Past exam of the laser science and quantum information processing course of the University of Oxford / 2005 / 7 and 8 (DPHY 2752, DPHP 2752)

Source extract: users.physics.ox.ac.uk/~Foot/C2/pdf/C22005Q7.pdf

This source extract contains questions 7 and 8.

The year is attributed from the preserved Oxford support-page record; it is not printed on the held page.

TODO: The remainder of this historical paper is not held. Only the supplied question material is covered.

The printed opening block offers a choice of two protocols for 8 marks in total. All three choices are solved here. Sections 1 and 2 below identify the subsequent unlettered blocks, worth 10 and 7 marks respectively; these section numbers are navigation labels.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

In entanglement swapping, start with independent Bell pairs $|{\varphi }^{+}{⟩}_{12}|{\varphi }^{+}{⟩}_{34}$. Send qubits 2 and 3 to a central Bell state analyzer, retaining 1 and 4 at separate endpoints. Reordering the tensor product factors to group 23 and 14 gives the identityConsequently each ideal Bell-basis measurement outcome has probability $1/4$ and projects 14 onto the corresponding Bell state, even though those two photons have never interacted. Communicating the two-bit result permits an endpoint Pauli gate to convert every outcome into the same chosen Bell pair. Without that record, the endpoints have ${\rho }_{14}=I_4/4$, a product state, rather than a known entangled state.

The optical Bell state analyzer overlaps photons 2 and 3 at a balanced beam splitter, with matched spatial modes, spectra and arrival times, and resolves the output photon polarizations using polarizing beam splitters and single-photon detectors. Two-photon interference maps Bell states to different output patterns. The antisymmetric $|{\psi }^{-}⟩$ gives one photon in each output with opposite polarizations. The symmetric $|{\psi }^{+}⟩$ gives two opposite-polarization photons in the same output; the two polarization detectors there can resolve this pattern. The $|{\varphi }^{\pm }⟩$ states give same-polarization pairs and remain mutually ambiguous in this basic setup. The detected pattern therefore heralds a particular endpoint Bell state. With ideal detection this arrangement succeeds unambiguously on two of four equally likely inputs, with probability $1/2$; the successful events can be selected. A complete analyzer would accept all four outcomes. The endpoint quantum entanglement is conditional on the measurement record, so it enables no instantaneous signalling.^[ref]

For quantum key distribution with Bell pairs, a source distributes one photon to Alice and its partner to Bob. One implementation is the Ekert protocol, using the spin singlet state $|{\psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$. Each party independently chooses a polarization measurement setting after distribution. Equal settings give opposite outcomes; Bob reverses his outcome-to-bit assignment so these rounds give matching random bits. Other setting pairs test a Bell inequality, with ideal spin singlet state correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$ for the corresponding Bloch sphere axes. The optical polarizer angles correspond to twice those angles on the appropriate Bloch sphere great circle.

Over an authenticated public classical channel, Alice and Bob announce settings and detection times, select coincident detection rounds, and perform key sifting. They reveal a test sample to estimate the quantum bit error rate and the chosen Bell inequality signal, discarding all revealed bits. Excess disturbance causes an abort. Otherwise information reconciliation makes the retained strings agree and privacy amplification shortens them to remove an adversary's residual information, including information disclosed during reconciliation. The tested quantum entanglement supplies correlations unavailable to an independently prepared classical mixture; a security claim also requires stated detector and sampling assumptions.^[ref]

Each photon polarization analyzer uses a setting-dependent waveplate followed by a polarizing beam splitter. A single-photon detector on each output converts the routed photon into an electrical click, recording the outcome and its time. For example, a single-photon avalanche diode is a reverse-biased photodiode operated above breakdown: absorption creates a charge carrier, impact ionization triggers an avalanche, and the resulting pulse is registered. A quenching circuit lowers the bias to stop the avalanche and then resets the device. The detector dead time during recovery limits the count rate; dark counts produce false clicks, and imperfect detection loses rounds. The polarization information comes from the preceding analyzer, while the detector registers which output fired. Coincidence timing associates Alice's and Bob's records with the same emitted pair. ^[ref]

Let the unknown input qubit be $|\psi {⟩}_S=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared Bell pair is $|{\varphi }^{+}{⟩}_{AB}$. All three-qubit strings below use the order $S,A,B$. Initially,The Bell-basis teleportation identity givesThus a perfect Bell state analyzer on Alice's two qubits identifies the Pauli operator affecting Bob's quantum state.

