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For bipartite pure source and target states with positive target entanglement entropy, the optimal rate of LOCC conversion with vanishing output error is the ratio of their entanglement entropies. Here vanishing error means that the trace distance between the output and $|\varphi ⟩⟨\varphi {|}^{\otimes m}$ tends to zero; a recorded failure probability must also tend to zero. Entanglement concentration first provides $nS({\rho }_A^{\psi })-o(n)$ Bell pairs, and entanglement dilution needs $mS({\rho }_A^{\varphi })+o(m)$ such pairs. This achieves the ratio. Conversely, refine the output into pure branches. Average monotonicity of pure-state entanglement entropy bounds their average entropy by the input entropy; continuity bound for quantum conditional entropy, with trivial conditioning, makes the output entropy loss per target tend to zero. This forbids a larger rate. The convention permits approximations for finite blocks: exact deterministic conversion of every finite block can obey stricter majorization constraints. The reversible concentration construction is given in the original concentration paper.

Let distinct nonorthogonal pure states have overlap magnitude $0<c<1$, and let one quantum channel take $n$ identical inputs to $m$ outputs. Let ${ϵ}_u,{ϵ}_v$ be the respective probabilities that a test for the ideal $m$-copy pure state fails. Trace distance contraction, its triangle inequality and the pure-target upper bound on trace distance give $\sqrt{1-c^{2m}}\le \sqrt{{ϵ}_u}+\sqrt{1-c^{2n}}+\sqrt{{ϵ}_v}$. Squaring with the Cauchy-Schwarz inequality gives the displayed strategy-independent positive bound. It establishes both a worst-case bound and an equal-prior average bound. Averaging rotated pairs establishes a positive Haar-average bound for unknown qubit inputs, without assuming independent clone errors.

A mixture of two orthogonal pure qubit states has Von Neumann entropy equal to the binary entropy of its weights. Its entropy difference from the first pure state is exactly $h_2(u)$ and its trace distance from that state is $u$. The elementary bound $-(1-u)\ln (1-u)\le u$ gives $h_2(u)\le u{\log }_2(e/u)$; entropy also never exceeds one bit. For small $u$, the leading term is $u{\log }_2(1/u)$.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

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]

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

If $\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1\to 0$, thenIn finite dimension this follows from multipartite superadditivity of quantum relative entropy, additivity of quantum relative entropy, contraction of trace distance under partial trace, and uniform continuity on the compact state space.

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.

Assume the three physical phase-flip channels act independently, with the same probability $p$ on each qubit, and encoding, recovery and decoding are ideal. Zero or one physical error is corrected. HenceThe probability of an uncorrectable two- or three-site error is $p_L=3p^2(1-p)+p^3=3p^2-2p^3$.

One must also choose the logical convention consistently with the output phase error specified here. The usual phase-flip repetition code with codewords $|+++⟩$ and $|---⟩$ has residual operator $Z_1{Z}_2{Z}_3$ after two or three physical errors: their error syndrome equals that of the complementary zero- or one-error pattern. This operator exchanges those codewords, so plain inverse encoding would give logical $X$, not $Z$.

To obtain the stated logical phase-flip convention, use the logical Hadamard convention for a phase-flip repetition code. Put a logical Hadamard gate before that encoding and after its decoding. The encoded state iswhere the plus and minus states in these codewords are the real Hadamard basis states. The uncorrectable exchange is now conjugated to $HXH=Z$. With that convention the decoded density operator and its trace distance are exactlyFor $0<p<1/2$, $p-p_L=p(1-p)(1-2p)>0$, so coding reduces the trace distance. The displayed recovery probability means success of correction as a quantum operation on an arbitrary unknown input. If the particular input is a $Z$ eigenstate, even the logical phase-error branch represents the same physical state, and the probability of recovering that particular state is one.

Independence is an extra noise assumption needed for the numerical probability. If all three physical qubits flip together with probability $p$, each marginal still has the given phase-flip channel, but the logical failure probability is $p$, not $3p^2-2p^3$. More generally, for a classical joint phase-error distribution with a random number $K$ of flips, $P_{\mathrm{rec}}=\mathrm{Pr}(K\le 1)$ and $p_L=\mathrm{Pr}(K\ge 2)$; the same decoded mixture holds in the stated logical convention. Thus marginal error probabilities alone do not determine the requested recovery probability. A complex environment overlap likewise requires the coherent rotation to be retained or separately compensated before this stochastic phase-error analysis applies.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

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.

