Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 (Quantum Information Theory)

The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

For a purification $|{\Psi }_{\rho }{⟩}_{RA}$ of $\rho$, the entanglement fidelity isUsing a Kraus representation $\Lambda (X)={\sum }_k{A}_{k}X{A}_k^{†}$ givesIn a Schmidt decomposition of the purification,Therefore

Embed the input qubit as the span of $|0⟩,|1⟩$ in the three-dimensional output. Because the input is pure, its only purification has a trivial reference and entanglement fidelity reduces to its survival probability:Since $|e⟩$ is orthogonal to the embedded qubit subspace,

The state is a convex combination of product density operators and is therefore a separable quantum state. Every product density operator has a decomposition into product pure states, so

The Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ has two nonzero Schmidt coefficients. Since the density operator is pure, every ensemble decomposition uses vectors in the same one-dimensional support, and hence

The four Bell states form an orthonormal basis, so their uniform mixture isIt is a product state and therefore

The linear extension of the stated k-reduction map isIt suffices to consider a pure state $|v⟩$ of Schmidt rank $r\le k$, because positivity is preserved by sums. WriteThenFor every $|x⟩={\sum }_{ij}{x}_{ij}|ij⟩$, the Cauchy-Schwarz inequality givesThus the operator is a positive semidefinite operator. Applying this to every vector in a Schmidt-number-$k$ ensemble proves

Take Schmidt decompositions of the two purifications across $A:R$. Their squared Schmidt coefficients and their $A$-side eigenspaces are fixed by the same reduced state ${\rho }_A$. The reference-side Schmidt vectors are two orthonormal families, so a unitary $U_R$ maps one family to the other, including arbitrary choices inside degenerate subspaces. Hence the unitary freedom of purification givesIf the reference supports have different dimensions, the corresponding statement uses an isometry.

Fororthonormality of the $|b_i⟩$ givesNormalization follows from ${\sum }_i{\lambda }_i=1$, so this is a purification of a density operator.

With $|\Omega {⟩}_{RA}=n^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_A$,The partial trace over $R$ isIts norm is $\mathrm{Tr}{\rho }_A=1$. The positive square root of an operator $\sqrt{{\rho }_A}$ is unique, proving the claim.

The states ${\rho }_A^i=|a_i⟩⟨a_i|$ are pure, so a purification isTracing out the orthonormal reference labels removes all cross terms and recovers the stated classical-quantum mixture. Its rank is the number of nonzero $p_{ij}$, so that is the minimum reference dimension for a particular state. The smallest dimension that can purify every state of the stated form is

Construct ensemble purificationsThey have the same reduced state exactly when ${\rho }_1={\rho }_2$. By the unitary freedom of purification, this holds exactly when $|{\Phi }_2⟩=(I\otimes U)|{\Phi }_1⟩$ for a unitary $U$. Comparing reference-basis coefficients gives the Hughston–Jozsa–Wootters theorem relationConversely, substituting this relation and using $U^{†}U=I$ immediately gives ${\rho }_2={\rho }_1$.

Yes. Uhlmann's theorem says that, for the fixed purification $|{\Psi }^{\rho }{⟩}_{AR}$,Choose a maximizing purification of ${\sigma }_A$ on the same reference space. In the unsquared fidelity convention,

The finite-dimensional Stinespring dilation theorem states that every quantum channel $\Lambda :\mathcal{B}(H_A)\to \mathcal{B}(H_B)$ has an environment $H_E$ and an isometry $V:H_A\to H_B\otimes H_E$ such thatConversely, every map of this form is completely positive and trace preserving. From Kraus operators $A_j$, one may take $V={\sum }_j{A}_j\otimes |j{⟩}_E$.

Conjugation by ${\sigma }_z$ maps $(r_x,r_y,r_z)$ to $(-r_x,-r_y,r_z)$. The phase-flip channel therefore acts on the Bloch vector as

In the computational basis,At $p=1/2$, the two opposite off-diagonal contributions cancel, giving

A computational-basis projective measurement has projections $P_0=|0⟩⟨0|$ and $P_1=|1⟩⟨1|$. Discarding its result produces the nonselective projective measurementThusComplete phase randomization and unread computational-basis measurement implement the same dephasing on this qubit.

The four index pairs giveThe last global phase cancels under conjugation. Hence the Pauli channel probabilities are

Conjugation by one Pauli matrix preserves its parallel Bloch component and reverses the other two. Therefore

The trace distance and quantum fidelity, in the unsquared convention, are

Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be the positive-negative decomposition of a Hermitian operator. Since $\mathrm{Tr}\Delta =0$,For $0\le P\le I$,Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the variational characterization of trace distance

Distinct Bell states are orthogonal. For orthogonal pure states, the trace distance is one and the fidelity is the absolute overlap, zero. Hence

The difference $\tau ={\rho }_1-{\rho }_2$ determinesso it suffices for trace distance.

It does not determine fidelity. For $0<t<1$, both pairsandhave the same difference $\mathrm{diag}(t,-t)$. Their fidelities are respectively $\sqrt{1-t^2}$ and $\sqrt{1-t}$, which are generally unequal.

For the binary POVM $\{E,I-E\}$ and equal priors,Maximizing with the variational characterization of trace distance gives the equal-prior Holevo–Helstrom theorem

For priors $p_1,p_2$,If $\Delta =p_1{\rho }_1-p_2{\rho }_2$, the maximum is $p_2+\mathrm{Tr}{\Delta }_{+}$. Sinceone obtains

The generalized measurement postulate specifies operators $\{M_i\}$ satisfyingOn state $\rho$, outcome $i$ has probabilityand, when $p_i>0$, conditional stateThe effects $E_i=M_i^{†}{M}_i$ form a positive operator-valued measure.

Introduce a quantum ancilla with basis $|i⟩$ and defineThe completeness relation gives $V^{†}V=I$, so $V$ is an isometry and extends to a unitary operator on a sufficiently large system-plus-ancilla space. Prepare the ancilla in a fixed state, apply that unitary, and perform the projective measurement $\{I\otimes |i⟩⟨i|\}$. Outcome $i$ has probability $\Vert M_i|\psi ⟩{\Vert }^2$ and leaves the system in the normalized state $M_i|\psi ⟩$. By linearity the same holds for mixed states, implementing the generalized measurement.

LetThe three effectsform a POVM, because the largest eigenvalue of the sum of the two rank-one projectors is $1+\cos \theta$. Outcome 1 never occurs on $|\theta ⟩$, while outcome 2 never occurs on $|0⟩$. Thus conclusive outcomes are never wrong; $E_{?}$ records failure. This is unambiguous quantum state discrimination.

Arrange the amplitudes of $|{\varphi }_1⟩$ as the matrixThe second state has $C_2=C_{1}X$, where $X$ swaps the $A_2$ basis states. If Bob receives $A_1$, his two reduced density matrices areand are identical. No measurement on $A_1$ contains any information about the shared state.

If Bob instead receives $A_2$,They differ because $a^2+c^2<b^2+d^2$. Bob can therefore distinguish them with better-than-random success. For equal priors,so the optimal success probability is $\frac{1}{2}(1+D)$. It is generally below one, so a single copy does not permit certain identification.

Discarding the outcome of the complete projective measurement givesFor $m$ projectors, let $\omega =e^{2\pi i/m}$ and $U={\sum }_j{\omega }^j{P}_j$. Then the pinching identity isThe Concavity of Von Neumann entropy and its invariance under unitary operators implyEquality holds exactly when every conjugate in the average is the same, equivalentlyThus equality holds when the input already has no coherence between distinct measurement subspaces.