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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.

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Factor the coefficients into a tensor product:This is already a Schmidt decomposition with one term. ConsequentlyEach reduced density matrix is the rank-one orthogonal projector onto $(|0⟩-|1⟩)/\sqrt{2}$, so its other eigenvalue is zero. The joint pure state is a product state, with no entanglement.

The coefficient matrix in the computational orthonormal bases, with Alice's index labelling rows, isSince the entries are real, the reduced density matrices areBoth have trace one and determinant $1/9$. Their characteristic polynomial is ${\lambda }^2-\lambda +1/9$, so the nonzero eigenvalues areBoth are positive. Thus the Schmidt rank, or the Schmidt number of this pure state, isIf “Schmidt numbers” denotes the individual positive Schmidt coefficients instead, those areTheir squares are ${\lambda }_{\pm }$; giving both the Schmidt rank and the Schmidt coefficients removes this terminology ambiguity. The pure state is entangled because its Schmidt rank exceeds one.

Yes: a deterministic LOCC protocol exists for every allowed value of the parameter. Here is an explicit implementation, so existence does not depend on merely quoting a pure-state conversion theorem.

First convert the input Bell state to $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ by local unitary operations:Write $c=\cos \varphi$ and $s=\sin \varphi$. Alice performs the two-outcome measurement in quantum mechanics with Kraus operatorsThey define a physical measurement in quantum mechanics becauseActing on $|{\Phi }^{+}⟩$, the unnormalized branch vectors areBoth branches have probability $1/2$. Alice communicates the outcome to Bob. If it is one, both apply the Pauli X gate; this turns the normalized second branch into $c|00⟩+s|11⟩$. If it is zero, neither applies this correction. Finally Bob applies a Pauli X gate in either branch, obtainingEvery outcome produces the same target pure state, so this is deterministic local operations and classical communication, not a successful branch selected by postselection. This two-outcome LOCC dilution of a Bell pair remains valid at $\varphi =0$, where the output is a product state, and at $\varphi =\pi /4$, where it is another Bell state.

The target Schmidt coefficients are $c,s$, while those of the input are both $1/\sqrt{2}$. For an interior value $0<\varphi <\pi /4$, local unitary operations alone could not change these Schmidt coefficients; the local measurement in quantum mechanics is the step that reduces the entanglement.

Expand an initial bipartite pure state in local orthonormal bases asBy the Schmidt decomposition, or equivalently the singular value decomposition of $C$, its Schmidt rank is $\mathrm{rank}C$. In a specified local measurement branch, the product Kraus operator $A\otimes B$ changes the coefficient matrix toIndeed the coefficient of $|k⟩|l⟩$ is ${\sum }_{i,j}{A}_{ki}{C}_{ij}{B}_{lj}$. The elementary matrix rank inequality givesNormalizing a nonzero branch multiplies $C^{\prime }$ by a scalar and therefore leaves its matrix rank unchanged.

Now refine a complete LOCC transcript to record every individual Kraus operator. Although later choices depend on the earlier classical outcomes, fixing a transcript fixes all those choices. Composing the local operations therefore gives a single product Kraus operator $A_{\tau }\otimes B_{\tau }$ in that branch. Each nonzero branch obeys the same Schmidt-rank contraction under product operators. Local auxiliary systems prepared independently are included by local isometries and do not add entanglement. Discarding a local auxiliary system is included by resolving its partial trace in an orthonormal basis; this again yields refined product Kraus operators.

If the protocol produces a specified pure state deterministically, its output density operator is a sum of positive rank-one branch density operators. Every nonzero branch vector must be proportional to that output vector: its squared overlap with any vector orthogonal to the output sums to zero, so each such overlap is separately zero. Thus the output Schmidt rank equals that of each surviving branch, andThis proves monotonicity of Schmidt rank under LOCC, even for a nonzero branch selected by postselection. If the classical transcript is instead discarded and the result is mixed, the refined branch vectors supply an ensemble with Schmidt ranks at most $r_{\mathrm{in}}$; by definition its Schmidt number is at most $r_{\mathrm{in}}$. This is the correct mixed-state extension, rather than a bound on the ranks of its reduced density matrices. In finite dimensions, any limiting exact protocol obeys the same pure-state bound because the set of coefficient matrices of rank at most $r_{\mathrm{in}}$ is closed, being defined by vanishing minors.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Let $D={\mathcal{C}}_{\mathrm{ev}}$, and write ${\mathcal{C}}_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in {\mathcal{C}}_H$. The parity functional on the four-dimensional binary linear code ${\mathcal{C}}_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\dots ,1)$ belongs to the standard length-seven Hamming code and has odd Hamming weight. Denote the two logical basis states by $|{\psi }_0⟩$ and $|{\psi }_1⟩$, corresponding to these two cosets. The eight computational orthonormal basis vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them orthogonal.

