No-cloning theorem

Remarks

Suppose such a $U$ exists, and let $\psi$, $\varphi$ be unit vectors. Since $U$ preserves inner products, $\langle \varphi, \psi \rangle \langle e, e\rangle = \langle \varphi, \psi \rangle^2$. So $\langle \varphi, \psi \rangle$ is $0$ or $1$ for all unit vectors, which fails for $\psi = |0\rangle$ and $\varphi = (|0\rangle + |1\rangle)/\sqrt2$, whose inner product is $1/\sqrt2$.

Allowing global phases, $U(\psi \otimes e) = e^{i\alpha(\psi)} \psi \otimes \psi$, does not help: the same argument gives $\abs{\langle\varphi,\psi\rangle} \in \set{0, 1}$. The theorem also holds for quantum channels: no completely positive trace-preserving map sends every pure state $|\psi\rangle\langle\psi|$ to $|\psi\rangle\langle\psi| \otimes |\psi\rangle\langle\psi|$.

The requirement "for every $\psi$" is essential. Any single orthonormal basis can be cloned: on qubits, the controlled-NOT gate maps $|0\rangle|0\rangle$ to $|0\rangle|0\rangle$ and $|1\rangle|0\rangle$ to $|1\rangle|1\rangle$.

Going deeper

For a student

No-cloning separates quantum from classical information. It is the basis of the security of quantum key distribution, since an eavesdropper cannot copy the transmitted qubits and keep a copy. It is consistent with teleportation, where the original state is destroyed at the sender's side. It also explains why quantum error correction must protect information without copying it, by spreading it over entangled qubits.