 Solution

A universal quantum gate set can implement any unitary operator on any finite number of qubits, exactly when arbitrary continuous gate parameters are allowed, or to arbitrary accuracy with a suitable finite gate alphabet. More precisely, for every target $V$ and $ϵ>0$ there is a finite quantum circuit $W$ with ${\min }_{\gamma }\Vert W-e^{i\gamma }V\Vert <ϵ$ in operator norm. This ignores only an overall global phase of the complete operation. Universality lets a fixed collection of controllable elementary operations run arbitrary quantum algorithms rather than requiring a different physical interaction for each algorithm.

All single-qubit unitary gates, together with the controlled-NOT gate between selectable pairs, form an exact universal quantum gate set. One way to see sufficiency is to eliminate entries of a finite unitary matrix by two-level rotations between computational basis vectors; these rotations decompose into controlled single-qubit rotations, which in turn decompose into single-qubit gates and controlled-NOT gates. Single-qubit gates alone preserve product states and cannot supply general quantum entanglement. A finite approximate universal example is the Hadamard gate, the T gate $T=\mathrm{diag}(1,e^{i\pi /4})$, and the controlled-NOT gate; finite exact words over a finite alphabet cannot represent every continuously parameterized unitary operator exactly.

Distinguish the Hamiltonian operator $H_{\mathrm{int}}$ from the Hadamard gate $H_{\mathrm{Had}}$. For the printed collisional phase gate,As written, $\Omega$ has units of energy. The Schrödinger equation givesbecause only $|01⟩$ has a nonzero energy eigenvalue. In the ordered computational basis $(00,01,10,11)$, defineIf an interaction strength is instead quoted as an angular frequency ${\Omega }_{\mathrm{f}}$, the Hamiltonian operator is $\hslash {\Omega }_{\mathrm{f}}|01⟩⟨01|$ and $\varphi ={\Omega }_{\mathrm{f}}\tau$. An alternative definition with $e^{+i\varphi }$ would have $\varphi =-\Omega \tau /\hslash$. The minus sign above follows the usual time-evolution operator convention. At $\varphi =\pi$ these conventions give the same $U_{\pi }$.

For the quantum circuit, use $X$ on the first qubit to move the selective minus sign from $01$ to $11$:Conjugating the target of this Controlled-Z gate by the Hadamard gate gives the required controlled-NOT gate:The rightmost matrix acts first. Thus the network applies $X_1,H_2$, then $U_{\pi }$, then $X_1,H_2$, and maps $|a,b⟩$ to $|a,b\oplus a⟩$, where $\oplus$ is exclusive or.