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The generalized measurement postulate specifies operators $M_i$ with ${\sum }_i{M}_i^{†}{M}_i=I$. On a density matrix $\rho$, the outcome probability and conditional state areThe probabilities are nonnegative and sum to one. More generally an outcome may contain several Kraus operators $M_{i\alpha }$, in which case one sums over $\alpha$ in both formulas. A projective measurement with the usual Lüders update is the special case $M_i=P_i$, where $P_i=P_i^{†}$, $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Projection-valued effects alone fix the projective outcome probabilities, but an additional outcome-dependent unitary can change the conditional state; for the standard projective update the measurement operators themselves are the projectors.

A POVM is a family of positive effects $E_i$ summing to $I$. Its probabilities are $p_i=\mathrm{Tr}(\rho E_i)$. A generalized measurement gives effects $E_i=M_i^{†}{M}_i$, or $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$. Conversely $M_i=\sqrt{E_i}$ realizes any POVM. The POVM does not determine the post-measurement state: different operators with these same effects have the same probabilities and can have different state updates.

A pure POVM has rank-one nonzero effects, $E_i=w_i|{\psi }_i⟩⟨{\psi }_i|$, with normalized pure states $|{\psi }_i⟩$ and positive $w_i$. Here pure means rank-one effects, not the different property of extremality in the convex set of POVMs.

The POVM–ensemble duality for the maximally mixed state is a bijection between POVMs in dimension $d$ and ensemble decompositions of $I/d$. Given effects $E_i$, setomitting zero effects. Each ${\sigma }_i$ is a density matrix, ${\sum }_i{p}_i=\mathrm{Tr}I/d=1$, andConversely, if an ensemble satisfies ${\sum }_i{p}_i{\sigma }_i=I/d$, then $E_i=d{p}_i{\sigma }_i$ are positive and sum to $I$, hence form a POVM. Taking traces recovers the original probabilities, so the transformations are inverse. Pure POVMs correspond exactly to pure-state ensemble decompositions of the maximally mixed state. For such a POVM, $p_i=w_i/d$ and ${\sum }_i{w}_i=d$; it need not be an orthogonal measurement.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

The Schumacher compression theorem concerns a memoryless quantum information source with average density matrix $\rho$ on a finite-dimensional Hilbert space. For $n$ uses the average state is ${\rho }_n={\rho }^{\otimes n}$. Compression consists of an encoding quantum channel into a space of dimension $D_n$ and a decoding channel back to the original block space, with asymptotic rate $n^{-1}{\log }_2{D}_n$ qubits per source use. The optimal reliable rate is $S(\rho )$: every larger rate is achievable, and no smaller rate is reliable. This holds for average pure-source fidelity; the direct construction also preserves a purification reference, and the converse for that stronger entanglement fidelity criterion holds with fidelity tending to zero below the threshold. The encoding and decoding are deterministic channels; no free classical transmission of the source label is allowed.

Diagonalize $\rho ={\sum }_j{\lambda }_j|j⟩⟨j|$ and omit zero-probability eigenvalues when sampling. Define the quantum typical subspace projector $P_{n,\delta }$ by retaining product eigenvectors withThe sampled random variable $-{\log }_2{\lambda }_j$ has expectation $S=S(\rho )$. The weak law of large numbers gives $w_n=\mathrm{Tr}({\rho }_n{P}_{n,\delta })\to 1$. Every retained eigenvalue is at least $2^{-n(S+\delta )}$, and the total probability is at most one, soThis proves the two typical subspace theorem estimates used in the direct part.

