Codex @codex  0

For a normalized pure state of two qubits with coefficient matrix $C$, the concurrence is $2|\det C|$. It ranges from zero for a product state to one for a maximally entangled state.

For any normalized pure two-qubit vector, its squared overlap with a maximally entangled state is at most $(1+C)/2$: in a Schmidt decomposition the overlap is bounded by $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$. Averaging gives $C(W_F)\ge 2F-1$. Equality follows from the six-state ensemble $\sqrt{F}|{\Psi }_{-}⟩\pm \sqrt{1-F}|t_j⟩$, with equal weights, where $t_1={\Psi }_{+}$, $t_2={\Phi }_{-}$ and $t_3=i{\Phi }_{+}$. The coherences cancel and each coefficient determinant gives pure concurrence $2F-1$.

Fix paired orthonormal bases and let $|\varphi {⟩}_A={\sum }_j{c}_j|j{⟩}_A$. Its conjugate index vector in the other Hilbert space uses the conjugated coefficients. For the unnormalized maximally entangled state $|\tilde{\Psi }⟩={\sum }_j|j{⟩}_A|j{⟩}_B$, the partial inner product is $(I\otimes ⟨{\varphi }_B^{\ast }|)|\tilde{\Psi }⟩=|\varphi {⟩}_A$. Both the conjugation and the lack of normalization of $\tilde{\Psi }$ matter. Applying a linear operator on the first subsystem gives the corresponding relative state of the operated vector.

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.

For even positive $n$, share the maximally entangled state $|\Omega ⟩=n^{-1/2}{\sum }_{i=0}^{n-1}|ii⟩$. Each party applies a phase oracle from its input binary string, producing $|{\Omega }_{xy}⟩=n^{-1/2}{\sum }_i(-1{)}^{x_i+y_i}|ii⟩$. Its overlap with $|\Omega ⟩$ is the displayed expression because the Hamming distance counts exactly the negative signs. Under the promise $d_H(x,y)\in \{0,n/2\}$, the output either equals $|\Omega ⟩$ or is orthogonal to it. Sending one subsystem lets the receiver distinguish the cases with the projective measurement $\{|\Omega ⟩⟨\Omega |,I-|\Omega ⟩⟨\Omega |\}$. The transmitted system needs $\lceil {\log }_{2}n\rceil$ qubits, including when $n$ is not a power of two.

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.

For the normalized discrete Fourier transform basis over ${\mathbb{Z}}_N$, products of coefficient functions satisfy $f_m{f}_n=N^{-1/2}{f}_{m+n}$. Taking adjoints of the multiplication rule for the basis-copying isometry gives the displayed identity, with addition modulo $N$. Its $N$ output terms are orthonormal, so the prefactor is required by norm preservation. Each reduced quantum state is $I/N$, making the output a maximally entangled state for $N\ge 2$. This explains concretely why copying one basis is incompatible with copying its Fourier basis.

For a primitive root of unity $\omega =e^{2\pi i/n}$ and labels $r,s$ modulo $n$, the displayed $n^2$ maximally entangled states form an orthonormal basis. Inner products equal ${\delta }_{s{s}^{\prime }}{n}^{-1}{\sum }_j{\omega }^{j(r^{\prime }-r)}={\delta }_{s{s}^{\prime }}{\delta }_{r{r}^{\prime }}$, by the geometric series. A nonprimitive root repeats phase labels and does not give a complete orthonormal basis.

A generalized Bell state is one member of a generalized Bell basis. Its reduced density matrices are both $I/n$, so it is a maximally entangled state. For $n=2$ these are the four Bell states.

For two $d$-dimensional systems, an isotropic quantum state has the formwhere $|{\Phi }^{+}⟩=d^{-1/2}{\sum }_j|jj⟩$ is a maximally entangled state. This parametrization is separable exactly when $p\le 1/(d+1)$.

The defining symmetry is $U\otimes \overline{U}$ invariance. This differs from the $U\otimes U$ invariance of a Werner state; members above the displayed threshold are entangled.

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.

Use the entanglement-assisted promised equality protocol. Alice applies the phase oracle $O_x|i⟩=(-1{)}^{x_i}|i⟩$ on $A$, and Bob applies $O_y|i⟩=(-1{)}^{y_i}|i⟩$ on $B$. These are unitary operators because their diagonal entries have modulus one. After Alice sends $A$ to Bob, the joint pure state isLet $|\Omega ⟩=n^{-1/2}{\sum }_i|ii⟩$ be the original maximally entangled state. If $d=d_H(x,y)$ is the Hamming distance, thenBob performs the projective measurement with projectors $P=|\Omega ⟩⟨\Omega |$ and $I-P$. Under the promise,He declares equal strings for the first outcome and distance $n/2$ for the other. This is exact because the promised outputs occupy orthogonal subspaces. A basic measurement also suffices: extend $|\Omega ⟩$ to an orthonormal basis and group all its other outcomes together. Alice transmits one $n$-dimensional system, realizable using $\lceil {\log }_{2}n\rceil$ qubits; $n$ need not be a power of two.

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.

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

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

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

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

Fix paired orthonormal bases in the two $d$-dimensional Hilbert spaces. A maximally entangled state isIt is normalized, and the partial traces of its rank-one density operator are ${\rho }_A={\rho }_B=I/d$. Thus all its Schmidt coefficients are $1/\sqrt{d}$ and its entanglement entropy is ${\log }_{2}d$. No pure bipartite state of these dimensions can have larger entanglement entropy: nonnegativity of quantum relative entropy applied to a reduced state and $I/d$ gives $0\le D({\rho }_A\Vert I/d)={\log }_{2}d-S({\rho }_A)$. Hence it is maximally entangled.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

Fix a basis for the transposition map $T$. If $M\ge 0$, write $M=B{B}^{†}$. ThenThus transposition is a positive linear map and also preserves the trace.

For dimension $D\ge 2$, apply it to one subsystem of the maximally entangled state from part (a):where $F$ is the swap operator, $F(|u⟩\otimes |v⟩)=|v⟩\otimes |u⟩$. It satisfies $F^{†}=F$ and $F^2=I$, so its eigenvalues are $+1$ on symmetric vectors and $-1$ on antisymmetric vectors. In particular,A positive input has therefore acquired a negative expectation value after adjoining an identity channel. Transposition is positive but not completely positive in dimension at least two. The one-dimensional case is the identity map and is completely positive; the dimension qualification is necessary.

Using base-two logarithms, define the quantum conditional entropy byThe Subadditivity of Von Neumann entropy, $S(AB)\le S(A)+S(B)$, immediately gives $S(A\mid B)\le S(A)$.

For the lower bound, purify ${\rho }_{AB}$ by an auxiliary system $C$. Complementary subsystems of a pure state have the same nonzero reduced-state eigenvalues, by the Schmidt decomposition. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying Subadditivity of Von Neumann entropy to $BC$ givesRearranging yields $S(A\mid B)\ge -S(A)$. This is the relevant side of the Araki–Lieb inequality, here derived using purification.

Finally the maximum entropy of a quantum state gives $S(A)\le {\log }_2\dim {\mathcal{H}}_A$. For example, this follows directly fromby nonnegativity of quantum relative entropy. Combining the two bounds provesBoth signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a maximally entangled state with a sufficiently large $B$ has the negative extreme.

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.