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The Bell states are joint eigenstates of commuting $Z_A{Z}_B$ and $X_A{X}_B$. Two shared Bell state meter pairs realize their two entanglement-assisted nondemolition parity measurements using spacelike local circuits. Their projector products give the four rank-one Bell state outcomes and preserve each input Bell state. The globally identified outcome requires later local operations and classical communication.

Average the four correlated local Pauli operations to remove coherences in the Bell states basis. A one-party Pauli operation can permute the largest of the four resulting weights into the singlet sector; that weight is at least $1/4$. Correlated common local unitaries then permute the three triplet projectors, while collective-unitary covariance of the two-qubit singlet keeps the singlet projector unchanged. Averaging all six permutations makes the triplet weights equal, producing a two-qubit Werner state. Shared randomness or local operations and classical communication coordinates these local operations. Common unitaries alone cannot increase an initially small singlet weight, because that weight is invariant under them.

Share a Bell state meter pair, apply local system-to-meter CNOT gates, and measure each meter in the computational basis. The parity of the local meter records measures system $Z_A{Z}_B$ while preserving all coherence within either parity sector. The complete tuple of local records has Kraus operator $P_{u\oplus v}/\sqrt{2}$. Each local record is uniform, and the nonlocal result requires local operations and classical communication to compare records. A local Hadamard gate conjugation supplies the corresponding $X_A{X}_B$ measurement.

Entanglement dilution uses local operations and classical communication to prepare partially entangled states from maximally entangled states, such as Bell pairs. For many identical bipartite pure targets, the asymptotic cost in Bell pairs per target, allowing a vanishing approximation error, is their entanglement entropy in bits. The quantum typical subspace of the target's Schmidt coefficients has almost all its weight and dimension at most $2^{m(S+\delta )}$. A maximally entangled state of this dimension can be converted to the normalized truncated target by Nielsen's pure-state conversion theorem. The resulting state approaches the full target as the retained weight tends to one. Average monotonicity of pure-state entanglement entropy and entropy continuity give the converse lower bound. This is the reverse asymptotic task to entanglement concentration.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

An entanglement monotone quantifies entanglement without increasing on average under local operations and classical communication. The average is over recorded measurement in quantum mechanics outcomes: individual successful branches can have more entanglement, provided their probabilities compensate for that increase. This distinction makes an entanglement monotone useful for bounding probabilistic state conversion. A pure-state entanglement monotone need only satisfy this inequality on pure input states and pure conditional branches, with local measurement outcomes refined to individual Kraus operators.

A shared resource enables deterministic exact transmission of an arbitrary unknown qubit using only local operations and classical communication if and only if it can produce a shared Bell pair deterministically by such operations. Sufficiency follows from quantum teleportation. For necessity, prepare a Bell pair locally at the sender and apply the purported transmission protocol to one half: exact identity-channel action leaves a shared Bell pair between the retained reference and the receiver. A separable shared resource cannot satisfy this condition because local operations and classical communication preserve separability.

The resource here is one copy of a two-qubit Werner state, with the physical range $1/2<F\le 1$. The impossibility must include successful conditional branches of local operations and classical communication; average monotonicity alone would not exclude probabilistic enhancement.

For a normalized two-qubit pure state $|\chi ⟩={\sum }_{i,j}{M}_{ij}|ij⟩$, define its concurrence as $C(\chi )=2|\det M|$. Its mixed-state concurrence of two qubits is the convex roof extensionCombining near-optimal decompositions proves convexity. A separable quantum state has zero concurrence, because it has a decomposition into product vectors with zero coefficient determinant.

First establish the concurrence of a two-qubit Werner state. A pure state with squared Schmidt coefficients $\lambda ,1-\lambda$ has $C=2\sqrt{\lambda (1-\lambda )}$. In this Schmidt basis a maximally entangled state has coefficient matrix $V/\sqrt{2}$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$, because $|V_{ii}|\le 1$. Thus the singlet overlap satisfiesAveraging any decomposition of $W_F$ gives $C(W_F)\ge 2F-1$.