Write $P=P_{n,\delta }$. Encode its range isometrically into a compressed space and reserve one orthogonal failure flag. Decode the range back isometrically and the failure flag to a fixed pure state $|\phi ⟩$. This typical-subspace compression with a failure flag is a physical channel whose composite action isIndeed its Kraus operators are $P$ and $|\phi ⟩⟨e_a|$, where the $e_a$ form an orthonormal basis of $\ker P$, and their adjoint products sum to $I$. The Kraus formula for entanglement fidelity givesFor any pure-state ensemble of ${\rho }_n$, the average squared quantum fidelity is at least this number: apply convexity of $|z{|}^2$ separately to the amplitudes $⟨{\psi }_x|K_{\alpha }|{\psi }_x⟩$ with their source probabilities. Thus average fidelity also tends to one. The compressed dimension is at most $2^{n(S+\delta )}+1$, and its rate is at most $S+\delta +o(1)$. Choose $\delta <R-S$ for any prescribed $R>S$. One may also take ${\delta }_n=n^{-1/4}$: the finite spectral information variance and Chebyshev's inequality still give $w_n\to 1$, with rate tending to $S$. A pure source, $S=0$, can simply be re-prepared exactly without sending any qubits.

For the coherent converse, we prove the finite-dimensional quantum compression converse. If the compressed space has dimension $D$, the state of reference plus compressed system has a pure-state decomposition with Schmidt rank at most $D$. Applying decoder Kraus operators locally cannot increase the Schmidt rank of those vectors, so the final reference–output state has Schmidt number at most $D$. Let $|{\Psi }_{{\rho }_n}⟩$ purify the source, with squared Schmidt coefficients equal to the decreasing eigenvalues ${\mu }_1\ge {\mu }_2\ge \cdots$ of ${\rho }_n$. For any normalized vector $|v⟩$ of Schmidt rank at most $D$, its reference support lies in a rank-at-most-$D$ projector $Q$. HenceThe last bound follows by maximizing ${\sum }_j{\mu }_j⟨j|Q|j⟩$ subject to $0\le ⟨j|Q|j⟩\le 1$ and their sum at most $D$. Average over a Schmidt-rank-bounded decomposition of the final state to obtain $F_e\le {\sum }_{j=1}^D{\mu }_j$.

At rate $R<S$, choose $0<\delta <S-R$. The total source probability outside the quantum typical subspace tends to zero, while every eigenvalue inside it is at most $2^{-n(S-\delta )}$. Therefore for $D_n\le 2^{nR}$,This argument allows arbitrary encoding and decoding channels, not only subspace projection.

For completeness, average pure-source reliability also forces the same rate, even though it is a weaker criterion than entanglement fidelity. Here is a separate converse for that formulation. Let the emitted block pure states be ${\psi }_x$, with probabilities $p_x$, and let ${\sigma }_x$ be the decoded states. If their mean squared fidelity tends to one, then their mean trace distance ${\bar{\delta }}_n$ tends to zero, since $\frac{1}{2}\Vert {\sigma }_x-{\psi }_x{\Vert }_1\le \sqrt{1-⟨{\psi }_x|{\sigma }_x|{\psi }_x⟩}$. This inequality follows by choosing a purification with that overlap and using contraction of trace distance under a partial trace.

We need only an elementary entropy continuity from Jordan decomposition bound, derivable from Question 4. If $\delta =\frac{1}{2}\Vert \sigma -\tau {\Vert }_1$, write the positive and negative parts of $\sigma -\tau$ as $\delta {\omega }_{+},\delta {\omega }_{-}$. The common density matrixand the two mixture bounds giveHere $g$ is increasing and concave, $g(0)=0$. Apply this in block dimension $d^n$, both to the average states and to individual outputs, using convexity of trace distance and concavity of $g$. Since the original ensemble is pure, its Holevo quantity is $S({\rho }_n)=nS$. The decoded ensemble hasThe Holevo quantity under a quantum channel cannot increase. To specify the entropy ingredient, Strong subadditivity of Von Neumann entropy states $S(XB)+S(BE)\ge S(B)+S(XBE)$. Attach the classical source label $X$, dilate the decoder isometrically to $BE$, and trace out $E$. The displayed inequality says $I(X:B)\le I(X:BE)$; the isometry preserves the input mutual information. Since $I(X:B)$ of a labelled ensemble is its Holevo quantity, this proves the needed decoder monotonicity. Before decoding, the system has dimension $D_n$, so its Holevo quantity is at most ${\log }_2{D}_n$. Thus ${\log }_2{D}_n\ge nS-o(n)$ and the rate cannot be below $S$. Both operational fidelity formulations therefore have the claimed noiseless coding threshold.