For a binary vector $w\in {\mathbb{F}}_2^7$, define the product of Pauli Z gateswhere the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, isIf $w\in D^{\perp }$, every summand is one. If $w\notin D^{\perp }$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. ThusOff-diagonal logical overlaps vanish for every $w$:because a diagonal Pauli operator cannot move a computational basis word from one coset to the disjoint other coset.

The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\dots ,7$. Set $e_0=0$. Each phase flip is its own adjoint and inverse, soIf $i\ne j$, the vector $e_i+e_j$ is nonzero and has Hamming weight one or two. By the given dual code identity $D^{\perp }={\mathcal{C}}_H$ and the minimum Hamming distance three of the Hamming code, it cannot belong to $D^{\perp }$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are ${\delta }_{ab}$. Combining both cases givesThis proves the Knill--Laflamme condition with an identity error-overlap matrix, not merely a possibly degenerate scalar matrix. In particular the eight two-dimensional error subspaces $E_i{\mathcal{X}}_{\mathrm{Steane}}$ are mutually orthogonal. This is precisely nondegenerate phase-flip correction in the Steane code.

To exhibit recovery rather than only the criterion, let $P$ be the orthogonal projector onto the code and $P_i=E_{i}P{E}_i^{†}$. The projective measurement consisting of these eight mutually orthogonal projectors, completed by $I-{\sum }_i{P}_i$, determines which error occurred. For an arbitrary encoded pure state $|\chi ⟩=\alpha |{\psi }_0⟩+\beta |{\psi }_1⟩$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi ⟩$ unchanged. Applying $E_j$ again recovers $|\chi ⟩$. The outcome gives the error syndrome and does not reveal $\alpha$ or $\beta$, so recovery preserves logical quantum superpositions.

An equivalent explicit error syndrome uses three commuting X-type stabilizer generators. Choose a parity-check matrix for ${\mathcal{C}}_H$ with the seven different nonzero binary columns:Its rows span ${\mathcal{C}}_H^{\perp }=D$. For a row $h_l$, the operator $X(h_l)={⨂}_j{X}_j^{(h_l{)}_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A phase flip at position $j$ changes its eigenvalue to $(-1{)}^{H_{lj}}$, since $X(h_l)Z_j=(-1{)}^{H_{lj}}{Z}_{j}X(h_l)$. Measuring the three commuting stabilizer generators gives the $j$th column of $H$ as a binary syndrome; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these stabilizer generators never distinguishes the two logical states.

Finally, if a noise Kraus operator is a coherent linear combination $F={\sum }_j{c}_j{E}_j$, the same syndrome measurement separates its mutually orthogonal components $c_j{E}_j|\chi ⟩$. Conditional recovery returns $|\chi ⟩$ in every nonzero branch. Linearity then also corrects any quantum channel whose Kraus operators lie in this span. The argument therefore corrects arbitrary unknown single-site phase flips nondegenerately, including coherent phase-error amplitudes, rather than only a known error on a basis codeword.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

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.

Successive uses of the same quantum channel mean coupling to a fresh vacuum environment each time, or resetting the environment between uses. They do not mean repeatedly applying the dilation unitary to one unreinitialized two-state environment. Direct composition givesInductively, with $q_n=1-(1-p{)}^n$,For $0<p\le 1$, this tends to the pure ground state $|0⟩⟨0|$, whose Von Neumann entropy is zero. For $p=0$, the channel is the identity: the state remains $\rho$ and its Von Neumann entropy remains $S(\rho )$. Thus the repeated amplitude damping limit must include this no-decay exception.

A zero-Von Neumann entropy output means there is no uncertainty about the final atomic state. It does not mean that the receiver has gained information about the initial preparation: all initial states have the same limiting output, so their distinguishability has been erased from the atom. Under the global unitary evolution, information is carried into the environment. Without observing an emission record, this is a nonselective dissipative process, not an inference about which input was sent. Indeed, starting from the excited pure state, one intermediate output has Von Neumann entropy equal to the binary entropy of $p$, which is positive for $0<p<1$; the atom's Von Neumann entropy need not decrease monotonically during the approach to the final ground state.

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.