No-cloning Theorem - Proof

Proof

Suppose the state of a quantum system A, which we wish to copy, is (see bra-ket notation). In order to make a copy, we take a system B with the same state space and initial state . The initial, or blank, state must be independent of, of which we have no prior knowledge. The composite system is then described by the tensor product, and its state is

There are only two ways to manipulate the composite system. We could perform an observation, which irreversibly collapses the system into some eigenstate of the observable, corrupting the information contained in the qubit. This is obviously not what we want. Alternatively, we could control the Hamiltonian of the system, and thus the time evolution operator U (for a time independent Hamiltonian, where is called the generator of translations in time) up to some fixed time interval, which yields a unitary operator. Then U acts as a copier provided that

for all possible states in the state space (including ). Since U is unitary, it preserves the inner product:

\langle e|_B \langle \phi|_A |\psi\rangle_A |e\rangle_B = \langle e|_B \langle \phi|_A U^{\dagger} U |\psi\rangle_A |e\rangle_B = \langle \phi|_B \langle \phi|_A |\psi\rangle_A |\psi\rangle_B,

and since quantum mechanical states are assumed to be normalized, it follows that

This implies that either (in which case ) or is orthogonal to (in which case ). However, this is not the case for two arbitrary states. While orthogonal states in a specifically chosen basis, for example:

and

fit the requirement that, this result does not hold for more general quantum states. Apparently U cannot clone a general quantum state. This proves the no-cloning theorem.

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