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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.

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

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

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

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

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.

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.

The normalized ray produces a secure key; the literal printed vector is not normalized. The squared norm of the coefficients in the PDF is $8/9+1/81=73/81$. Thus the physical pure state on that ray has normalized coefficients $6\sqrt{2}/\sqrt{73}$ and $1/\sqrt{73}$. This is a genuine entangled state with two nonzero Schmidt coefficients, and Eve is independent of it by purity decouples a subsystem from its purification.

For an explicit way to obtain uniform key bits, Alice applies a two-outcome local measurement in quantum mechanics with Kraus operatorsTheir squared products sum to the identity. The successful unnormalized state is $(|00⟩+|11⟩)/\sqrt{73}$, soAlice announces only success or failure. Success leaves an exact Bell pair, yielding one uniform secret bit; failure leaves $|00⟩$ and is discarded. With many copies the successful fraction is positive. Alternatively, direct computational basis measurements give a shared biased bit with probabilities $72/73$ and $1/73$, from which privacy amplification can extract uniform secret bits. The coefficient $1/9$ has not been silently replaced by $1/3$.

The Schmidt decomposition of a normalized bipartite pure state isThe two sets of local vectors are orthonormal. The Schmidt coefficients are the nonnegative numbers $\sqrt{{\lambda }_j}$, and the Schmidt number for this pure state, also called its Schmidt rank, is $r$, the number of nonzero coefficients. It equals the rank of either reduced density operator and of the coefficient matrix in any product bases. The general mixed-state Schmidt number specializes to this same value on a pure state.

The two displayed amplitudes have modulus $1/\sqrt{2}$. Absorb their phases into Bob's basis rather than use complex Schmidt coefficients. Take $|u_0⟩=|0⟩$, $|u_1⟩=|1⟩$, andBoth local bases are orthonormal. A Schmidt decomposition is thereforeThe two Schmidt coefficients are $1/\sqrt{2}$, the Schmidt number is two, and both reduced density operators are $I/2$. Their equal eigenvalues make this decomposition nonunique.

Introduce the orthonormal local vectors $|+⟩=(|1⟩+|2⟩)/\sqrt{2}$ and $|-⟩=(|1⟩-|2⟩)/\sqrt{2}$. Reexpressing the second state's factors in this basis gives its Schmidt decompositionThe squared Schmidt coefficients are therefore $1/3,2/3$, whereas those of the first state are $1/2,1/2$.

To see directly why these distinguish the states, a local unitary operation $U_A\otimes U_B$ changes the reduced density matrix toThis is local-unitary invariance of reduced-state spectra: the unitary on the traced subsystem cancels, and conjugation on $A$ leaves its eigenvalues unchanged. Here the reductions areTheir spectra differ. Hence no pair of local unitary operations carries the first state to the second. More generally, normalized bipartite pure states are locally unitarily equivalent exactly when their squared Schmidt coefficients agree, including multiplicities; sufficiency follows by mapping one pair of Schmidt bases to the other.

Put $s={\sin }^2\alpha$ and $t={\sin }^2\beta$. Then $0<s<t<1/2$, and the ordered squared Schmidt coefficients of the source and target are $(1-s,s)$ and $(1-t,t)$.

Nielsen's pure-state conversion theorem states that a deterministic exact conversion between bipartite pure states by LOCC exists precisely when the source's vector of squared Schmidt coefficients is majorized by the target's vector, with zero padding if needed. In the convention $x\prec y$, majorization requires ${\sum }_{j=1}^k{x}_j^{\downarrow }\le {\sum }_{j=1}^k{y}_j^{\downarrow }$ for every $k$, and equality of total sums. Here the first inequality would require $1-s\le 1-t$, but $s<t$ gives the reverse strict inequality. Thus exact conversion with certainty is impossible.

An upper bound follows directly from the Schmidt-tail entanglement monotonewhere ${\rho }_A^{\chi }$ is the reduced density matrix of a bipartite pure state. On these two-qubit states, this is the smaller squared Schmidt coefficient, so $E_2(\psi )=s$ and $E_2(\varphi )=t$.

Here is the average-monotonicity argument needed for all LOCC strategies. The largest eigenvalue is convex: for any unit vector $v$, $⟨v|{\sum }_r{p}_r{\rho }_r|v⟩\le {\sum }_r{p}_r{\lambda }_{\max }({\rho }_r)$, and maximizing the left side gives the claim. Hence $1-{\lambda }_{\max }$ is concave. If Alice makes a local measurement, Bob's conditional reductions average to his original reduction. Concavity therefore gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. If Bob measures, use Alice's reductions instead. The two reductions of a pure state have the same nonzero eigenvalues by the Schmidt decomposition. Refining all measurement outcomes into individual Kraus operators keeps conditional states pure; iterating the inequality covers adaptive rounds of classical communication. The same bound persists for limits of such protocols.

If success produces the target with probability $p$, its contribution to the final average is $pt$. Failure branches have nonnegative $E_2$. ThusThis is a bound on arbitrary local protocols, not merely on a particular filtering measurement.

Nielsen's pure-state conversion theorem gives the exact deterministic LOCC criterion. Let the Schmidt decompositions be $|\psi ⟩={\sum }_j\sqrt{{\lambda }_j}|a_j{b}_j⟩$ and $|\varphi ⟩={\sum }_j\sqrt{{\mu }_j}|a_j^{\prime }{b}_j^{\prime }⟩$. The vectors $\lambda ,\mu$ consist of squared Schmidt coefficients, equivalently the eigenvalues of either reduced density operator. Order each in decreasing order and pad with zeros to a common length $d$.

Then deterministic conversion is possible if and only ifThis is majorization, with the input vector majorized by the output vector. The direction matters: a maximally entangled state has a uniform vector, which is majorized by a product state's vector $(1,0,\dots )$, so entanglement can be discarded by LOCC. The criterion is for certain exact conversion, without catalysts or postselection on a successful branch.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.