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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 optimal three-to-one qubit random access code. For Alice's bits $(b_1,b_2,b_3)$, put $s_j=(-1{)}^{b_j}$ and send the pure state with Bloch vectorThese are the eight vertices of a cube inscribed in the Bloch sphere. If an explicit ket is wanted, choose $\cos {\theta }_b=s_3/\sqrt{3}$ and $e^{i{\phi }_b}=(s_1+i{s}_2)/\sqrt{2}$, and prepare $\cos ({\theta }_b/2)|0⟩+e^{i{\phi }_b}\sin ({\theta }_b/2)|1⟩$.

Bob wanting bit $j$ performs a Pauli measurement of ${\sigma }_x,{\sigma }_y,{\sigma }_z$ respectively, reporting zero for eigenvalue $+1$ and one for $-1$. Its success for every input and every requested bit isOnly one chosen bit is extracted; the protocol does not recover all three simultaneously and uses no shared entanglement.

To prove that no one-qubit protocol gives a greater guaranteed probability, allow Alice arbitrary mixed encoding vectors $|r_b|\le 1$ and Bob any binary positive operator-valued measure. Write his decision difference as $D_j=E_{j,0}-E_{j,1}=t_{j}I+v_j\cdot \sigma$. Its eigenvalues lie in $[-1,1]$, implying $|v_j|\le 1-|t_j|\le 1$. On input $b$, success for bit $j$ is $[1+s_j(t_j+v_j\cdot r_b)]/2$. Average over all eight strings and three choices. The terms in $t_j$ cancel, leavingThe Cauchy-Schwarz inequality and cancellation of cross terms over all sign choices giveHence $p_{\mathrm{avg}}\le p_{\ast }$. Any worst-case guarantee is no greater than this average; the cube protocol attains it uniformly, proving optimality both for guaranteed success and for the uniform-input average. Biased prior information about the bits would define a different optimization problem.

A density operator represents a pure state exactly when it has rank one, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$. The spectra just found giveTheir wavefunctions are defined up to a global phase. The other two are mixed states, with purity $(3/4{)}^2+(1/4{)}^2=5/8$, and have no single-wavefunction representation as rank-one projectors.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.

c1 is true for density-state controllability. Every qubit density operator can be written $\rho ={\lambda }_{2}I+({\lambda }_1-{\lambda }_2)|\psi ⟩⟨\psi |$ using its eigenvalues. A control taking one pure orthogonal projector to any other therefore also connects every pair of qubit density operators with the same spectrum. In Lie-algebra language, $\mathfrak{sp}(1)=\mathfrak{su}(2)$; there is no larger-dimensional symplectic exception. This does not supply arbitrary global phase: $SU(2)$ alone is already a counterexample to an implication involving strict operator controllability.

c2 is false. For every even $N=2\ell \ge 4$, the defining $Sp(\ell )$ action is transitive on unit vectors, so a system generating this algebra is pure-state controllable. It still preserves additional mixed-state invariants and lacks density operator controllability, as the preceding explicit four-dimensional example demonstrates. pure states have fewer orbit requirements than general mixed states.

The linear operator $M_A$ acts on the first factor, while the partial inner product acts on the second. They therefore commute in the contraction:Thus the operated vector is exactly the relative state obtained from the operated bipartite vector using the same conjugate index vector. For a physical normalized pure state, divide by $\Vert M_A{\varphi }_A\Vert$ when that norm is nonzero. If $M_A{\varphi }_A=0$, the vector identity still holds, but there is no normalized output state for that zero-probability branch. An arbitrary operator is not assumed unitary or norm preserving.

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

Include all fixed resource states and all measurement records in an environment, so the entire physical protocol has an linear isometry $U$ from the unknown input to a larger pure output. If every recipient receives a perfect copy, its reduced density matrix is the pure projector $|\psi ⟩⟨\psi |$. A pure marginal forces factorization: write a joint vector in a basis beginning with $|\psi ⟩$; its squared components along every orthogonal basis vector vanish because their marginal probabilities are zero. Applying this successively to all recipients giveswhere $|R⟩$ is the fixed initial resource and $|E_{\psi }⟩$ may depend on the input. This includes Alice's discarded systems and every possible outcome record.

Choose two distinct nonorthogonal pure states, for example $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, and write their overlap as $s$. Preservation of the inner product by the linear isometry requiresNormalized environment vectors have overlap of absolute value at most one. Thus $0<|s|<1$ would imply $|s|\le |s{|}^N$, impossible for $N>1$. Hence perfect transmission of a copy to every recipient is impossible for $N>1$. This proves the no-cloning theorem obstruction even though Alice keeps no copy herself.