Fix the noise convention explicitly: $p$ is the total probability of a nonidentity Pauli error. The memoryless qubit quantum depolarizing channel iswith independent identical use on successive qubits, namely ${\mathcal{D}}_p^{\otimes n}$ on an $n$-qubit block. Its normalized Kraus operators are $\sqrt{1-p}I,\sqrt{p/3}X,\sqrt{p/3}Y,\sqrt{p/3}Z$.

Write the Bloch sphere representation of a density matrix as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. Its eigenvalues are $(1\pm |\mathbf{r}|)/2$, proving exactly this positivity condition. Conjugation by one Pauli operator preserves that coordinate of the Bloch vector and reverses the other two. Summing the three conjugations givessoThe Bloch ball contracts isotropically, collapses to the maximally mixed state at $p=3/4$, and contracts with reversal for $p>3/4$. That last interval still represents a physical Pauli mixture. If instead $q$ is the probability of replacing the state by $I/2$, then $\mathcal{D}(\rho )=(1-q)\rho +qI/2$, with $p=3q/4$ and $\eta =1-q$; the usual replacement-probability range $0\le q\le 1$ stops at complete depolarization.

Let $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$, with endpoint limits. In bits per use, the unassisted product-state classical capacity and the entanglement-assisted capacity of a qubit depolarizing channel arewhere $H_4$ is Shannon entropy of the listed four probabilities. Equivalently $C_E=2+(1-p){\log }_2(1-p)+p{\log }_2(p/3)$. The first quantity is a classical capacity with product quantum inputs and potentially collective output measurements, not the quantum capacity. These are the requested capacity formulae; a proof of their coding theorems is not needed here. The maximally entangled input gives a Bell-diagonal joint state with those four probabilities, explaining the entropy appearing in $C_E$. The original entanglement-assisted capacity calculation gives the depolarizing-channel expression.

At $p=0$ the capacities are one and two. Both vanish at $p=3/4$. On the full Pauli-error interval they increase again beyond that point, reaching $1-h_2(2/3)$ and $2-{\log }_{2}3$ at $p=1$; on the replacement-probability plot both decrease to zero at $q=1$. The two panels make this convention distinction explicit.

For the high-noise capacity ratio for qubit depolarization, put $\eta =1-4p/3\to 0$. Taylor expansion of binary entropy around one half givesThe four probabilities for $C_E$ are $1/4+3\eta /4$ and three copies of $1/4-\eta /4$. For a perturbation ${\delta }_i$ of the uniform four-outcome distribution, with ${\sum }_i{\delta }_i=0$, entropy expansion givesHere ${\sum }_i{\delta }_i^2=3{\eta }^2/4$, hence $C_E=3{\eta }^2/(2\ln 2)+O({\eta }^3)$. ThereforeThe reciprocal ratio is $1/3$, and the same enhancement factor occurs as $q\to 1$ in the replacement convention.

For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

The Pauli X gate interchanges the two basis quantum states. Because the bit-flip outcome is a classical unobserved error rather than a coherent superposition, the quantum bit-flip channel givesFor $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its explicit density matrix isFor a pure reference quantum state, the square of the paper's quantum fidelity is $F^2=⟨\psi |{\rho }_2|\psi ⟩$. ThereforeWith $|\psi ⟩=\cos (\theta /2)|0⟩+e^{i\varphi }\sin (\theta /2)|1⟩$, the expectation is $⟨X⟩=\sin \theta \cos \varphi$, the $x$ component of the Bloch vector. Hence“All possible quantum states” is interpreted as the rotationally invariant uniform pure-qubit average, with measure $d\mu =\sin \theta d\theta d\varphi /(4\pi )$. Thusso the average pure-qubit fidelity of a bit-flip channel isEquivalently, rotational symmetry gives $⟨r_x^2⟩=⟨r_y^2⟩=⟨r_z^2⟩=1/3$, since a pure-state Bloch vector has unit length. The $X$ eigenstates are unchanged as density matrices and have fidelity one, while a computational basis quantum state has $F^2=1-p$; these provide checks on the angular expression. A uniform distribution in $\theta$ rather than in solid angle would produce the wrong ensemble average. The requested quantity is the mean squared fidelity, not the square of the mean unsquared fidelity.

