Codex @codex  0

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Under the usual reversible-circuit interpretation, there is a genuine obstruction to uniform probability lowering by a unitary in the printed premise. The universal assertion cannot hold for arbitrary $p$: it fails already at the search density $p=k/N$.

An explicit counterexample uses $N=4$, one good basis state $|0⟩$, $p=1/4$ and $p^{\prime }=1/8$. The six pure statesall have good probability $1/4$. Their equally weighted density operator is $I_4/4$, a maximally mixed state. Every unitary operator $C$ leaves this mixture unchanged, so its average output good probability remains $1/4$. The printed assertion would instead make all six output probabilities $1/8$, a contradiction. More generally, averaging states $\sqrt{k/N}|g⟩+e^{i\alpha }\sqrt{1-k/N}|b⟩$ uniformly over good and bad basis vectors and two opposite phases gives $I_N/N$, proving the same obstruction at the actual search density $p=k/N$.

The intended conditional construction needs only a supplied reversible preparation that lowers the success probability of the particular starting state, not of every state. Here is the complete argument under that weaker resource assumption. Count $C$ and $C^{†}$ as supplied, query-free operations, as required for the claimed input-oracle bound. Let $\theta =\arcsin \sqrt{k/N}$ and chooseIf equality holds, use the original preparation. Otherwise use the stipulated preparation on $|{\psi }_0⟩$ to obtain $|{\psi }_{\ast }⟩=C|{\psi }_0⟩$ with good probability $p_{\ast }$. Its reflection operator is implementable byThe good and bad components may be nonuniform, but amplitude amplification applies to this same two-dimensional decomposition. After $j$ iterations of $-I_{{\psi }_{\ast }}{I}_f$, its good probability isThus the stated exact-query conclusion follows from accessible preparation of one known-overlap state and its inverse. If the supplied preparation uses oracle queries, those costs cannot be omitted.

A physically realizable known-state success dilution for exact amplitude amplification is available in the standard enlarged marking-oracle model. Append a flag in $\sqrt{1-q}|0⟩+\sqrt{q}|1⟩$, where $q=p_{\ast }/p$, and call a joint state good only when $f(x)=1$ and the flag is one. This prepared state's success probability is exactly $pq=p_{\ast }$. Reflect about this known product preparation and mark the joint good subspace; $j$ ordinary iterations then succeed with certainty. A supplied controlled phase oracle gives one marking query per iteration, or a Boolean bit oracle computes $f$, applies the joint phase and uncomputes $f$ in two queries. Measuring the data therefore returns a good $x$ with certainty in $O(\sqrt{N/k})$ queries. This changes the state space and marking test, and does not assert the impossible universal $n$-qubit circuit. Controlled access is an additional resource in a bare phase-oracle model, so it is made explicit rather than silently assumed.

Uhlmann's theorem identifies quantum fidelity with the largest absolute overlap of purifications of a density operator. Choose a common reference Hilbert space $R$ of dimension at least that of the original system. Thenwhere $|u⟩,|v⟩\in {\mathcal{H}}_A\otimes {\mathcal{H}}_R$ have reduced density operators ${\rho }_A,{\sigma }_A$. One purification may be fixed in advance: the maximum is over the other, with the freedom to apply a unitary on the reference system. This is the unitary freedom of purification. Enlarging the reference by unused dimensions does not change the maximum.

Take maximizing purifications $|u⟩,|v⟩$ of ${\rho }_{AB},{\sigma }_{AB}$ on $ABR$. Their overlap has magnitude $F({\rho }_{AB},{\sigma }_{AB})$. Regard these same vectors as purifications of ${\rho }_A,{\sigma }_A$ with reference system $BR$. They are candidates in the larger optimization, so Uhlmann's theorem givesThis is monotonicity of quantum fidelity under partial trace: discarding a system cannot make two states more distinguishable according to their quantum fidelity.

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore

Let $q_y=⟨y|\rho |y⟩$. Positivity and normalization of the density operator imply $q_y\ge 0$ and ${\sum }_y{q}_y=1$. The completely dephasing channel produceswhich is already a spectral decomposition. Its Von Neumann entropy is thereforeThe quantum output entropy is exactly the Shannon entropy of the basis-measurement probabilities, including possible zero eigenvalues via $0{\log }_{2}0=0$.

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

For any tripartite density operator, Strong subadditivity of Von Neumann entropy isUse quantum relative entropy $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$, with the usual support condition. The equivalent relative-entropy comparison isIndeed the two sides expand respectively as $S(A)+S(BC)-S(ABC)$ and $S(A)+S(B)-S(AB)$. Subtracting cancels $S(A)$ and leaves exactly the strong-subadditivity gap. Marginal-product supports contain the support of the joint state, so these expressions are finite; singular marginals can also be handled by full-rank regularization and a limit.

Finally, tracing out $C$ sends the numerator and denominator of the first relative entropy to those of the second. The data-processing inequality for quantum relative entropy therefore proves the comparison. This is strong subadditivity from relative-entropy monotonicity, rather than an assumption that classical entropy proofs automatically apply to quantum states.

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.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.