No-communication theorem

Remarks

By the Kraus representation, $\Phi(\tau) = \sum_k K_k \tau K_k^*$ with $\sum_k K_k^* K_k = I$. The partial trace over $A$ is unchanged by moving $K_k^*$ from the right to the left inside it, since $K_k$ acts on $A$ only. So the left side equals $\tr_A \big( (\sum_k K_k^* K_k \otimes I) \rho_{AB} \big) = \tr_A(\rho_{AB})$.

A measurement by Alice whose outcome Bob does not learn is such a channel, $\Phi(\tau) = \sum_j M_j \tau M_j^*$. For example, if $\rho_{AB}$ is the Bell state $(|00\rangle + |11\rangle)/\sqrt2$, Bob's reduced state is $\tfrac12 I$ before and after any operation of Alice.

The theorem concerns the average state. Conditioned on Alice's outcome $j$, Bob's state does change, but Bob can only use this after Alice tells him $j$ by a classical message.

Going deeper

For a student

The theorem shows that quantum theory, despite Bell nonlocality, does not allow signalling faster than light. It is the reason why teleportation and superdense coding need a classical or quantum channel in addition to shared entanglement. In the language of channels, it says that the partial trace over $A$ is invariant under every channel acting on $A$.