= Solution
A <measurement in quantum mechanics> with outcomes $m$ is specified, in the single-operator-per-outcome description, by <linear operators> $M_m$ satisfying $\sum_mM_m^\dagger M_m=I$. Its effects $F_m=M_m^\dagger M_m$ form a <positive operator-valued measure>. The two measurement postulates are the <Born rule> for outcomes and the conditional state update:
$$
\boxed{p_m=\langle\psi|M_m^\dagger M_m|\psi\rangle,\qquad
|\psi\rangle\longmapsto\frac{M_m|\psi\rangle}{\sqrt{p_m}}\quad(p_m>0).}
$$
For a <density operator> the corresponding formulas are $p_m=\operatorname{Tr}(M_m\rho M_m^\dagger)$ and $\rho_m=M_m\rho M_m^\dagger/p_m$. More general <quantum instruments> have multiple operators for each outcome and sum these terms. For a <projective measurement> of an <observable> $A=\sum_m a_mP_m$, 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> $|\Phi^+\rangle_{AB}=(|00\rangle+|11\rangle)/\sqrt2$. Alice also holds the input $|\psi\rangle_C=\alpha|0\rangle+\beta|1\rangle$, 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 $|\psi\rangle_C|\Phi^+\rangle_{AB}$, direct expansion gives
$$
\frac12\left(
|00\rangle_{CA}|\psi\rangle_B+
|01\rangle_{CA}X|\psi\rangle_B+
|10\rangle_{CA}Z|\psi\rangle_B+
|11\rangle_{CA}XZ|\psi\rangle_B
\right).
$$
For example, the four receiver vectors are $\alpha|0\rangle+\beta|1\rangle$, $\beta|0\rangle+\alpha|1\rangle$, $\alpha|0\rangle-\beta|1\rangle$, and $-\beta|0\rangle+\alpha|1\rangle$. 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 $X^bZ^a|\psi\rangle$. Alice sends the two outcome bits. Bob applies the inverse, $Z^aX^b$, and hence
$$
\boxed{Z^aX^bX^bZ^a|\psi\rangle=|\psi\rangle.}
$$
The 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 $|\psi\rangle\mapsto X^bZ^a|\psi\rangle/2$, 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 $\tfrac14\sum_{a,b}X^bZ^a\rho Z^aX^b=I/2$, 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>.
Back to article page