For the reverse bound use the orthonormal triplet vectorsEach is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C({\chi }_{j,\pm })=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence

Every fully specified record of an LOCC protocol gives a product Kraus operator $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output asFor invertible $A,B$, the concurrence scaling under invertible local filters isHere is a derivation of that scaling. Each pure coefficient matrix transforms to $AM{B}^{\mathsf{T}}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized concurrence is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_j{p}_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of ${\rho }^{\prime }$, so taking both infima proves equality. If either local filter has rank one, the output is a separable quantum state and the desired bound is immediate.

It remains to show that the filter normalization cannot make this ratio exceed one. Use the Pauli matrices representationWrite the positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, so $a_0\ge |a|$, $b_0\ge |b|$. Taking traces givesFor completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$, and both sides are nonnegative. This proves the single-copy Werner filtering normalization bound, and hence

Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F^{\prime }}$ would therefore obey $2F^{\prime }-1\le 2F-1$, contradicting $F^{\prime }>F$. Thus even heralded single-copy LOCC cannot produce the specified improved Werner state. This proves that single-copy Werner filtering cannot increase concurrence. It does not forbid collective entanglement distillation from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.

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.

First explain the stated local conversion. The four equally weighted correlated local Pauli operations $I\otimes I$, $X\otimes X$, $Z\otimes Z$, and $Y\otimes Y$ remove coherences in the Bell states basis, since those states have distinct joint parity eigenvalues. This leaves four Bell probabilities. One-party Pauli operations permute the Bell projectors, so the largest probability can be placed on the singlet ${\Phi }_{-}$. Call it $p$; necessarily $p\ge 1/4$.

The two common unitaries found above generate permutations of the three triplet projectors: $S\otimes S$ exchanges ${\Psi }_{+}$ and ${\Psi }_{-}$, and $HSH\otimes HSH$ exchanges ${\Psi }_{+}$ and ${\Phi }_{+}$. Averaging the six triplet permutations replaces their weights by their common mean, while collective-unitary covariance of the two-qubit singlet fixes the singlet projector. This is Bell twirling followed by triplet symmetrization, producing the specified two-qubit Werner state. Shared randomness coordinates the local choices, so the procedure is allowed by local operations and classical communication. Common unitaries alone would preserve the original singlet weight, which need not initially be at least $1/4$; the preceding one-party permutation is what ensures that range.

For the eigenvalue calculation, put $P_s=|{\Phi }_{-}⟩⟨{\Phi }_{-}|$, $a=(1-p)/3$, and $b=p-a=(4p-1)/3$. Then $\rho =aI+b{P}_s$. In the ordered computational basis $00,01,10,11$, transposing the second subsystem givesThe $00,11$ block has eigenvalues $a-b/2$ and $a+b/2$, while the other two basis vectors also have eigenvalue $a+b/2$. Thus the partial transpose spectrum isThe last three are positive throughout the given range, so the partial transpose is not positive exactly when $p>1/2$. A separable quantum state is a convex sum of product density operators, each still positive after transposing one factor. The negative eigenvalue therefore proves entanglement.

For completeness, separability on the other side can be proved explicitly rather than inferred solely from a theorem name. Let $P_{j,\pm }=(I\pm {\sigma }_j)/2$ be the one-qubit Pauli eigenprojectors. At $p=1/2$, the mixture of six antiparallel product states isusing $P_s=(I-{\sum }_j{\sigma }_j\otimes {\sigma }_j)/4$. It is manifestly separable. For $1/4\le p\le 1/2$,with nonnegative coefficients summing to one; $I/4$ is also separable. ConsequentlyAt the boundary $p=1/2$ the partial transpose has a zero eigenvalue and the explicit decomposition shows separability.

Yes, the conversion can be deterministic using local operations and classical communication. Put $s=\sin \theta$ and $c=\cos \theta$. The initial state's squared Schmidt coefficients are $(1/2,1/2)$, while the target's decreasingly ordered squared Schmidt coefficients are $(c^2,s^2)$. Since $0<\theta <\pi /4$, one has $1/2<c^2$ and both pairs sum to one. ThusThe direction of this majorization is the allowed one in Nielsen's pure-state conversion theorem: maximal entanglement can be reduced by deterministic LOCC.

