 Solution

Let $|\psi ⟩=a|0⟩+b|1⟩$ and $|\varphi ⟩=c|0⟩+d|1⟩$ be normalized qubit states. By linearity, the SWAP gate givesThe output remains a product state. Single-qubit unitary operators also preserve product states, so an induction over any quantum circuit made from these operations shows that it cannot create quantum entanglement from a product state. Equivalently, $\mathrm{SWAP}(A\otimes B)=(B\otimes A)\mathrm{SWAP}$ moves every exchange through the local operations, leaving only local gates and a permutation of wires. This is the product-state obstruction to SWAP universality.

Consequently SWAP gates and arbitrary single-qubit gates do not form a universal quantum gate set. For example, they cannot map $|00⟩$ to the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, whereas a Hadamard gate followed by a CNOT gate does so.