To display the intermediate three-qubit states explicitly, realize that Bell-basis measurement by a controlled-NOT gate $S\to A$, a Hadamard gate on $S$, and two computational-basis measurements. After the controlled-NOT gate,After the Hadamard gate,Measuring $S,A$ gives $(m,n)$ with probability $1/4$ for each outcome, independently of $\alpha ,\beta$. Conditional on that result, the normalized three-qubit post-measurement state isIf the Bell state analyzer is instead regarded as an ideal direct projective measurement preserving the measured qubits, replace $|mn{⟩}_{SA}$ here by the corresponding Bell state; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's photons.

Alice sends the two classical bits $(m,n)$ to Bob. During transmission the above conditional three-qubit state is unchanged. Bob then applies the inverse Pauli operator $Z^m{X}^n$:The final conditional three-qubit state is therefore $|mn{⟩}_{SA}\otimes |\psi {⟩}_B$. The protocol transmits two classical bits per teleported qubit, consuming one shared Bell pair. Without Alice's record Bob averages over four Pauli operators and has density matrix $I/2$, so quantum teleportation respects the no-communication theorem. Alice's input has been measured and is no longer an independent copy, consistent with the no-cloning theorem.

For teleportation with an ambiguous Bell-state phase label, write $\rho =|\psi ⟩⟨\psi |$. The Bell-basis measurement outcomes ${\varphi }^{+}$ and ${\varphi }^{-}$ each occur with probability $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the Pauli Z gate and gives $Z\rho Z$. The other two Bell states are identified correctly. Since Alice guesses each ambiguous label with probability $1/2$, the overall wrong-correction probability is $(1/2)(1/2)=1/4$. Averaging over the actual quantum measurement outcome and her guess gives the phase-flip channelConditional only on an ambiguous analyzer result, the output is $(\rho +Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.

Use the teleportation fidelity for a pure state target, namely the squared quantum fidelityThis compares the input with the resulting mixed state; averaging state vectors before forming a density matrix would give a different, incorrect answer. Direct substitution givesThus the maximum is $1$, attained by $|0⟩$ and $|1⟩$ up to a global phase, because these are eigenstates of the Pauli Z gate. The minimum is $3/4$, attained by every equal-weight quantum superpositionon the equator of the Bloch sphere. The unsquared quantum fidelity is $\sqrt{F_{\psi }}$ if that alternative convention is used. No average over different input quantum states is needed for the requested state-dependent answer.

The three printed unlettered blocks are identified below by navigation labels 1, 2 and 3. Their mark allocations are 10, 10 and 5 respectively.

A universal quantum gate set can implement any unitary operator on any finite number of qubits, exactly when arbitrary continuous gate parameters are allowed, or to arbitrary accuracy with a suitable finite gate alphabet. More precisely, for every target $V$ and $ϵ>0$ there is a finite quantum circuit $W$ with ${\min }_{\gamma }\Vert W-e^{i\gamma }V\Vert <ϵ$ in operator norm. This ignores only an overall global phase of the complete operation. Universality lets a fixed collection of controllable elementary operations run arbitrary quantum algorithms rather than requiring a different physical interaction for each algorithm.