Here is an explicit protocol, so no conversion theorem need be assumed. The sender first applies a Pauli Z gate to remove the initial Bell-state minus sign. Starting from $|{\Phi }^{+}⟩$, she performs a two-outcome local measurement with Kraus operatorsThey define a valid measurement because $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. The two unnormalized states after measurement areEach outcome has probability $1/2$. For outcome zero, the normalized state is already $s|00⟩+c|11⟩$. For outcome one, both parties apply Pauli X gates, obtaining the same state. The sender communicates only which outcome occurred. Finally the receiver applies a Pauli X gate, giving $s|01⟩+c|10⟩$ in either branch. This is deterministic two-qubit entanglement dilution, with total success probability one; it does not rely on postselection.

Write $a=|\alpha {|}^2$, $b=|\beta {|}^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party giveswhere $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate eigenspace, rather than measuring her four bits separately. Her projective measurement has projectors ${\Pi }_k={\sum }_{z(x)=k}|x⟩⟨x|$, and the Born rule givesAlice sends $k$ to Bob. Up to the common phase of ${\alpha }^k{\beta }^{4-k}$, their conditional pure state isThis is Schmidt projection entanglement concentration: its equal Schmidt coefficients give a maximally entangled state of Schmidt rank $d_k$.

For $k=0$ or $4$, $d_k=1$, and the branch is a product state. No Bell pair can be extracted using local operations and classical communication. The respective probabilities are $b^4$ and $a^4$.

For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a local unitary operation extending the basis relabellingSuch a unitary exists because these are four distinct orthonormal basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leavesThus these branches give exactly two Bell pairs, with probabilities $4a{b}^3$ and $4a^{3}b$, and no further measurement in quantum mechanics is needed.

For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local POVM gives two Bell pairs with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the Kraus operatorsOn the six-dimensional support, each label is retained by two operators, so ${\sum }_{r=0}^2{K}_r^{†}{K}_r={\Pi }_2$. Add $K_{\perp }=I-{\Pi }_2$ to define a complete measurement in quantum mechanics on Alice's four qubits; that extra outcome has probability zero on this branch. Each nonzero outcome has conditional probability $(1/2)(4/6)=1/3$, and gives the normalized stateAlice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00⟩,|01⟩,|10⟩,|11⟩$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two Bell pairs. The joint probability of $(k,r)=(2,r)$ is $p_2/3=2a^2{b}^2$ for each $r$.

For an explicit local implementation of this POVM, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a deterministic reduction of maximally entangled Schmidt rank from six to four.

The complete output distribution for this strategy is consequentlyThe mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has Schmidt rank at most six, whereas three independent Bell pairs require Schmidt rank eight. In any refined LOCC branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its rank.

No secure key can be produced from this state alone. Factoring the printed amplitudes gives the normalized product stateEve is uncorrelated with the joint pure state, but Alice and Bob also have no correlations with each other: local outcome probabilities factor. They may generate independent private random bits locally, but local operations and classical communication cannot make those bits shared and secret using only public messages. A product state supplies no shared secret resource.

They need a pre-shared entanglement resource that permits deterministic production of a Bell pair. In particular, one shared Bell pair and two classical bits suffice for exact quantum teleportation. This condition is stronger than merely having some entanglement.

Let Alice's input be $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, and let her system $A$ and Bob's system $B$ share $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Use the Bell states $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$ and $|{\Psi }^{\pm }⟩=(|01⟩\pm |10⟩)/\sqrt{2}$. Direct expansion givesAlice performs a projective measurement in this Bell basis and sends its two-bit outcome. Each outcome has probability $1/4$. Bob applies respectively $I,Z,X,ZX$, recovering $|\psi ⟩$ exactly, without either party knowing $\alpha ,\beta$. Before he receives the outcome, his averaged state is $I/2$, so the protocol does not communicate without the classical message. Alice's measurement consumes the input and the shared entanglement.

For necessity, exact transmission of every pure qubit means the induced quantum channel is the identity: pure-state projectors span the operator space, and a physical channel is linear. Alice can therefore prepare a Bell pair locally and transmit one half by the proposed protocol, leaving a Bell pair shared with Bob. The protocol would itself be a deterministic local operations and classical communication conversion of their resource into a Bell pair. This proves the exact teleportation resource criterion. With no shared entanglement, it is impossible: local operations and classical communication preserve separable quantum states, whereas the resulting Bell pair would be entangled.

Quantum teleportation transfers an arbitrary unknown qubit state to a separated receiver while using local operations and classical communication, rather than sending that input qubit. Alice and Bob first share one maximally entangled state of two qubits, such as a Bell pair; a shared spin singlet state is equivalent after a known local unitary operation.

