no cloning theorem

The no-cloning theorem states that:

An unknown quantum system cannot be cloned by unitary transformations

Proof: Suppose there is a unitary transformation $\mathcal{U}$ that can clone a quantum system i.e. for any state $|\psi ⟩$$$\mathcal{U}(|\psi 0\rangle) = |\psi \psi \rangle$$Suppose $|\varphi ⟩$ is linearly independent to $|\psi ⟩$$$\begin{array}{ll}\mathcal{U}(|\psi 0\rangle) &= | \psi \psi \rangle \ \mathcal{U}(| \phi 0 \rangle) &= |\phi \phi \rangle \end{array}$$Consider the action of $\mathcal{U}$ on the composite system $|\gamma ⟩=\frac{1}{\sqrt{2}}(|\psi ⟩+|\varphi ⟩)$$$\mathcal{U}(|\gamma 0\rangle) = \frac{1}{\sqrt2}\mathcal{U}(|\psi 0\rangle + | \phi0 \rangle) = \frac{1}{\sqrt2}(|\psi \psi \rangle + |\phi\phi\rangle)$$But since $\mathcal{U}$ is a cloning transformation we also have$$\mathcal{U}(|\gamma0\rangle) = |\gamma \gamma \rangle = \frac{1}{2}(|\psi\psi\rangle + |\psi\phi\rangle + |\phi\psi\rangle + |\phi\phi\rangle)$$
Which is a contradiction to the previous result. Therefore no such unitary transformation exists.

Copying bit strings as an exception

Classical bit strings such as 10 or 101 can be encoded into quantum states using the computational basis. For example:$$|101\rangle = |1\rangle \otimes |0\rangle \otimes |1\rangle$$This is a product of known states (not entangled or superposition)

Since the standard states are orthogonal and known, we can indeed "copy" quantum encoded bit strings.

For example: to copy a single quantum state we can use the CNOT gate with the target bit set to $|0⟩$. Since the CNOT gate performs an XOR computation, for an arbitrary state $|\varphi ⟩$ $\text{CNOT}(|0⟩\otimes |\varphi ⟩)=|\varphi ⟩\otimes |\varphi ⟩$.

This does not violate the no-cloning theorem because:

• the states are known and orthogonal
• there is no quantum superposition or entanglement involved
• the operation is unitary and reversible