All single-qubit unitary gates, together with the controlled-NOT gate between selectable pairs, form an exact universal quantum gate set. One way to see sufficiency is to eliminate entries of a finite unitary matrix by two-level rotations between computational basis vectors; these rotations decompose into controlled single-qubit rotations, which in turn decompose into single-qubit gates and controlled-NOT gates. Single-qubit gates alone preserve product states and cannot supply general quantum entanglement. A finite approximate universal example is the Hadamard gate, the T gate $T=\mathrm{diag}(1,e^{i\pi /4})$, and the controlled-NOT gate; finite exact words over a finite alphabet cannot represent every continuously parameterized unitary operator exactly.

Distinguish the Hamiltonian operator $H_{\mathrm{int}}$ from the Hadamard gate $H_{\mathrm{Had}}$. For the printed collisional phase gate,As written, $\Omega$ has units of energy. The Schrödinger equation givesbecause only $|01⟩$ has a nonzero energy eigenvalue. In the ordered computational basis $(00,01,10,11)$, defineIf an interaction strength is instead quoted as an angular frequency ${\Omega }_{\mathrm{f}}$, the Hamiltonian operator is $\hslash {\Omega }_{\mathrm{f}}|01⟩⟨01|$ and $\varphi ={\Omega }_{\mathrm{f}}\tau$. An alternative definition with $e^{+i\varphi }$ would have $\varphi =-\Omega \tau /\hslash$. The minus sign above follows the usual time-evolution operator convention. At $\varphi =\pi$ these conventions give the same $U_{\pi }$.

For the quantum circuit, use $X$ on the first qubit to move the selective minus sign from $01$ to $11$:Conjugating the target of this Controlled-Z gate by the Hadamard gate gives the required controlled-NOT gate:The rightmost matrix acts first. Thus the network applies $X_1,H_2$, then $U_{\pi }$, then $X_1,H_2$, and maps $|a,b⟩$ to $|a,b\oplus a⟩$, where $\oplus$ is exclusive or.

Define the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and write $s=e^{-i\varphi }$. After $U_{\varphi }$ the two-qubit quantum state isApplying the Hadamard gate to the first qubit gives the normalized quantum stateFor $\varphi =n\pi$, $s=(-1{)}^n$, soThe first is a product state. The second is exactly the minus Bell state in the printed phase gate convention; its reduced density matrices are $I/2$.

For three qubits, let $U_{12}$ use the ordered pair $(1,2)$ and $U_{23}$ the ordered pair $(2,3)$. The two diagonal quantum gates commute. For $\varphi =n\pi$ their action before the final Hadamard gates isIf $n$ is even this is $|+++⟩$, and $H_1{H}_3$ gives $|0⟩|+⟩|0⟩$. If $n$ is odd, group the quantum superposition by the middle qubit:Since $H_{\mathrm{Had}}|+⟩=|0⟩$ and $H_{\mathrm{Had}}|-⟩=|1⟩$, the final states areThe odd case is a GHZ state: each individual qubit has reduced density matrix $I/2$, and it is related to $(|000⟩+|111⟩)/\sqrt{2}$ by the local Pauli gates $Z_1{X}_3$. Conversely, applying those gates converts $|G⟩$ to that standard form. They have not been applied in the printed network, so the actual output used in the next part is $|G⟩=(|001⟩-|110⟩)/\sqrt{2}$.

Keep the actual odd-$n$ GHZ state from the preceding calculation. Resolving its first qubit in the Hadamard basis givesThus the $+$ outcome leaves $|{\psi }^{-}{⟩}_{23}$ and the $-$ outcome leaves $|{\psi }^{+}{⟩}_{23}$, each with probability $1/2$. Conditional on a known result the remaining qubits share a maximally entangled state.

When the result is not revealed, the appropriate density matrix is the probability-weighted mixture from a nonselective quantum measurement:The cross terms cancel. The displayed convex combination of product states proves directly that this is a separable quantum state. It still has perfect computational-basis measurement anticorrelation, so it is not itself a product state. It also equals the partial trace ${\mathrm{Tr}}_1|G⟩⟨G|$, as required by the no-communication theorem: a local quantum measurement with an unobserved outcome cannot change the remote reduced density matrix.