Let $P_{\varphi }=|\varphi ⟩⟨\varphi |$. An arbitrary deterministic strategy, including auxiliary systems, intermediate measurements, adaptive actions and discarded records, is a quantum channel $\mathcal{E}$ from the two input qubits to the three returned qubits. Its output is $R_{\varphi }=\mathcal{E}(P_{\varphi }^{\otimes 2})$. The three local tests commute, and their joint all-pass projector is $P_{\varphi }^{\otimes 3}$. Thusis exactly the probability of at least one failed test. No independence of the three returned qubits is assumed.

We prove a uniform positive obstruction using trace distance, $D(\rho ,\sigma )=\Vert \rho -\sigma {\Vert }_1/2$. Two normalized pure kets obeyIndeed their projector difference is supported on their two-dimensional span, has trace zero, and has eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$.

Two further estimates will cover even mixed outputs. First, trace distance contracts under a quantum channel. To see this directly, the positive spectral projector of the trace-zero difference gives $D(\rho ,\sigma )={\max }_{0\le A\le I}\mathrm{Tr}[A(\rho -\sigma )]$. The adjoint of a channel is positive and unital, so it takes any output effect into an input effect. Maximizing over this subset cannot exceed the input maximum, proving contraction.

Second, for a mixed state $R={\sum }_j{q}_j|u_j⟩⟨u_j|$ and a pure target $P=|v⟩⟨v|$, the trace-norm triangle inequality and weighted Cauchy-Schwarz inequality give the pure-target upper bound on trace distance:In particular $D(R_{\varphi },P_{\varphi }^{\otimes 3})\le \sqrt{{ϵ}_{\varphi }}$.

Now take $|{\varphi }_0⟩=|0⟩$ and $|{\varphi }_1⟩=|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The two-copy input distance is $D_2=\sqrt{1-1/4}=\sqrt{3}/2$, whereas the ideal three-copy distance is $D_3=\sqrt{1-1/8}=\sqrt{7/8}$. Writing $R_j=R_{{\varphi }_j}$, contraction and the triangle inequality yieldLet $\delta =D_3-D_2>0$. Then $\sqrt{{ϵ}_0}+\sqrt{{ϵ}_1}\ge \delta$, and squaring givesThis is the copy-number trace-distance obstruction. It applies to every Bob strategy, irrespective of its internal complexity. If Alice chooses these two possible inputs with equal probabilities, the left side is her unconditional failure probability. For every device, at least one of the two inputs also has failure probability at least ${\delta }^2/4$, establishing the worst-case guarantee.

The same positive bound holds if Alice chooses a uniformly random pure-qubit state. Apply the preceding inequality to $U|0⟩$ and $U|+⟩$ for every single-qubit unitary $U$, since their overlap is unchanged. Average over normalized Haar measure. Both rotated kets have the same uniform distribution on the Bloch sphere, so each of the two average failures is the uniform-input average. Consequently that average is also at least ${\delta }^2/4$. A constant satisfying the strict requested inequality in either this average or the worst-case interpretation isThe proven bound is $2p$, hence strictly greater than $p$. This proof establishes a quantitative gap, rather than relying only on the impossibility of exact quantum cloning.

There is a necessary quantifier qualification because the PDF does not specify Alice's input distribution. A positive lower bound for every fixed state against every strategy would be false: a device hardwired to return $|000⟩$ never fails when the input is $|0{⟩}^{\otimes 2}$. The unknown-state assertion is therefore meaningful as a worst-case guarantee over possible inputs, or as an average for a specified hidden input ensemble. Both the equal-prior example and the uniform-input version have been proved here, with a constant independent of Bob's strategy.

Yes: distinct density operators have a projective measurement with different outcome probabilities. Let $\rho ,\sigma$ be the two density operators and let $D=\rho -\sigma$. It is a nonzero Hermitian operator with trace zero. Its nonzero eigenvalues cannot all have the same sign, since their sum is zero. Let $P_{+}$ be the orthogonal projection onto its positive-eigenvalue subspace. Then the two-outcome projective measurement $\{P_{+},I-P_{+}\}$ satisfiesThe outcome therefore carries statistical information about which quantum ensemble was used.

