconsistency visible already at the level of these deterministic parity constraints.

The transmitted branch is the Kraus operatorFor nonzero loss it is not trace preserving: $\mathrm{tr}(E\rho E^{†})$ is the transmission probability. A complete channel includes an orthogonal vacuum or erasure flag, with a loss operator $L=\sqrt{R}|\mathrm{vac}⟩⟨0|$ and $E^{†}E+L^{†}L=I$. Postselecting only the first outcome gives the single-operator description in the question.

For $|\pm ⟩$, the unnormalized transmitted vector is $(\sqrt{1-R}|0⟩\pm |1⟩)/\sqrt{2}$. Its norm squared and normalized conditional state areThe reduced density matrix of either member of the initial pair has eigenvalues $\alpha ,1-\alpha$. Since the joint state is pure, its Von Neumann entropy is zero. With logarithms to base two and $h_2$ the binary entropy function,The two nonzero Schmidt coefficients show that this pure pair is entangled exactly when $0<\alpha <1$. Its mutual information rises from zero to two bits at $\alpha =1/2$ and returns to zero.

For this known pure family, every positive mutual information signals entanglement. For an otherwise unspecified two-qubit state, positive mutual information can also be classical. The mutual information bound for separable states gives the stronger general witness $I(A:B)>1$ bit. On this curve it holds for approximately $0.110028<\alpha <0.889972$; the rest of the open interval is still entangled, but that generic one-bit witness alone would not certify it.

If Bob's qubit passes through the channel, the unnormalized retained state is $\sqrt{\alpha }|11⟩+\sqrt{(1-\alpha )(1-R)}|00⟩$. ThusFor $0\le R<1$, equal Schmidt coefficients require $\alpha =(1-\alpha )(1-R)$, givingThe retained state is then $(|00⟩+|11⟩)/\sqrt{2}$. At $R=1$, this limiting compensation has zero transmission probability and yields no maximally entangled pairs. Conditioning can increase the entanglement of retained pairs, while the discarded fraction accounts for the cost.