No cloning theorem

Given two arbitrary states $\ket\psi_A, \ket \chi_B$ in a space $V$ , there is no unitary operator $U \in \mathcal L(V \otimes V)$ such that

\begin{align*} U \ket \psi_A \otimes \ket \chi_B = e^{i\theta} \ket \psi_A \otimes \ket \psi_B. \end{align*}

For any $\theta$ .

Proof follows by contradiction. Let $\ket \phi$ be some arbitrary state and assume such an operator $U$ exists.

\begin{align*} \braket{\psi|\phi}_A \braket{\chi|\chi}_B &= \bra{\psi}_A \otimes \bra{\chi}_B \, \ket \phi_A \otimes \ket \chi_B \\ &= \bra{\psi}_A \otimes \bra{\chi}_B U \adj U \ket \phi_A \otimes \ket \chi_B \\ &= e^{i\theta_1} e^{i\theta_2} \bra{\psi}_A \bra{\psi}_B \, \ket{\phi}_A \ket {\phi}_B \\ &= e^{i\theta} \braket{\psi|\phi}^2. \end{align*}

The states are normalized, so $\braket{\chi|\chi} = 1$ .

\begin{align*} \braket{\psi|\phi} &= e^{i\theta} \braket{\psi|\phi}^2 \\ | \braket{\psi|\phi} | &= | \braket{\psi|\phi} | ^2 \\ \braket{\psi|\phi} &= \{ 0, 1 \} \\ \ket{\phi} &= \{ e^{i\theta} \ket{\psi}, \ket 0 \}. \end{align*}

We reach the conclusion that $\ket\phi$ is not arbitrary, which contradicts our assumption that $U$ exists.

You cannot clone a quantum state.