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For a quantum state ensemble of pure vectors and an orthogonal projection $P$, measure $\{P,I-P\}$, preserve the successful subspace, and decode failure to a fixed normalized vector $\phi$. The resulting quantum channel is $\Lambda (X)=PXP+\mathrm{Tr}[(I-P)X]|\phi ⟩⟨\phi |$. For ${\alpha }_k=\Vert P{\psi }_k\Vert$, its squared quantum fidelity on the $k$th source vector is ${\alpha }_k^4+(1-{\alpha }_k^2)|⟨{\psi }_k|\phi ⟩{|}^2$. The inequality $a^2\ge 2a-1$ on $[0,1]$ proves the displayed average bound. It depends only on the source density operator's weight in $P$ and does not require the emitted vectors to be orthogonal.

A two-qubit unitary gate can approximately clone the known pair $|0⟩,|+⟩$ with global cloning fidelity ${\cos }^2(\pi /24)$ for each input, using no additional workspace. To construct it, put $a=|00⟩$, $b=|++⟩$, $u=(a+b)/\sqrt{3}$ and $v=a-b$. The vectors $u,v$ are orthonormal, and $a,b$ have coefficients $(\cos (\pi /6),\pm \sin (\pi /6))$ in that basis. Choose output vectors with coefficients $(\cos (\pi /8),\pm \sin (\pi /8))$. Their inner product is $\cos (\pi /4)=1/\sqrt{2}$, exactly the overlap of the inputs $|00⟩,|+0⟩$. Their linear matching therefore defines a linear isometry on the input span and extends to a two-qubit unitary operator by unitary extension of a finite-dimensional isometry. Each output's overlap with its ideal copy is $\cos (\pi /6-\pi /8)=\cos (\pi /24)$, proving the stated squared quantum fidelity.

For density operators $\rho ,\sigma$ on the same finite-dimensional Hilbert space, quantum fidelity measures how close their quantum states are. This article uses the unsquared convention:Each square root is the positive square root of an operator, and $\Vert \cdot {\Vert }_1$ is the trace norm. Quantum fidelity is symmetric in its inputs; it is one exactly when $\rho =\sigma$, and zero exactly when $\rho \sigma =0$, meaning that the support of a positive operator for one state is orthogonal to that for the other. Larger quantum fidelity means greater similarity. The separate squared quantum fidelity is $F_{\mathrm{sq}}=F^2$. ^[ref]

For two normalized pure states, quantum fidelity is the magnitude of their inner product, $F=|⟨\psi |\phi ⟩|$. Thus $|0⟩$ compared with itself has $F=1$, while $|0⟩$ and $|1⟩$ have $F=0$. For $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, comparison with $|0⟩$ gives $F=1/\sqrt{2}$, $F_{\mathrm{sq}}=1/2$, and trace distance $D=1/\sqrt{2}$ by the pure-state trace distance and fidelity identity.

Why compare density operators rather than their entries? A change of coordinates should not make the same physical preparations look more or less alike. Unitary invariance of trace distance and fidelity makes quantum fidelity independent of a common basis choice. Multiplying a pure state vector by a global phase leaves its density operator unchanged, so its quantum fidelity with the original state is one. These features make quantum fidelity useful for checking whether quantum channels, state preparations, and Schumacher compression preserve a target quantum state.

The connection to Uhlmann's theorem supplies another beautiful interpretation: quantum fidelity is the largest absolute inner product between pure vectors obtained from a purification of a density operator for each state, using a sufficiently large common reference space. Uhlmann's 1976 paper extended transition probability to algebraic states, including those of a C-star algebra; for density operators its transition probability corresponds to our squared quantum fidelity. The measurement characterization of quantum fidelity gives a complementary description directly in terms of observable outcome statistics. ^[ref]

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

For a quantum cloning device with pure state output $U|\varphi ,0⟩$, the global cloning fidelity is its squared quantum fidelity with the desired joint two-copy pure state. It is the probability that a joint projector test for two ideal copies accepts the actual output. It differs from a one-clone marginal fidelity: correlations between the output qubits affect the joint overlap. For a controlled-NOT gate with input $|+⟩|0⟩$, the output is a Bell state and its global cloning fidelity with $|++⟩$ is $1/2$; for input $|0⟩|0⟩$ it is one.

