= Solution
The target states here are <qubit> states, as appropriate to sharing <singlet states> of two <qubits>. Write $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ and choose $|\psi^\perp\rangle=-\beta^*|0\rangle+\alpha^*|1\rangle$. These form an <orthonormal basis>, and the shared <Bell state> can be expressed as
$$
|\Psi^-\rangle_{AB}
=\frac{|0\rangle_A|1\rangle_B-|1\rangle_A|0\rangle_B}{\sqrt2}
=\frac{|\psi\rangle_A|\psi^\perp\rangle_B
-|\psi^\perp\rangle_A|\psi\rangle_B}{\sqrt2}.
$$
Alice knows the classical description of $|\psi\rangle$, so she can perform the <projective measurement> in this basis. The outcome $|\psi^\perp\rangle$ has <probability> $1/2$ and leaves Bob in $|\psi\rangle$, 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>:
$$
\boxed{p_{\mathrm{success}}=\frac12,\qquad
\text{one singlet and one classical success flag}.}
$$
The other outcome leaves $|\psi^\perp\rangle$ 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 $|\psi_j\rangle,|\psi_j^\perp\rangle$. 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 $\bigotimes_{j=1}^n|\psi_j\rangle$. Thus <one-bit heralding of a remote state preparation block> gives
$$
\boxed{p_{\mathrm{success}}=2^{-n},\qquad
n\text{ singlets and one block-success flag}.}
$$
On 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,
$$
\mathbb P(J=j)=q(1-q)^{j-1},\qquad j\geq1.
$$
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 $0\leq R<M$. 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 $r=(1-q)^M$,
$$
\mathbb P(Q\geq k)=r^k,\qquad
\mathbb E Q=\sum_{k=1}^\infty r^k=\frac{r}{1-r}.
$$
Since $(1-1/M)^M\leq e^{-1}$,
$$
\boxed{\mathbb E L=n+\frac1{1-r}\leq n+\frac e{e-1},\qquad
\frac{\mathbb E L}{n}\longrightarrow1.}
$$
This version succeeds exactly with <probability> one and has the stated expected <classical communication> cost. It consumes $n\,\mathbb EJ=n2^n$ <singlet states> on average, compatible with the unlimited entanglement assumption.
There is also a fixed-message-length version. Try at most $K=n2^n$ candidate blocks and send the first successful index, or a zero codeword reporting that none succeeded. Encoding the $K+1$ possibilities uses $\lceil\log_2(K+1)\rceil=n+\log_2n+O(1)$ <bits>. Its failure <probability> is
$$
(1-2^{-n})^{n2^n}\leq e^{-n}.
$$
Thus \b[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>.
Back to article page