For example, with equal prior probabilities, guess $\rho$ after $+$ and $\sigma$ after the other outcome. The success probability isSince the positive and negative eigenvalue sums of $D$ have equal absolute value, $\mathrm{Tr}(P_{+}D)=\Vert D{\Vert }_1/2$. ConsequentlyThis is the equal-prior measurement in the Holevo–Helstrom theorem, expressed using the trace distance. The conclusion is some information, not necessarily perfect one-shot discrimination: nonorthogonal density operators generally cannot be distinguished with certainty.

Use the usual asymptotic convention: the output approaches the desired $m(n)$-copy pure target in trace distance, and any recorded failure probability tends to zero. Define the binary entropyThe entanglement entropies per source and target are $E_{\psi }=h_2(s)$ and $E_{\varphi }=h_2(t)$. The asymptotic pure-state entanglement conversion rate isBoth achievability and the upper bound matter here.

For entanglement concentration, expand $n$ sources in their correlated computational basis. Every pair of strings of Hamming weight $k$ has the same amplitude $(1-s{)}^{(n-k)/2}{s}^{k/2}$. Alice measures only this weight, without resolving the individual string. Bob's string has the same weight automatically. Conditional on $k$, the normalized state is maximally entangled on $D_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ correlated strings. Its outcome probability is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)s^k(1-s{)}^{n-k}$, a binomial distribution. By the weak law of large numbers, $k/n\to s$ in probability; Stirling's formula then gives $n^{-1}{\log }_2{D}_k\to h_2(s)$. A uniform Schmidt vector on $D_k$ entries is majorized by a uniform vector on $2^{\lfloor {\log }_2{D}_k\rfloor }$ entries padded with zeros. Thus Nielsen's pure-state conversion theorem converts the conditional state into that many Bell pairs exactly. For any fixed $\delta >0$, one obtains at least $n(E_{\psi }-\delta )$ Bell pairs with probability tending to one.

For entanglement dilution, consider the Schmidt strings of $m$ targets with probabilities $q_x$. The quantum typical subspace retains strings satisfyingThe retained probability $Q_m$ tends to one by the weak law applied to the independent variables $-{\log }_2{q}_{x_j}$. The number of retained strings is at most $2^{m(E_{\varphi }+\delta )}$, because their probabilities sum to at most one. Normalize the target after truncation to those strings. Its squared overlap with the full target is $Q_m\to 1$. A uniform Schmidt vector of dimension $2^L$, with $L=\lceil m(E_{\varphi }+\delta )\rceil$, is majorized by the truncated target's Schmidt vector padded with zeros: the sum of its largest $k$ entries is at least $k/2^L$. Hence $L$ Bell pairs prepare the truncated target exactly by LOCC. Its trace distance from the true target is $\sqrt{1-Q_m}\to 0$. Combining concentration and dilution achieves any rate $R<E_{\psi }/E_{\varphi }$. Choosing margins tending sufficiently slowly to zero attains the ratio as a limiting rate.

For the converse, average monotonicity of pure-state entanglement entropy follows from Concavity of Von Neumann entropy: a measurement by one party leaves the other party's reduction unchanged on average, so its average conditional entropy cannot increase. Refine local measurement records, including local discarded systems, to obtain pure output branches $\{p_r,|{\chi }_r⟩\}$ on the $m$ output qubits of each party. Iterating the concavity argument givesLet $|{\Phi }_m⟩=|\varphi {⟩}^{\otimes m}$ and let the squared-overlap error of the average output ${\rho }_{\mathrm{out}}$ be $\epsilon =1-⟨{\Phi }_m|{\rho }_{\mathrm{out}}|{\Phi }_m⟩\to 0$. Define $d_r=\sqrt{1-|⟨{\Phi }_m|{\chi }_r⟩{|}^2}$, the pure-state trace distance. Concavity of the square root gives ${\sum }_r{p}_r{d}_r\le \sqrt{\epsilon }$. Partial trace cannot increase trace distance, so the reduced states are also within $d_r$. The continuity bound for quantum conditional entropy, with trivial conditioning and local dimension $2^m$, impliesConsequentlywhich excludes a limiting rate exceeding $E_{\psi }/E_{\varphi }$.

The vanishing-error convention is essential. If exact deterministic conversion were required for every finite block, majorization would also require the largest squared Schmidt coefficient to satisfy $(1-s{)}^n\le (1-t{)}^m$, henceThis can be strictly smaller than the entropy ratio: at $s=0.1,t=0.2$, the two bounds are approximately $0.4722$ and $0.6496$. Thus the boxed rate answers the standard asymptotic task, rather than imposing unmentioned exact zero-error conditions on every finite block.