Alice performs a joint Bell-basis measurement on the input qubit and her half of the pair. There are four possible outcomes. Bob's conditional qubit is the input state transformed by a known Pauli operator determined by that outcome. Alice sends the outcome as two classical bits, and Bob applies the inverse Pauli operator. Thus one shared Bell pair and two classical bits, together with local measurement and correction, transmit one unknown qubit exactly. The entangled resource is consumed.

The procedure does not produce a classical description of the unknown amplitudes and does not leave a second copy at Alice, so it respects the no-cloning theorem. Before receiving the classical outcome, Bob's reduced density matrix is $I/2$, independent of the input; this respects the no-communication theorem. The next part proves the full identity-channel action, including inputs entangled with another system.

Write the system's $z$ basis as $|0⟩,|1⟩$, and use a separate pair of meter qubits in the Bell state $|{\Phi }^{+}⟩$. Alice applies a CNOT gate from system $A$ to her meter qubit; Bob simultaneously applies a CNOT gate from $B$ to his meter qubit. Flipping neither or both meter qubits preserves $|{\Phi }^{+}⟩$, while flipping exactly one gives $|{\Psi }^{+}⟩$. Hence the entanglement-assisted nondemolition parity measurement interaction produceswhere $P_{\mathrm{e}}=|00⟩⟨00|+|11⟩⟨11|$ and $P_{\mathrm{o}}=|01⟩⟨01|+|10⟩⟨10|$. Each party now measures only their meter qubit in the $z$ basis. If their binary records are $u,v$, the system Kraus operator isUnequal records verify zero total $z$ spin, since $S_z^{\mathrm{tot}}=\hslash (Z_A+Z_B)/2$ vanishes precisely on the odd sector. The probability of success is $⟨\psi |P_{\mathrm{o}}|\psi ⟩$, and the successful conditional state is $P_{\mathrm{o}}|\psi ⟩/\sqrt{⟨P_{\mathrm{o}}⟩}$. Every zero-total-$z$-spin state is left unchanged, including any coherent superposition of $|01⟩$ and $|10⟩$. Similarly the even-sector coherence is preserved. This is a quantum nondemolition measurement of the parity, rather than separate measurements of both system spins.

All quantum operations and local meter measurements can finish within the spacelike time window. Nevertheless each local meter record is individually uniform: $K_{u0}^{†}{K}_{u0}+K_{u1}^{†}{K}_{u1}=I/2$. The verification result is obtained only by comparing the records using local operations and classical communication. Thus “instantaneous” refers to the local completion of the joint measurement instrument, not instant access to its nonlocal outcome; quantum no-signalling remains intact.

Apply the modulo operation to the eigenvalues of $Z_A+Z_B$. The product eigenstates $|00⟩,|01⟩,|10⟩,|11⟩$ have ordinary eigenvalues $2,0,0,-2$, respectively, and residues $2,0,0,2$ modulo $4$. The resulting observable isUse the entanglement-assisted nondemolition parity measurement from part (a). Equal local meter records give $O=2$; unequal records give $O=0$. Its conditional quantum measurement maps are $\rho \mapsto P_{\mathrm{e}}\rho P_{\mathrm{e}}$ and $\rho \mapsto P_{\mathrm{o}}\rho P_{\mathrm{o}}$, with normalization by their probabilities. The quantum nondemolition measurement preserves every vector within each degenerate eigenspace, including superpositions of $|00⟩$ and $|11⟩$. Measuring the two system spins separately would destroy that even-sector coherence and would therefore not realize the same Lüders rule instrument. The nonlocal eigenvalue again becomes known only after local operations and classical communication compares the local records.

At $\theta =0$ the four eigenstates are the computational product basis, with irrelevant signs on two vectors. Alice and Bob measure their own system qubits in $Z$. These local projective measurements preserve each product eigenstate; their pair of records identifies the global outcome after local operations and classical communication.

At $\theta =\pi /4$ the four states are the Bell states. They are simultaneous eigenstates of the commuting Pauli operators $Z_A{Z}_B$ and $X_A{X}_B$: ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ have respective pairs $(+,+),(+,-),(-,+),(-,-)$. Their nonlocal parity measurements can be performed without directly distinguishing the local system spins.

