Codex @codex  0

For a pure state with two descending squared Schmidt coefficients ${\lambda }_0\ge {\lambda }_1$, this quantum instrument acts in one party's Schmidt basis. The first outcome balances the two Schmidt amplitudes when $\tan \theta =\sqrt{{\lambda }_1/{\lambda }_0}$, with probability $2{\lambda }_0{\lambda }_1$. This is not the optimal unrestricted success probability $2{\lambda }_1$ in single-copy entanglement concentration. An embedding in a larger subsystem requires a POVM on the occupied support or a full-space completion.

For the Schmidt-basis Pauli correlation tensor, choose Alice's axes ${\mathbf{e}}_z,{\mathbf{e}}_x$ and Bob's $({\mathbf{e}}_z\pm s{\mathbf{e}}_x)/\sqrt{1+s^2}$. The CHSH inequality expression is $2\sqrt{1+s^2}$. It exceeds two exactly when both Schmidt coefficients are nonzero, reaching $2\sqrt{2}$ for maximal quantum entanglement. A product state gives $s=0$ and saturates rather than violates the inequality.

For two-qubit pure states with smaller squared Schmidt coefficients $s<t$, optimal stochastic conversion of a two-qubit pure state succeeds independently with probability $s/t$. The asymptotic pure-state entanglement conversion rate of collective LOCC is instead $h_2(s)/h_2(t)$, where $h_2$ is binary entropy. The difference is strict: differentiating gives $\frac{d}{dx}[h_2(x)/x]={\log }_2(1-x)/x^2<0$. Hence $h_2(s)/s>h_2(t)/t$. Independent filtering loses entanglement in its product-state failure branches, whereas collective concentration and dilution preserve entanglement asymptotically.

Let $|{\Phi }_d⟩=d^{-1/2}{\sum }_{j=0}^{d-1}|j,j⟩$ and $1\le m\le d$. It can be converted deterministically by LOCC to $|{\Phi }_m⟩$. For each $m$-element subset $S$ of the local basis labels, let ${\Pi }_S$ be its projector and let Alice's Kraus operator be $K_S={\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)}$. Every label occurs in $\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)$ subsets, so ${\sum }_S{K}_S^{†}{K}_S=I$. Each outcome has probability $m/[d(\genfrac{}{}{0ex}{}{d-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d}{m}\right)$ and leaves the maximally entangled state $m^{-1/2}{\sum }_{j\in S}|j,j⟩$. Alice reports $S$, and both parties relabel these $m$ basis states to $0,\dots ,m-1$ by local unitary operations.

For $d=6$, $m=4$, only three outcomes are needed: partition the six labels into disjoint pairs $T_0,T_1,T_2$ and use $K_r=(I-{\Pi }_{T_r})/\sqrt{2}$. Each label is retained twice, proving completeness, and each outcome has probability $1/3$. Every branch leaves four equal Schmidt coefficients, hence two independent Bell pairs after binary relabelling.

An auxiliary entangled pure state can enable an otherwise forbidden deterministic bipartite LOCC transformation while being returned exactly and uncorrelated with the output. Nielsen's pure-state conversion theorem reduces this to the displayed majorization condition on sorted squared Schmidt coefficients. It does not follow from a scalar comparison of entanglement entropies: tensoring changes which products occupy the largest cumulative-weight sums. An exact returned catalyst can be reused; merely restoring its marginal while allowing correlations is a different resource task.^[ref]

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.

The entanglement spectrum is the spectrum of $-\log {\rho }_A$, equivalently the negative logarithms of squared Schmidt coefficients. A nontrivial projective virtual symmetry can enforce degeneracies throughout this spectrum.

Every entangled pure state of two qubits violates a CHSH inequality for a suitable choice of local measurement axes. If its two Schmidt coefficients are $c_0,c_1>0$, the optimal construction already gives a value $2\sqrt{1+(2c_0{c}_1{)}^2}>2$.

Two bipartite pure states on fixed subsystem spaces are local-unitary equivalent when related by the displayed product of unitary operators. Identity operators give reflexivity, inverse operators give symmetry, and products of local operators give transitivity. Thus this is an equivalence relation, whose classes are the orbits under local unitaries.

Every pair of normalized product states is local-unitary equivalent: extend each factor to an orthonormal basis and use a unitary operator carrying one basis to the other. Local operators preserve a product factorization, and their inverses do too, so quantum entanglement cannot be created or removed this way. More generally, the tensor-product coefficient matrix changes as $C\mapsto UC{V}^T$, which preserves its singular values and therefore its Schmidt coefficients. The transpose on the second factor is ordinary, not a conjugate transpose, because coefficients of a tensor product depend bilinearly on the two input vectors.

