Consider a [separable pair](Separable%20states.md) of [pure states](Pure%20state.md) $|\psi_1\rangle$ and $|\psi_2\rangle$ that we want to attempt to [clone](No-cloning%20theorem.md) onto a receiving initial state, $|\phi\rangle$ using some hypothetical [unitary operator](Unitary%20transformations%20in%20quantum%20mechanics.md#Unitary%20operators%20in%20quantum%20mechanics) $\hat{U},$ such that $\hat{U}(|\psi_1\rangle\otimes|\phi\rangle)=|\psi_1\rangle\otimes|\psi_1\rangle$ and $\hat{U}(|\psi_2\rangle\otimes|\phi\rangle)=|\psi_2\rangle\otimes|\psi_2\rangle.$ ^56fd11 Taking the inner product of the [above equations](Proof%20of%20the%20no-cloning%20theorem#^56fd11), we find that $(\langle\phi|\otimes\langle\psi_1|)\hat{U}\hat{U}^\dagger(|\psi_2\rangle\otimes|\phi\rangle) = \langle\psi_1|\otimes\langle\psi_1|\psi_2\rangle\otimes|\psi_2\rangle$$\langle\psi_1|\psi_2\rangle = (\langle\psi_1|\psi_2\rangle)^2$ ^3a81f9 The equality, $\langle\psi_1|\psi_2\rangle = (\langle\psi_1|\psi_2\rangle)^2$ would imply either that $|\psi_1\rangle=|\psi_2\rangle$ or that $|\psi_1\rangle$ and $|\psi_2\rangle$ are [orthogonal.](state%20vector%20normalization.md) Therefore, a cloning operator $\hat{U}$ would not be universally applicable. Rather, if we wanted to clone some state $|\psi\rangle$, this would require us to know which other states it is orthogonal to and thus $|\psi\rangle$ can't be an unknown state.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^50a44c #QuantumMechanics/QuantumInformation #QuantumMechanics/FoundationsOfQuantumMechanics #QuantumMechanics/MathematicalFoundations/QuantumInformation #Proofs