Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 10D (Quantum Information and Computation)

The no-signalling theorem states that any trace-preserving quantum operation performed locally by Bob leaves Alice's reduced density matrix unchanged when Bob's outcome is not communicated:Thus Bob's choice cannot change the statistics of any measurement made by Alice.

A unitary cloner with blank state $|0⟩$ would satisfyfor every $|\psi ⟩\in S$. For $|\psi ⟩,|\varphi ⟩\in S$, preservation of inner products givesHence every pair is either orthogonal or represents the same state. Conversely, an orthonormal set can be cloned by defining the map on its basis states and extending it to a unitary. Therefore the no-cloning theorem says that a set of pure states is unitarily clonable exactly when its distinct members are mutually orthogonal.

Let Alice and Bob share the Bell stateAlice may measure in the computational basis, remotely preparing Bob's ensemble $\{|0⟩,|1⟩\}$, or in the Hadamard basis, preparing $\{|+⟩,|-⟩\}$. Without her outcome, both ensembles have density matrix $I/2$, as no-signalling requires.

If Bob could clone all four states, the resulting two-qubit ensembles would beandThey are distinguishable: a $Z\otimes Z$ measurement is perfectly correlated for the first and only equally likely correlated or anticorrelated for the second. Bob could infer Alice's basis choice instantaneously, violating no-signalling. Thus no-signalling implies no-cloning for this set.