Use the first shared Bell state pair to perform the entanglement-assisted nondemolition parity measurement of $Z_A{Z}_B$. Use the second shared pair for $X_A{X}_B$: both parties apply a local Hadamard gate to their system qubit, execute the same local system-to-meter CNOT gates and $z$-meter measurements, then undo the Hadamard gates. This measures $X_A{X}_B$ because $HZH=X$. The two system parity projectors commute, since anticommutation at both sites cancels:Each $P_{z,x}$ is the corresponding rank-one Bell state projector. Every complete tuple of four local meter records has system Kraus operator $P_{z,x}/2$ for its two parities. Summing the four record tuples compatible with $(z,x)$ gives the ideal outcome map $\rho \mapsto P_{z,x}\rho P_{z,x}$. The protocol is a Bell-state nondemolition measurement: an input Bell state is preserved, while an arbitrary input is projected onto the reported Bell state with the Born rule probability.

The local circuits need no adaptive communication between the laboratories, so both parties can finish inside the specified time window. Global identification of $(z,x)$ still requires later local operations and classical communication. This endpoint protocol respects quantum no-signalling, unlike the hypothetical intermediate-angle instrument in part (i).

Alice performs a projective Bell-basis measurement on her qubits $C$ and $A$. Outcome $(x,z)$ occurs with probability $1/4$ and leaves Bob's qubit inAlice sends the two classical bits $(x,z)$ to Bob. He applies the inverse Pauli operationrecovering $|\alpha ⟩$ exactly, up to an irrelevant global phase. This is quantum teleportation: only local operations and classical communication are used, while the shared Bell pair supplies the entanglement resource.

Nielsen's pure-state conversion theorem concerns exact deterministic transformations of bipartite pure states by local operations and classical communication. Let $\lambda (\psi )$ be the decreasing vector of squared Schmidt coefficients, equivalently the eigenvalues of either party's reduced density matrix, padded with zeros when necessary. ThenThe majorization relation means ${\sum }_{j=1}^k{\lambda }_j(\psi )\le {\sum }_{j=1}^k{\lambda }_j(\chi )$ for every $k$, with equality for the total sum. The direction allows a maximally entangled state to become a product state, not the reverse. Local-unitary invariance of reduced-state spectra alone would require equality of all the weights; measurements and communicated outcomes permit this larger class of conversions. A comparison of entanglement entropies alone is not a sufficient test in higher dimension.^[ref]

For the supplied states, the weight vectors and cumulative sums areAt $k=2$, $0.8>0.75$, violating the required inequality. The desired conversion is impossible with certainty using LOCC alone. The reverse conversion also fails, already at $k=1$, so these states are incomparable under this deterministic ordering.

Taking a tensor product with the catalyst multiplies every original weight by each catalyst weight. Regrouping the two local systems on Alice's side and the two on Bob's side leaves a Schmidt decomposition across the Alice–Bob partition. After sorting, the resulting weight vectors areThe actual nonnegative Schmidt amplitudes are $\sqrt{x_j}$ and $\sqrt{y_j}$; the displayed numbers are their squares. The cumulative sums areEvery component obeys $X_k\le Y_k$, so $x\prec y$. HenceThe final catalyst is exactly its initial state and is uncorrelated with the desired output. It can therefore be returned and reused. This is entanglement catalysis; it is stronger than merely retaining the same catalyst marginal while leaving correlations with the output.^[ref]

For the maximally entangled two-level auxiliary state, the sorted input weights are $(0.2,0.2,0.2,0.2,0.05,0.05,0.05,0.05)$, while the output weights are $(0.25,0.25,0.125,0.125,0.125,0.125,0,0)$. Their first four sums are $0.8$ and $0.75$. This maximally entangled auxiliary state cannot catalyze the conversion while being returned unchanged.

Nevertheless, $(0.5,0.5)\prec (0.6,0.4)$, so it can first be converted deterministically into the nonmaximal catalyst. The desired conversion can then proceed, after which Alice and Bob can return $|\varphi ⟩$, rather than the borrowed maximally entangled state. Turning that returned pair alone back into a maximally entangled pair would increase its entanglement entropy and is forbidden by LOCC. No joint restoration procedure can both keep the exact desired $|{\psi }_2⟩$ and return the original maximally entangled auxiliary state deterministically: its composition with the preceding steps would contradict the failed four-weight inequality above.