No-Cloning Theorem

There exists no unitary operator UU that can clone a Qubit. Note that ⊗\otimes is the Tensor Product.

C(∣ψ⟩)=∣ψ⟩⊗∣ψ⟩C(\left\lvert \psi \right\rangle)=\left\lvert \psi \right\rangle\otimes\left\lvert \psi \right\rangle

Proof

Let

C∣0⟩→∣0⟩∣0⟩C\left\lvert 0 \right\rangle\rightarrow\left\lvert 0 \right\rangle\left\lvert 0 \right\rangle C∣1⟩→∣1⟩∣1⟩C\left\lvert 1 \right\rangle\rightarrow\left\lvert 1 \right\rangle\left\lvert 1 \right\rangle

Then

C∣+⟩→=C∣0⟩+∣1⟩2C\left\lvert + \right\rangle\rightarrow=C\frac{\left\lvert 0 \right\rangle+\left\lvert 1 \right\rangle}{2}\\ ≠12(∣0⟩∣0⟩+∣1⟩∣1⟩)\neq\frac{1}{\sqrt{2}}(\left\lvert 0 \right\rangle\left\lvert 0 \right\rangle+\left\lvert 1 \right\rangle\left\lvert 1 \right\rangle)\\ ≠∣+⟩∣+⟩\neq \left\lvert + \right\rangle\left\lvert + \right\rangle

This breaks the linearity of operators.