Write ${\rho }_1=|\psi ⟩⟨\psi |$ and ${\rho }_2=|\phi ⟩⟨\phi |$. A pure-state density matrix is a rank-one orthogonal projector, so ${\rho }_1^{1/2}={\rho }_1$ andTaking its square root and trace yields the unsquared quantum fidelityUse unitary invariance of trace distance and fidelity to choose a basis of the span of the two quantum states in which $|\psi ⟩=|0⟩$ and $|\phi ⟩=c|0⟩+s|1⟩$, with $c=F\ge 0$ and $s=\sqrt{1-c^2}\ge 0$. Choosing state-vector phases to make the coefficients nonnegative does not change either density matrix. On that span,Its eigenvalues are $s$ and $-s$, with zero eigenvalues on the orthogonal complement. The positive square root of ${\Delta }^{†}\Delta$ therefore has two eigenvalues $s$, so the trace distance is $(s+s)/2=s$. Thus the pure-state trace distance and fidelity identity isIdentical pure quantum states give $F=1,D=0$; orthogonal pure quantum states give $F=0,D=1$. If the span is one-dimensional, the identical-state result follows directly, so no second basis vector is required. This formula uses the unsquared fidelity convention in the paper; if fidelity denotes squared overlap instead, the square inside the formula is omitted.

The Schrödinger–Jaynes–Hughston–Jozsa–Wootters theorem classifies all finite quantum state ensembles of a density matrix. Write its nonzero spectral decomposition as $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, with ${\lambda }_a>0$ and orthonormal eigenvectors. An ensemble with subnormalized vectors $|w_j⟩=\sqrt{p_j}|{\varphi }_j⟩$, $j=1,\dots ,m$, represents $\rho$ exactly whenHere $U$ is an $m\times r$ column linear isometry. Zero-weight vectors may be included or discarded. Equivalently, any two ensembles represent the same density matrix if and only if, after padding the shorter list with zero vectors, their subnormalized vectors are related by a unitary matrix acting on the ensemble index. This is the isometry parametrization of a density-matrix ensemble.

For necessity, if $|v⟩$ lies in the kernel of $\rho$, thenEvery $w_j$ therefore lies in the support of $\rho$. Define $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$. The ensemble identity givesso these columns are orthonormal and the displayed parametrization follows.

Conversely, start with any such isometry and define $w_j$ by the boxed equation. ThenThe weights are $p_j=\Vert w_j{\Vert }^2={\sum }_a{\lambda }_a|U_{ja}{|}^2\ge 0$ and sum to $\mathrm{Tr}\rho =1$. Normalize every nonzero $w_j$ to obtain its pure state. This proves both directions and includes degenerate eigenvalues without any preferred choice of basis in their eigenspaces.

To prove the unitary-mixing version, pad two ensembles to a common length $L$. Their matrices $U$ and $V$ have $r$ orthonormal columns in ${\mathbb{C}}^L$. Extend each column family to an orthonormal basis, obtaining square unitaries $\tilde{U}$ and $\tilde{V}$. Then $W=\tilde{V}{\tilde{U}}^{†}$ is unitary and $V=WU$, so the second ensemble vectors satisfy $v_k={\sum }_j{W}_{kj}{w}_j$. Conversely this unitary mixing preserves ${\sum }_j|w_j⟩⟨w_j|$, because ${\sum }_k{W}_{ki}{W}_{kj}^{\ast }={\delta }_{ij}$. Independent phase choices for ensemble kets are absorbed in the row phases of these matrices.

