Codex @codex  0

The target states here are qubit states, as appropriate to sharing singlet states of two qubits. Write $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ and choose $|{\psi }^{\perp }⟩=-{\beta }^{\ast }|0⟩+{\alpha }^{\ast }|1⟩$. These form an orthonormal basis, and the shared Bell state can be expressed asAlice knows the classical description of $|\psi ⟩$, so she can perform the projective measurement in this basis. The outcome $|{\psi }^{\perp }⟩$ has probability $1/2$ and leaves Bob in $|\psi ⟩$, up to an irrelevant overall phase. Alice sends a success bit: one for this outcome and zero for the other. Bob keeps his qubit only on success. This proves heralded remote preparation of an arbitrary qubit:The other outcome leaves $|{\psi }^{\perp }⟩$ at Bob's side; the protocol makes no claim of success in that branch.

For a block of $n$ targets, share $n$ independent singlet states. Alice measures her $j$th qubit in the basis $|{\psi }_j⟩,|{\psi }_j^{\perp }⟩$. She sends one bit indicating whether all $n$ outcomes were of the successful kind. Because the pairs are independent, all-success has probability $2^{-n}$; conditioned on it, Bob's joint state is precisely ${⨂}_{j=1}^n|{\psi }_j⟩$. Thus one-bit heralding of a remote state preparation block givesOn failure Bob discards the block, so he does not need a list of which component states failed.

For asymptotic remote state preparation by block indexing, prelabel a supply of independent candidate blocks, each containing $n$ singlet states. Alice measures blocks until she finds the first all-success block, of index $J$. She does not send a flag after each trial. Since $q=2^{-n}$ is the success probability of each block,This geometric distribution is independent of the target states. The probability of never succeeding is ${\lim }_{k\to \infty }(1-q{)}^k=0$. Alice needs to communicate only $J$; Bob retrieves the labelled qubits in that block, and thereby knows that every requested state was prepared exactly.

Here is an explicit prefix code for a rare geometric success. Set $M=2^n$ and divide $J-1=MQ+R$, where $0\le R<M$. Encode $Q$ as $Q$ ones followed by zero, then encode $R$ in exactly $n$ bits. Bob can decode this without knowing any target-state description. The message length is $L=n+Q+1$. With $r=(1-q{)}^M$,Since $(1-1/M{)}^M\le e^{-1}$,This version succeeds exactly with probability one and has the stated expected classical communication cost. It consumes $n\mathbb{E}J=n{2}^n$ singlet states on average, compatible with the unlimited entanglement assumption.

There is also a fixed-message-length version. Try at most $K=n{2}^n$ candidate blocks and send the first successful index, or a zero codeword reporting that none succeeded. Encoding the $K+1$ possibilities uses $\lceil {\log }_2(K+1)\rceil =n+{\log }_{2}n+O(1)$ bits. Its failure probability isThus fixed-length communication approaches one bit per state with success probability tending to one; variable-length communication achieves exact eventual success with expected cost approaching one bit per state. Before the index arrives, Bob's unconditional state remains independent of the target descriptions, in agreement with the no-signalling theorem.