Under a product unitary operator $U_A\otimes U_B$, the reduced density matrix of $A$ becomes $U_A{\rho }_A{U}_A^{†}$. The unitary on $B$ cancels under the partial trace. Unitary conjugation preserves eigenvalues, so the squared Schmidt coefficients of a bipartite pure state are unchanged. Conversely, two normalized bipartite pure states with the same squared Schmidt coefficients, including multiplicities and zero padding, have Schmidt decompositions carried into each other by local unitaries mapping their Schmidt bases. Thus reduced-state spectra completely classify bipartite pure states up to local unitary operations.

An operation that applies a unitary operator to one party's subsystem and the identity to the other is a local unitary operation. Several separated parties may each apply a known unitary, resulting in a product of their local unitaries. Such operations preserve the squared Schmidt coefficients across the relevant bipartition.

Martin Bodo Plenio and Daniel Jonathan showed that a shared auxiliary entangled state can enable a deterministic LOCC conversion that is impossible for the original pair of pure quantum states alone. The auxiliary quantum state is returned unchanged and uncorrelated with the output, so it acts as a catalyst for entanglement catalysis. The key distinction is between consuming an extra resource and using one that can be recovered: a majorization obstruction for the original Schmidt coefficients can disappear after including the catalyst's Schmidt coefficients.^[ref]

A normalized bipartite pure state of Schmidt rank $r$ has quantum fidelity at most $\sqrt{r/d}$ with a fixed maximally entangled state on $d\otimes d$. Its coefficient matrix has matrix rank $r$ and Hilbert-Schmidt norm one. Its overlap is the trace of that matrix divided by $\sqrt{d}$; trace duality and the Cauchy-Schwarz inequality bound this by $\sqrt{r/d}$. Uniform Schmidt coefficients on $r$ matching basis pairs attain equality.

An exact deterministic LOCC conversion of a bipartite pure state to another is possible exactly when the input vector of squared Schmidt coefficients is majorized by the output vector. Sort the two probability vectors decreasingly and pad with zeros to a common length before applying majorization. A maximally entangled state may therefore be converted to a product state, while the reverse conversion is forbidden. The criterion does not assert conversion by postselection or with a catalyst.

Taking partial inner products with $⟨0|$ and $⟨1|$ on the first wire gives $(|0⟩+|1⟩)/2$ and $(-|0⟩+|1⟩)/2$. They are nonzero and orthogonal, hence linearly independent. This contradicts the necessary product-state property proved in part (a), so the second state is entangled.

Equivalently, its tensor-product coefficient matrix is $C=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$, with $\det C=1/2\ne 0$. The two-qubit product-state determinant criterion gives the same conclusion. In fact $C{C}^{†}=I/2$, so its Schmidt coefficients are both $1/\sqrt{2}$: it is maximally entangled.

Grouping by the first qubit givesThe vectors on both sides form orthonormal bases, so this is a Schmidt decomposition with two nonzero Schmidt coefficients $1/\sqrt{2}$. Hence the state is maximally entangled. Independently, its coefficient matrix has determinantwhich excludes a product state by the two-qubit product-state determinant criterion. Each reduced density matrix is $I/2$.

Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with qubit 1 written first. The half-time gate requires a branch convention. Even stationary Hamiltonian evolution with $W(2\tau )=\mathrm{SWAP}$ does not uniquely specify $W(\tau )$: different accumulated eigenstate phases can agree at $2\tau$ and differ at $\tau$. In particular, both $S$ and $S^{†}$ below square to SWAP gate and leave $|00⟩,|11⟩$ unchanged. Adopt the conventional continuous isotropic exchange interaction branch, with a relative singlet phase increasing from $0$ to $\pi$ during the full pulse. This is the exchange square-root branch convention.

The symmetric and antisymmetric eigenstateshave SWAP gate eigenvalues $+1,-1$. Let $P_{-}=|{\psi }_{-}⟩⟨{\psi }_{-}|$ and $P_{+}=I_4-P_{-}$ be their antisymmetric and symmetric orthogonal projectors. The adopted exchange evolution isFor effective spin-one-half qubits, $H_{\mathrm{ex}}=\pi \hslash (X_1{X}_2+Y_1{Y}_2+Z_1{Z}_2-I_4)/(8\tau )$, an isotropic exchange interaction plus an energy offset that fixes the common phase. At half time choose $+1$ on the symmetric sector and $+i$ on the antisymmetric sector, the conventional principal square-root value of $-1$. Hence the square root of the SWAP gate isThus $S|00⟩=|00⟩$, $S|11⟩=|11⟩$, $S|01⟩=a|01⟩+b|10⟩$ and $S|10⟩=b|01⟩+a|10⟩$, where $a=(1+i)/2$ and $b=(1-i)/2$. The identities $a^2+b^2=0$ and $2ab=1$ verify $S^2=\mathrm{SWAP}$, and the projector form proves unitarity. The endpoint truth table alone would allow other matrix square roots; the stated exchange branch supplies the additional physical assumption.

