Consider the following protocol:

  • Alice and Bob share the state \begin{equation} |\Phi^+\rangle=\frac{1}{\sqrt{2}}(|0\rangle|0\rangle \pm |1\rangle|1\rangle) \end{equation}
  • Alice has to teleport to Bob the state (which can be unknown even to her) \begin{equation} |\psi\rangle = c_0|0\rangle + c_1|1\rangle \end{equation} So she appends this state to her part of the system in this way: \begin{equation} |\psi\rangle\otimes|\Phi^+\rangle= \frac{1}{\sqrt{2}}(c_0|00\rangle_A|0\rangle_B+c_0|01\rangle_A|1\rangle_B+c_1|10\rangle_A|0\rangle_B+c_1|11\rangle_A|1\rangle_B)=\end{equation} \begin{equation} = \frac{1}{2}(|\Phi^+\rangle_A(c_0|0\rangle + c_1|1\rangle)+|\Phi^-\rangle_A(c_0|0\rangle - c_1|1\rangle)+|\Psi^+\rangle_A(c_1|0\rangle + c_0|1\rangle)+|\Psi^-\rangle_A(c_1|0\rangle - c_0|1\rangle))\end{equation} (where $|\Psi^\pm\rangle$ and $|\Phi^\pm\rangle$ are defined in the next stage of the protocol)
  • Alice performs a Bell measurement, with projectors obtained from the Bell basis \begin{equation} |\Phi^\pm\rangle = \frac{1}{\sqrt{2}}(|0\rangle|0\rangle \pm |1\rangle|1\rangle) \end{equation} \begin{equation} |\Psi^\pm\rangle = \frac{1}{\sqrt{2}}(|0\rangle|1\rangle \pm |1\rangle|0\rangle) \end{equation}
  • Alice sends two bits to Bob in order to communicate the the result of her measurement (e.g. 00 for $|\Phi^+\rangle$ and so on) -Now Bob applies a Pauli transformation on his part of the system depending on the result of the measurement of Alice and recovers the original state $|\psi\rangle$.

Here comes the question: why this protocol DOES NOT violate the no-cloning theorem?


1 Answer 1


You could think that quantum teleportation violates the no-cloning theorem, but it is actually not the case.

Indeed, the original state $|\psi\rangle$ has not been duplicated, since after the teleportation process only the target qubit is left in the state $|\psi\rangle$, while the original qubit ends up in one of the computational basis state, namely $|0\rangle$ or $|1\rangle$, depending on the measurement result.

  • $\begingroup$ So, the no-cloning theorem scenario implies that both the target and the register cannot be left at the same time in the state $|\psi\rangle$? $\endgroup$
    – Hub One
    Commented Jul 22, 2020 at 17:22
  • $\begingroup$ Yes. If it was possible, the state $|\psi\rangle$ would indeed be duplicated, violating the no-cloning theorem. $\endgroup$
    – A. Bordg
    Commented Jul 22, 2020 at 18:23

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