# How does the quantum teleportation protocol not violate the no-cloning theorem?

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?

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.
• So, the no-cloning theorem scenario implies that both the target and the register cannot be left at the same time in the state $|\psi\rangle$? Jul 22 '20 at 17:22
• Yes. If it was possible, the state $|\psi\rangle$ would indeed be duplicated, violating the no-cloning theorem. Jul 22 '20 at 18:23