For the network, fix the Pauli Z gate, Hadamard gate and S gate conventionsThe letter $P$ distinguishes this local S gate from the two-qubit exchange gate $S$. In particular $P^2=Z$; replacing $P$ by $P^{†}$ changes the network. Subscripts specify the acted-on qubit: $Z_1=Z\otimes I_2$, $P_2=I_2\otimes P$ and $H_2=I_2\otimes H$. Operators act rightmost first, so the printed quantum circuit is

First multiply the central two exchange pulses. Since $a^2-b^2=i$, their off-diagonal terms cancel:This reduces the entire network to $U=H_{2}D{H}_2$. Its action on all four computational basis states, retaining every relative quantum phase, isEquivalently,It flips the second qubit when the first is zero and contributes $-i$ when the first is one. That conditional phase cannot be discarded as a global phase.

To obtain the standard CNOT gate controlled by qubit 1, append an $X$ gate to qubit 2 and $P$ to qubit 1. They commute, andFor control zero, $X_2$ undoes the original flip. For control one, it supplies the desired flip and $P_1$ cancels the phase because $i(-i)=1$. This phase-corrected exchange network for a CNOT gate realizes the standard gate exactly with the declared conventions.

Finally, $S|01⟩=(e^{i\pi /4}|01⟩+e^{-i\pi /4}|10⟩)/\sqrt{2}$ has equal Schmidt coefficients and is maximally entangled: a half exchange creates quantum entanglement although a full SWAP gate does not. More constructively, the network above produces a CNOT gate from two half exchanges and local gates. Since CNOT gates and arbitrary single-qubit unitary operators form a universal quantum gate set, the square root of the SWAP gate and arbitrary single-qubit gates are universal. The alternative inverse exchange branch remains entangling and universal, but its phases and local correction must be recomputed consistently.

The transmitted branch is the Kraus operatorFor nonzero loss it is not trace preserving: $\mathrm{tr}(E\rho E^{†})$ is the transmission probability. A complete channel includes an orthogonal vacuum or erasure flag, with a loss operator $L=\sqrt{R}|\mathrm{vac}⟩⟨0|$ and $E^{†}E+L^{†}L=I$. Postselecting only the first outcome gives the single-operator description in the question.

For $|\pm ⟩$, the unnormalized transmitted vector is $(\sqrt{1-R}|0⟩\pm |1⟩)/\sqrt{2}$. Its norm squared and normalized conditional state areThe reduced density matrix of either member of the initial pair has eigenvalues $\alpha ,1-\alpha$. Since the joint state is pure, its Von Neumann entropy is zero. With logarithms to base two and $h_2$ the binary entropy function,The two nonzero Schmidt coefficients show that this pure pair is entangled exactly when $0<\alpha <1$. Its mutual information rises from zero to two bits at $\alpha =1/2$ and returns to zero.

For this known pure family, every positive mutual information signals entanglement. For an otherwise unspecified two-qubit state, positive mutual information can also be classical. The mutual information bound for separable states gives the stronger general witness $I(A:B)>1$ bit. On this curve it holds for approximately $0.110028<\alpha <0.889972$; the rest of the open interval is still entangled, but that generic one-bit witness alone would not certify it.

If Bob's qubit passes through the channel, the unnormalized retained state is $\sqrt{\alpha }|11⟩+\sqrt{(1-\alpha )(1-R)}|00⟩$. ThusFor $0\le R<1$, equal Schmidt coefficients require $\alpha =(1-\alpha )(1-R)$, givingThe retained state is then $(|00⟩+|11⟩)/\sqrt{2}$. At $R=1$, this limiting compensation has zero transmission probability and yields no maximally entangled pairs. Conditioning can increase the entanglement of retained pairs, while the discarded fraction accounts for the cost.

Taking the partial trace over either photon of the pure pair state givesThe two Schmidt coefficients are both $1/\sqrt{2}$, so the entanglement entropy is one bit for every $\phi$. A local phase gate on Alice maps $|{\Psi }^0⟩$ to $|{\Psi }^{\phi }⟩$, proving local-unitary equivalence. In strict four-vector Bell-basis terminology only $\phi =0,\pi$ give the standard $|{\Psi }^{+}⟩,|{\Psi }^{-}⟩$; every phase nevertheless gives a maximally entangled Bell-type state.

The eigenpolarizations of the requested observable areA polarizing beam splitter with its transmitted polarization axis at angle $\alpha /2$ to horizontal separates these two orthogonal states. Assign outcome $+1$ to its first detector and $-1$ to the other. Equivalently, use a fixed horizontal/vertical splitter preceded by a half-wave plate whose fast axis is at $\alpha /4$; the diagram shows this equivalent implementation. Bob uses $\beta /2$ for his splitter axis, or $\beta /4$ for the preceding half-wave plate.

The pair correlations are $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=\cos \phi$, and $⟨Z\otimes X⟩=⟨X\otimes Z⟩=0$. Expanding the two observables therefore givesThe half-angle between the Bloch observable parameter and the physical linear-polarization analyser angle is essential.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.