$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

No-Cloning Theorem

^{First created in June 2018}

The No-Cloning Theorem states that a single universal unitary transformation cannot clone a general quantum state.

Theorem: There is no unitary operator $U$ on $H\otimes H$ such that for all normalised states $\Ket{\phi}_A, \Ket{e}_B\in H,$\ $U(\Ket{\phi}_A\Ket{e}_B)=e^{i\alpha(\phi e)}\Ket{\phi}_A\Ket{\phi}_B$ for some real number $\alpha$ depending on $\phi$ and $e$.

Proof

Let us try to prove without the phase factor and show that $U(\Ket{\phi}_A\Ket{e}_B)=\Ket{\phi}_A\Ket{\phi}_B$ is not possible.

In fact, the linearity of the unitary transformation implies no-cloning theorem.

Consider $U(\Ket{\alpha}_A\Ket{e}_B)=\Ket{\alpha}_A\Ket{\alpha}_B$ and $U(\Ket{\beta}_A\Ket{e}_B)=\Ket{\beta}_A\Ket{\beta}_B$.

Let $\phi=c(\Ket{\alpha}_A+\Ket{\beta}_A)$, where $c$ is a trivial normalisation factor, unitary transformation would arrive at inconsistent result: \vspace{-0.3cm} \begin{enumerate}

\item $U(\Ket{\phi}_A\Ket{e}_B) =U(c(\Ket{\alpha}_A\Ket{e}_B+\Ket{\beta}_A\Ket{e}_B)) =c(U(\Ket{\alpha}_A\Ket{e}_B)+U(\Ket{\beta}_A\Ket{e}_B)) =c(\Ket{\alpha}_A\Ket{\alpha}_B+\Ket{\beta}_A\Ket{\beta}_B) $.

\item $U(\Ket{\phi}_A\Ket{e}_B) =\Ket{\phi}_A\Ket{\phi}_B =c(\Ket{\alpha}_A+\Ket{\beta}_A)\otimes c(\Ket{\alpha}_B+\Ket{\beta}_B) =c^2(\Ket{\alpha}_A\Ket{\alpha}_B+\Ket{\alpha}_A\Ket{\beta}_B+\Ket{\beta}_A\Ket{\alpha}_B+\Ket{\beta}_A\Ket{\beta}_B) $.

\end{enumerate}

What is cloning

The opening statement states that a single universal unitary transformation cannot clone a general quantum state. By "single" it means a unitary transformation independent of the state being cloned. By "general" it means an unknown state.

Cloning is to make an arbitrary qubit of unknown state and produce the same state on another qubit without affecting the first one, and the two can evolve independently afterward.

Reconstruction of a known state is not cloning

We can of course initialise a qubit into a basis state, like $\Ket0$. In fact, after each measurement, the qubit is in a known state. It is not hard to repeat the same measurement on another arbitrary qubit to obtain the same known state after sufficient number of trials.

This is not cloning, but simply construct the same known state twice.

Quantum teleportation is not cloning

On the other hand, quantum teleportation allows measuring an arbitrary qubit together with one of the two pre-entangled qubits, the result is a 2-bit classical information, which can ben sent to the recipient to reconstruct the original qubit state from the other pre-entangled qubit.

This is not cloning as the original qubit has been destroyed after the measurement at the sender's end.

Entanglement is not cloning

As in the theorem, the result of the cloning is $\Ket{\phi}_A\Ket{\phi}_B$, which is in state space $A\otimes B$. Subsequent manipulation of $\Ket{\phi}_A$ for example does not affect $\Ket{\phi}_B$.

On the other hand, entangled cannot be factorised into two independent spaces. The two qubits are in the same space of twice the number of dimensions. The evolution of one will affect the other instantaneously.