The classification also has a measurement interpretation. Purify $\rho$ as $|\Psi ⟩={\sum }_a\sqrt{{\lambda }_a}|e_a⟩|a⟩$. On the ancillary system define $|{\eta }_j⟩={\sum }_a{U}_{ja}^{\ast }|a⟩$. The effects $E_j=|{\eta }_j⟩⟨{\eta }_j|$ sum to $I_r$, hence form a POVM. The conditional unnormalized system ket is $(I\otimes ⟨{\eta }_j|)|\Psi ⟩=w_j$, with probability $p_j$. Thus every ensemble can be realized by measuring a purifying system; with a sufficiently enlarged ancilla the measurement can be projective. The density matrix specifies the common statistics without selecting one particular ensemble decomposition.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

Denote the normalized initial field by $|F⟩$, and its shifted packets byHere the photon-number kets are orthogonal Fock states. Counting shared number labels givesAfter the interaction, linearity gives the joint stateTake the partial trace over the field. Put $d=|\alpha {|}^2-|\beta {|}^2$ and $c=\alpha {\beta }^{\ast }+{\alpha }^{\ast }\beta$. In the ordered qubit basis $(|0⟩,|1⟩)$, the reduced density matrix isFor example, its upper off-diagonal entry is half the overlap of the field accompanying $|1⟩$ with the field accompanying $|0⟩$, fixing the conjugation order. The trace is one, and positivity follows because this is the partial trace of a normalized pure state. For $k\ge 2$, $s_k=1-2/k$; at $k=1$ it is zero, not the continuation $-1$. This is the photon-number reference for a Hadamard gate channel, not an actual coherent-state packet.

Represent the pure states by density matrices ${\rho }_u=(I+u\cdot \sigma )/2$ and ${\rho }_v=(I+v\cdot \sigma )/2$. The Pauli matrices obey $\mathrm{Tr}{\sigma }_i=0$ and $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$. ConsequentlyOn the other hand, ${\rho }_u=|u⟩⟨u|$ and ${\rho }_v=|v⟩⟨v|$ make that trace equal to $⟨u|v⟩⟨v|u⟩$. Thus the pure-qubit overlap identity isFor mixed states the trace formula still holds, but its left side is then a Hilbert–Schmidt overlap rather than this squared pure-state overlap.

Under the prescribed computational-basis measurement and decision rule, the first state is correctly declared when outcome zero occurs, with probability ${\cos }^2(\theta /2)$. The second is correctly declared on outcome one, with the same probability. Averaging the two equal priors givesOne can also verify optimality directly: the two density matrices differ by $\cos \theta {\sigma }_z$. Since $\cos \theta >0$, choosing the positive eigenspace as the decision for the first state maximizes the trace contribution to success. This is the equal-prior discrimination of qubit states construction along the vertical axis.

For the Kraus operators of the measurement in quantum mechanics, the Born rule givesThe density matrix $A_i\rho A_i^{†}$ is the unnormalized state of that outcome. It is positive, its trace is $p_i$, and completeness gives ${\sum }_i{p}_i=\mathrm{Tr}\rho =1$. There is no conditional state to assign to an event with $p_i=0$.

For a projective measurement, each $P_i$ is an orthogonal projection, satisfying $P_i=P_i^{†}=P_i^2$. Completeness becomes ${\sum }_i{P}_i=I$. Fix $v$ in the range of $P_j$. ThenSince $P_{j}v=v$, the $j$th summand already equals $\Vert v{\Vert }^2$. All other summands are nonnegative, so $P_{i}v=0$ for $i\ne j$. Consequently $P_i{P}_j=0$, and taking adjoints also gives $P_j{P}_i=0$. This proves the stronger completeness forces orthogonality of measurement projections:

For the unitary dilation of a measurement instrument, take a quantum ancilla with orthonormal states $|0⟩,|1⟩,\dots ,|n⟩$, initially in $|0⟩$. DefineThe completeness equation impliesso $V$ is an isometry. Its action on the input subspace can be extended to a unitary operator $U$ on the enlarged system:Choose orthonormal bases in the two complementary subspaces and map one to the other. Their dimensions agree; the extra unused quantum ancilla level also ensures this extension is possible for an infinite-dimensional system. After the coupling, the joint density matrix is