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Solutions to all three questions.^[ref] System labels specify the order of each tensor product of quantum systems. In an operator product the rightmost operator acts first. In Question 3(d), the stated fidelity is the squared quantum fidelity of the full two-qubit output with the desired two-copy pure state.

For input $|0{⟩}_S$, the controlled-NOT gate fixes $|00{⟩}_{SC}$, which is the ideal two-copy pure state. Thus its global cloning fidelity is $F_0=1$. For input $|+{⟩}_S$, the actual output is the Bell state $|{\varphi }_{+}⟩$, whereas the desired output is $|++⟩$. Their overlap isWith the given equal input probabilities,These values use the specified squared quantum fidelity for the full output, rather than a one-copy marginal fidelity.

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.

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 desired output is $|{\psi }_d⟩=((\alpha +\beta )|0⟩+(\alpha -\beta )|1⟩)/\sqrt{2}$. To evaluate its squared quantum fidelity, write the input Bloch vector as $(x,y,z)$, with $x=c$, $y=2\mathrm{Im}({\alpha }^{\ast }\beta )$ and $z=d$. The desired vector after the Hadamard is $(z,-y,x)$. From part (a), the actual vector isFor $k\ge 2$ this simplifies to $(1-1/k)(z,-y,x)-(x/k,0,0)$. The pure state overlap is one half of one plus the dot product with the desired unit vector, soSince $x^2+y^2+z^2=1$, $|xz|\le 1/2$. Therefore the infidelity is uniformly bounded by $3/(4k)$, and the entanglement effect on the implemented gate is negligible for large $k$, uniformly over all initial pure states. The shifted field packets then have overlaps tending to one, so they retain negligible information about the qubit transitions. A large fixed photon number without this broad number coherence would not suffice.

The printed overlap silently requires $k\ge 2$. If the allowed packet has $k=1$, the exact result instead isFor instance, $\alpha =\cos (\pi /8)$ and $\beta =\sin (\pi /8)$ give actual overlap $3/8$, whereas substituting $k=1$ into the printed expression gives $1/4$. This boundary correction follows directly from the two-shift overlap above.

Use the rotation-invariant distribution over pure initial states, namely the uniform pure-qubit average with area element $\sin \theta d\theta d\phi /(4\pi )$. Reflection symmetry gives $\mathbb{E}[xz]=0$, and rotational symmetry together with $x^2+y^2+z^2=1$ gives $\mathbb{E}[y^2]=1/3$. Averaging the squared quantum fidelity from part (b) therefore yieldsIf the single-number packet $k=1$ is included, its corrected average isUniform polar angle without the sine weight would not be the Haar-uniform average over all pure qubit states.

A compression scheme consists of quantum channelswhere each is a CPTP map. The Hilbert space ${\mathcal{K}}_n$ is the compressed register. Its block rate in qubits per source qubit is $R_n=n^{-1}{\log }_2\dim {\mathcal{K}}_n$, with asymptotic rate $R={\mathrm{limsup}}_n{R}_n$ (or the limit when it exists).

Use the squared quantum fidelity convention for the ensemble overlap:This definition tests preservation of the emitted vectors themselves, including nonorthogonal ones; preserving only their classical labels would be a different task. The scheme is reliable when $F_n\to 1$ as $n\to \infty$. This is the ensemble-average criterion for reliable quantum source compression. A square root is not taken in this definition, even though an unsquared convention is also used for quantum fidelity.

The Euclidean norm distance between normalized pure state vector representatives in a common Hilbert space. Its square is $2-2\mathrm{Re}⟨\psi |\varphi ⟩$. If vectors differing by a global phase are identified, minimize over that phase to obtain $\sqrt{2-2|⟨\psi |\varphi ⟩|}$. This is a different distance convention from trace distance or one minus squared quantum fidelity; stating the convention is necessary when defining $\epsilon$-closeness.

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.

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The University of Oxford Second Public Examination paper is dated Trinity Term 2022, Friday 3 June, 14:30–16:30. Its archive member is 2021/T/a16608w1.pdf; the printed cover establishes the year. Question 1 occupies printed pp. 2–3, Question 2 pp. 4–5, and Question 3 pp. 6–7. The final cloning task uses full-output squared quantum fidelity.

