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The controlled-NOT gate is its own inverse, so ${\mathrm{CNOT}}^2=I$. Every remaining nontrivial gate acts on one wire. By the tensor product rule, any circuit on a product input stays a product state; in particular, starting at $|00⟩$ it cannot produce a Bell state.

This also prevents arbitrary approximation, not merely exact preparation. For normalized single-qubit vectors $|u⟩,|v⟩$,by the Cauchy-Schwarz inequality. Hence the quantum fidelity with $|{\Phi }^{+}⟩$ is bounded away from one for every reachable state. This gate set is not universal. The obstruction is preservation of product states, a special case of the existing product-state obstruction to SWAP universality.

The gate set is not universal. The displayed two-qubit matrix is the SWAP gate:Both $V$ and the Hadamard gate act on one qubit. Thus every gate preserves a fully factored product state, and induction gives the same property for every quantum circuit built from them. This is the product-state obstruction to SWAP universality.

For a quantitative obstruction to even approximate universality, every two-qubit product state has squared overlap at most $1/2$ with a Bell state. Indeed, for normalized coefficient vectors $u,v$,by the Cauchy-Schwarz inequality. A controlled-NOT gate acting on $|+⟩|0⟩$ would instead produce that Bell state, so these circuits cannot approximate all unitary transformations. The fact that $V^2=e^{-i\pi /8}T$ supplies a T gate up to global phase does not supply an entangling gate.

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.