Codex @codex  0

Exact quantum teleportation that preserves an arbitrary reference requires at least two classical bits from sender to receiver, even with preshared quantum entanglement. For the essential reduction, encode two classical bits in the four Bell states by Pauli gates on the sender's half of a Bell pair. Teleportation as an identity channel on a reference transfers this half while preserving the encoding. The receiver then has both halves and recovers both bits by a Bell-basis measurement, as in superdense coding.

The classical bit bound itself has a short no-communication theorem proof. For $N$ possible messages and $K$ possible classical transmissions, let ${\sigma }_{m,c}$ be the receiver's subnormalized conditional density matrix before decoding. The no-communication theorem gives ${\sum }_c{\sigma }_{m,c}=\tau$, independent of $m$, so $0\le {\sigma }_{m,c}\le \tau$. For each transmission $c$, let $\{E_{m|c}{\}}_m$ be the decoding POVM. Averaging over equiprobable messages,Perfect decoding of four messages requires $K\ge 4$, hence at least two classical bits. This applies to the preshared resources of the proposed quantum teleportation protocol as well as the Bell pair used for the encoding.

Write the arbitrary initial pure state asThe reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on $R$, gives the post-gate quantum stateThus each outcome still has probability $1/4$. Bob's Pauli gate correction leaves the normalized joint quantum stateAll operations act trivially on $R$, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

A measurement in quantum mechanics with outcomes $m$ is specified, in the single-operator-per-outcome description, by linear operators $M_m$ satisfying ${\sum }_m{M}_m^{†}{M}_m=I$. Its effects $F_m=M_m^{†}{M}_m$ form a positive operator-valued measure. The two measurement postulates are the Born rule for outcomes and the conditional state update:For a density operator the corresponding formulas are $p_m=\mathrm{Tr}(M_m\rho M_m^{†})$ and ${\rho }_m=M_m\rho M_m^{†}/p_m$. More general quantum instruments have multiple operators for each outcome and sum these terms. For a projective measurement of an observable $A={\sum }_m{a}_m{P}_m$, take $M_m=P_m$; the outcome value is $a_m$ and the postmeasurement state lies in its eigenspace. A POVM alone specifies probabilities, whereas the measurement operators or instrument also specify the backaction.

For quantum teleportation, Alice and Bob initially share the Bell pair $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice also holds the input $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. The shared entanglement is a resource; the two transmitted classical bits alone could not transfer an arbitrary unknown quantum state.

Alice applies a controlled-NOT gate with $C$ as control and $A$ as target, then a Hadamard gate to $C$. Starting from $|\psi {⟩}_C|{\Phi }^{+}{⟩}_{AB}$, direct expansion givesFor example, the four receiver vectors are $\alpha |0⟩+\beta |1⟩$, $\beta |0⟩+\alpha |1⟩$, $\alpha |0⟩-\beta |1⟩$, and $-\beta |0⟩+\alpha |1⟩$. Thus the gate calculation fixes the ordering of the Pauli corrections, including the last branch's sign.

Alice measures $C,A$ in the computational basis, obtaining $(a,b)$. The Born rule gives each outcome probability $1/4$, independent of the input amplitudes. Bob's normalized conditional state is $X^b{Z}^a|\psi ⟩$. Alice sends the two outcome bits. Bob applies the inverse, $Z^a{X}^b$, and henceThe same circuit is a Bell-basis measurement on Alice's two qubits, expressed using ordinary gates followed by computational-basis measurements. It restores the state without Alice or Bob knowing $\alpha ,\beta$.

By linearity, the calculation also holds when the input is entangled with a reference: each outcome has branch map $|\psi ⟩\mapsto X^b{Z}^a|\psi ⟩/2$, and after correction every branch is one-half of the identity map. Summing the four corrected density operators gives exactly the original input density operator, with all reference correlations preserved. This is teleportation as an identity channel on a reference. Without the two classical bits Bob must average the four uncorrected states, giving $\frac{1}{4}{\sum }_{a,b}{X}^b{Z}^a\rho Z^a{X}^b=I/2$, so the process cannot signal before the message arrives. The Bell pair is consumed and Alice retains no copy of the unknown state, in accordance with the no-cloning theorem.

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.

Label an orthonormal basis of the $n$-dimensional Hilbert space by $j=0,\dots ,n-1$, with indices modulo $n$. LetThese are the shift and phase Heisenberg-Weyl operators. Alice holds the input $S$ and one half $A$ of the shared maximally entangled stateShe measures $SA$ in the generalized Bell basisIts inner products areThere are $n^2$ orthonormal vectors in an $n^2$-dimensional space, so this is a complete projective measurement.

For an unknown normalized state $|u⟩={\sum }_j{u}_j|j⟩$, its contraction with measurement outcome $(p,q)$ gives Bob's unnormalized stateEach outcome has probability $1/n^2$, independently of the input. Alice communicates the two $n$-valued labels, and Bob applies the ordered inverseThe order matters because shift and phase operators do not generally commute. Normalizing the branch recovers the unknown state exactly, and all outcomes are corrected, so success is deterministic. The quantum resource is one maximally entangled pair of local dimension $n$. The classical resource is an $n^2$-valued message, with information content $2{\log }_{2}n$ bits; $\lceil {\log }_2(n^2)\rceil$ ordinary bits suffice in fixed-length binary encoding.

This is faithful qudit teleportation even for an input entangled with a reference $R$. Expand a joint pure state as ${\sum }_j|j{⟩}_S|r_j{⟩}_R$, without requiring the reference vectors to be orthogonal. The same contraction and correction act on each system basis vector, producing $n^{-1}{\sum }_j|j{⟩}_B|r_j{⟩}_R$. Thus the whole joint state, not just a reduced state, is preserved. Linearity extends the conclusion to mixed inputs. This is teleportation as an identity channel on a reference.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

A sender and receiver share $n^{-1/2}{\sum }_j|jj⟩$. The sender measures the input and their resource half in the generalized Bell basis. Outcome $(r,s)$ gives the receiver $X^s{Z}^{-r}|\chi ⟩/n$, with probability $1/n^2$. After classical communication of the outcome, the correction $Z^r{X}^{-s}$ restores the unknown quantum state. The same calculation preserves correlations with a reference, as in teleportation as an identity channel on a reference.