= Solution
Let the unknown input <qubit> be $|\psi\rangle_S=\alpha|0\rangle+\beta|1\rangle$, with $|\alpha|^2+|\beta|^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared <Bell pair> is $|\phi^+\rangle_{AB}$. All three-<qubit> strings below use the order $S,A,B$. Initially,
$$
|\Psi_0\rangle=|\psi\rangle_S|\phi^+\rangle_{AB}
=\frac{\alpha|000\rangle+\alpha|011\rangle+\beta|100\rangle+\beta|111\rangle}{\sqrt2}.
$$
The <Bell-basis teleportation identity> gives
$$
|\Psi_0\rangle=\frac12\left[
|\phi^+\rangle_{SA}|\psi\rangle_B
+|\phi^-\rangle_{SA}Z|\psi\rangle_B
+|\psi^+\rangle_{SA}X|\psi\rangle_B
+|\psi^-\rangle_{SA}XZ|\psi\rangle_B\right].
$$
Thus a perfect <Bell state analyzer> on Alice's two <qubits> identifies the <Pauli operator> affecting Bob's <quantum state>.
To display the intermediate three-<qubit> states explicitly, realize that <Bell-basis measurement> by a <controlled-NOT gate> $S\to A$, a <Hadamard gate> on $S$, and two <computational-basis measurements>. After the <controlled-NOT gate>,
$$
|\Psi_1\rangle=\frac{\alpha|000\rangle+\alpha|011\rangle+\beta|110\rangle+\beta|101\rangle}{\sqrt2}.
$$
After the <Hadamard gate>,
$$
\begin{aligned}
|\Psi_2\rangle=\frac12\big[&\alpha(|000\rangle+|100\rangle+|011\rangle+|111\rangle)\\
&+\beta(|010\rangle-|110\rangle+|001\rangle-|101\rangle)\big]\\
=\frac12\big[&|00\rangle(\alpha|0\rangle+\beta|1\rangle)
+|01\rangle(\beta|0\rangle+\alpha|1\rangle)\\
&+|10\rangle(\alpha|0\rangle-\beta|1\rangle)
+|11\rangle(-\beta|0\rangle+\alpha|1\rangle)\big]_{SA,B}.
\end{aligned}
$$
Measuring $S,A$ gives $(m,n)$ with <probability> $1/4$ for each outcome, independently of $\alpha,\beta$. Conditional on that result, the normalized three-<qubit> <post-measurement state> is
$$
|\Psi_{mn}\rangle=|mn\rangle_{SA}\otimes X^nZ^m|\psi\rangle_B.
$$
If the <Bell state analyzer> is instead regarded as an ideal direct <projective measurement> preserving the measured <qubits>, replace $|mn\rangle_{SA}$ here by the corresponding <Bell state>; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's <photons>.
Alice sends the two classical <bits> $(m,n)$ to Bob. During transmission the above conditional three-<qubit> state is unchanged. Bob then applies the inverse <Pauli operator> $Z^mX^n$:
$$
\begin{array}{c|c|c|c}
\text{Bell result}&(m,n)&\text{Bob's state before correction}&\text{correction}\\\hline
\phi^+&00&|\psi\rangle&I\\
\psi^+&01&X|\psi\rangle&X\\
\phi^-&10&Z|\psi\rangle&Z\\
\psi^-&11&XZ|\psi\rangle&ZX
\end{array}
$$
The final conditional three-<qubit> state is therefore $|mn\rangle_{SA}\otimes|\psi\rangle_B$. The protocol transmits \b[two classical bits per teleported qubit], consuming one shared <Bell pair>. Without Alice's record Bob averages over four <Pauli operators> and has <density matrix> $I/2$, so <quantum teleportation> respects the <no-communication theorem>. Alice's input has been measured and is no longer an independent copy, consistent with the <no-cloning theorem>.
Back to article page