Here is a tripartite Schmidt projection entanglement concentration protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local qubits of each party giveswhere $|x|$ counts ones. Alice performs a projective measurement of that count, without learning the individual string. The orthogonal projector for outcome $k$ is ${\Pi }_k={\sum }_{|x|=k}|x⟩⟨x|$. By the Born rule, its probability and conditional pure state areAlice sends $k$ by classical communication; all three parties use local unitary operations to relabel the same ordered type class as $0,\dots ,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual qubits instead would have destroyed precisely this coherence.

To obtain an integer number $f$ of target GHZ states, choose $m=2^f\le d_k$. The following exact LOCC step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq \{0,\dots ,d_k-1\}$, Alice uses the Kraus operatorEvery label occurs in $\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)$ such subsets, proving ${\sum }_S{K}_S^{†}{K}_S=I$ on the occupied subspace. Thus this is a complete measurement in quantum mechanics. Each outcome has probability $m/[d_k(\genfrac{}{}{0ex}{}{d_k-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d_k}{m}\right)$, and leaves the normalized pure state $m^{-1/2}{\sum }_{r\in S}|r,r,r⟩$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into exactly $f$ independent copies of the target GHZ state. Local extensions of these operations handle unused basis vectors and do not affect the protocol.

For example, taking $f(k)=\lfloor {\log }_2{d}_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a binomial distribution with parameters $(n,q)$, the weak law of large numbers gives $K/n\to q$ in probability. Stirling's formula, uniformly when $k/n$ stays away from zero and one, givesBy continuity of the binary entropy, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.

A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, takeOn $|K-nq|\le n^{2/3}$, the bounded derivative of the binary entropy in a neighborhood of the fixed $q\in (0,1)$ and Stirling's formula give ${\log }_2{d}_K\ge n{h}_2(q)-C_q{n}^{2/3}-O(\log n)\ge f_n$. Chebyshev's inequality bounds the complementary probability by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset measurement in quantum mechanics with $m=2^{f_n}$. ThusThis matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

The construction is a convex roof extension. First establish its convexity, without assuming that the minimizing quantum state ensemble exists. If $\tau ={\sum }_j{w}_j{\tau }_j$, choose a pure state ensemble for each ${\tau }_j$ whose average value of $\mu$ is within $\epsilon$ of $F({\tau }_j)$. Combining these ensembles, with their weights multiplied by $w_j$, gives a pure state ensemble for $\tau$. Taking arbitrarily small $\epsilon >0$ provesAlso, $F(\pi )=\mu (\pi )$ for a pure state $\pi$. Indeed, if $\pi ={\sum }_j{p}_j{\pi }_j$ and $v$ is orthogonal to its one-dimensional support, then $0={\sum }_j{p}_j⟨v|{\pi }_j|v⟩$. All terms are nonnegative, so every positive-weight ${\pi }_j$ has that same support and equals $\pi$.

Represent outcome $k$ of the operation by a linear unnormalized map ${\mathcal{I}}_k$, as for a quantum instrument. Normalization is performed only after applying this map: ${\mathcal{I}}_k(\rho )=P_k{\rho }_k$. For any preparation ensemble $\rho ={\sum }_j{p}_j{\pi }_j$, putLinearity gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. For $P_k>0$, Bayes' theorem identifies the conditional preparation weightsOutcomes with zero probability contribute nothing; normalized states on such branches need not be defined. The conditional states ${\sigma }_{jk}$ may be mixed, so convexity of the convex roof is essential here.

Applying that convexity separately for each recorded outcome, and then the assumed average monotonicity for each pure preparation, yieldsThe left side depends only on the initial density operator and the operation, not on the chosen preparation ensemble. Taking the infimum of the last expression over all its pure state ensembles provesEquivalently, choose an ensemble within $\epsilon$ of the infimum and then let $\epsilon \downarrow 0$. This is the monotonicity of a convex roof under a quantum instrument. It uses linearity of the unnormalized branches, not a false assumption that normalized conditional states depend linearly on the input. As usual, the averages are assumed well defined; finite-valued roofs and finitely many outcomes suffice. If an input roof is $-\infty$, taking ensembles with averages tending to $-\infty$ gives the same conclusion in the extended-real interpretation whenever the output average is defined.