A qubit, or quantum bit, is a quantum system whose state space is a two-dimensional complex Hilbert space. Choose an orthonormal basis $|0⟩,|1⟩$, called the computational basis. A pure state has a normalized representative $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, where $\alpha ,\beta$ are complex probability amplitudes and $|\alpha {|}^2+|\beta {|}^2=1$. General qubit states, including mixed states, require a density matrix. ^[ref].

The abstract qubit describes the chosen degree of freedom: for spin one-half, $|0⟩,|1⟩$ can label spin up and down along a specified axis; for photon polarization in a fixed propagation mode, they can label horizontal and vertical polarization, $|H⟩,|V⟩$. These implementations share the same two-dimensional linear algebra, but their preparation and measurement in quantum mechanics use different apparatus. A qubit therefore need not be an entire two-level particle. ^[ref].

The usefulness of a qubit lies in controlling quantum superpositions and their relative quantum phases, then converting them into outcome probabilities with quantum gates and an appropriate measurement. One ideal computational-basis measurement returns one label, zero or one. It does not reveal both probability amplitudes, their relative quantum phase, or all the bits of a larger register's expansion. Quantum state tomography instead estimates a qubit state using many repetitions of the preparation and suitable measurement settings. ^[ref].

An actual preparation and measurement example is Dederick and Beck's 2014 teaching-laboratory experiment. A 405-nm laser pumped two nonlinear crystals, producing signal and idler photons near 810 nm. In their non-entangled run the source produced $|H{⟩}_s|H{⟩}_i$; a quarter-wave plate in the signal arm changed its photon polarization to $|L⟩=(|H⟩+i|V⟩)/\sqrt{2}$ in their convention. The plate supplies a relative quantum phase between polarization components. An idler detection selected the signal trials; signal detectors measured horizontal/vertical, diagonal, and circular polarization bases on separate runs. Before the polarization analyzer, a half-wave plate at $22.5$ degrees selected the diagonal basis, and a quarter-wave plate at $45$ degrees selected the circular basis. Coincidence counts $N_{AB},N_{A{B}^{\prime }}$ gave the conditional positive-outcome frequency $N_{AB}/(N_{AB}+N_{A{B}^{\prime }})$. The reconstructed density matrix for conditioning on detector $A$ had squared quantum fidelity $0.96$ with $|L⟩$ and purity of a density operator $0.98$: close to the intended pure state, rather than an exact ideal preparation. ^[ref].

The model diagram explains why those settings matter. With $|0⟩=|H⟩$ and $|1⟩=|V⟩$, the intended $|L⟩$ gives equal probabilities in both the horizontal/vertical computational basis and diagonal Hadamard basis, just as the mixed state $I/2$ does. In the circular basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, however, the ideal $|L⟩=|+i⟩$ always gives the plus outcome, whereas $I/2$ still gives equal probabilities. These are deductions from the Born rule, not the paper's measured counts. Unequal detector efficiencies and finite samples must be accounted for when interpreting experimental frequencies; the ideal comparison assumes equally efficient detection.

Compare the qubit pure statewith $|0⟩$, using the computational basis. Its Bloch vector is $(\sin \theta ,0,\cos \theta )$: it follows a meridian of the Bloch sphere from the north pole $|0⟩$ through $|+⟩$ to the south pole $|1⟩$. The inner product with $|0⟩$ is $\cos (\theta /2)$, nonnegative on this interval. Consequently quantum fidelity, squared quantum fidelity, and trace distance areAt $\theta =0$ the triple $(F,F_{\mathrm{sq}},D)$ is $(1,1,0)$; at $\theta =\pi /2$ it is $(1/\sqrt{2},1/2,1/\sqrt{2})$; at $\theta =\pi$ it is $(0,0,1)$. The angle between the Bloch vectors is $\theta$, while the quantum fidelity uses its half-angle. The pure-state trace distance and fidelity identity becomes the familiar relation ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$.

For equal prior probabilities, the Holevo–Helstrom theorem gives optimal quantum state discrimination success $P_{\mathrm{succ}}=(1+D)/2$. Along this meridian it rises from $1/2$ to one, with $(1+1/\sqrt{2})/2\approx 0.8536$ at $|+⟩$. This connects the geometric trace distance to the best possible decision from a single preparation; it is a different measurement task from the target-acceptance probability defining squared quantum fidelity.

// Target: quantum-information-theory.bigb

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.