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Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with qubit 1 written first. The half-time gate requires a branch convention. Even stationary Hamiltonian evolution with $W(2\tau )=\mathrm{SWAP}$ does not uniquely specify $W(\tau )$: different accumulated eigenstate phases can agree at $2\tau$ and differ at $\tau$. In particular, both $S$ and $S^{†}$ below square to SWAP gate and leave $|00⟩,|11⟩$ unchanged. Adopt the conventional continuous isotropic exchange interaction branch, with a relative singlet phase increasing from $0$ to $\pi$ during the full pulse. This is the exchange square-root branch convention.

The symmetric and antisymmetric eigenstateshave SWAP gate eigenvalues $+1,-1$. Let $P_{-}=|{\psi }_{-}⟩⟨{\psi }_{-}|$ and $P_{+}=I_4-P_{-}$ be their antisymmetric and symmetric orthogonal projectors. The adopted exchange evolution isFor effective spin-one-half qubits, $H_{\mathrm{ex}}=\pi \hslash (X_1{X}_2+Y_1{Y}_2+Z_1{Z}_2-I_4)/(8\tau )$, an isotropic exchange interaction plus an energy offset that fixes the common phase. At half time choose $+1$ on the symmetric sector and $+i$ on the antisymmetric sector, the conventional principal square-root value of $-1$. Hence the square root of the SWAP gate isThus $S|00⟩=|00⟩$, $S|11⟩=|11⟩$, $S|01⟩=a|01⟩+b|10⟩$ and $S|10⟩=b|01⟩+a|10⟩$, where $a=(1+i)/2$ and $b=(1-i)/2$. The identities $a^2+b^2=0$ and $2ab=1$ verify $S^2=\mathrm{SWAP}$, and the projector form proves unitarity. The endpoint truth table alone would allow other matrix square roots; the stated exchange branch supplies the additional physical assumption.

For the network, fix the Pauli Z gate, Hadamard gate and S gate conventionsThe letter $P$ distinguishes this local S gate from the two-qubit exchange gate $S$. In particular $P^2=Z$; replacing $P$ by $P^{†}$ changes the network. Subscripts specify the acted-on qubit: $Z_1=Z\otimes I_2$, $P_2=I_2\otimes P$ and $H_2=I_2\otimes H$. Operators act rightmost first, so the printed quantum circuit is

First multiply the central two exchange pulses. Since $a^2-b^2=i$, their off-diagonal terms cancel:This reduces the entire network to $U=H_{2}D{H}_2$. Its action on all four computational basis states, retaining every relative quantum phase, isEquivalently,It flips the second qubit when the first is zero and contributes $-i$ when the first is one. That conditional phase cannot be discarded as a global phase.

To obtain the standard CNOT gate controlled by qubit 1, append an $X$ gate to qubit 2 and $P$ to qubit 1. They commute, andFor control zero, $X_2$ undoes the original flip. For control one, it supplies the desired flip and $P_1$ cancels the phase because $i(-i)=1$. This phase-corrected exchange network for a CNOT gate realizes the standard gate exactly with the declared conventions.

Finally, $S|01⟩=(e^{i\pi /4}|01⟩+e^{-i\pi /4}|10⟩)/\sqrt{2}$ has equal Schmidt coefficients and is maximally entangled: a half exchange creates quantum entanglement although a full SWAP gate does not. More constructively, the network above produces a CNOT gate from two half exchanges and local gates. Since CNOT gates and arbitrary single-qubit unitary operators form a universal quantum gate set, the square root of the SWAP gate and arbitrary single-qubit gates are universal. The alternative inverse exchange branch remains entangling and universal, but its phases and local correction must be recomputed consistently.

Use the computational basis in the stated order and let the rightmost operation act first. The controlled phase gate is $|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes Z$. From the rotation definition,The control-zero block remains the identity, and the control-one block becomes $X$. ThusThis proves the relation for arbitrary superpositions, rather than only for computational-basis probabilities.

For the second relation write $S=U_{\mathrm{SW}}^{1/2}$. Its central block is $\left(\begin{array}{cc}a& b\\ b& a\end{array}\right)$, with $a=(1+i)/2$, $b=(1-i)/2$. Because $a^2=i/2$, $b^2=-i/2$ and $ab=1/2$, direct multiplication givesMeanwhile $R_z^{(1)}(\pi /2)R_z^{(2)}(-\pi /2)=\mathrm{diag}(1,-i,i,1)$. Multiplying these diagonal matrices and the overall factor $i$ givesThe supplied $S$ is indeed a square root of the SWAP gate: $S^2$ exchanges $|01⟩$ and $|10⟩$ while fixing the other basis states. It acts as one on the triplet subspace and as $i$ on the singlet subspace.

For an isotropic exchange interaction with Hamiltonian $H=J(t){\mathbf{\sigma }}_1\cdot {\mathbf{\sigma }}_2/4$, define $\phi =\int J(t)dt/\hslash$. Triplet and singlet energies differ by $J$, soAn exchange pulse with $\phi =\pi /2$ therefore implements $e^{-i\pi /8}S$. Two such pulses, interspersed with the specified local rotations, implement the controlled phase gate; the initial and final target rotations then turn it into a CNOT gate. Their common overall phase has no effect on an isolated gate's outcome probabilities; a subsequently controlled version must keep track of conditional phases.

The construction converts a physically available two-spin interaction into a standard entangling gate, which together with arbitrary one-qubit rotations supplies universal quantum computation. Practical operation needs calibrated exchange pulse area and timing, individually controllable rotations, and reproducible phase references. Charge noise can change $J$; magnetic-field gradients add unwanted single-spin evolution, and anisotropic exchange modifies the assumed Hamiltonian. Pulses must suppress leakage into unwanted orbital states while finishing before excessive decoherence. Readout, preparation, cross-talk and the extra operations in this decomposition all affect the achieved quantum gate fidelity. The algebraic identity alone does not establish a fault-tolerant quantum computation implementation.

With $S=P_{+}+i{P}_{-}$ the conventional square root of the SWAP gate, $P=\mathrm{diag}(1,i)$ the S gate, and operators acting rightmost first, $S{Z}_{1}S=\mathrm{diag}(1,i,-i,-1)$. Therefore $H_{2}S{Z}_{1}S{P}_2{H}_2=|0⟩⟨0|\otimes X-i|1⟩⟨1|\otimes I$. Appending $X_2$ changes the zero-controlled flip into a one-controlled flip, and $P_1$ removes the control-dependent phase. This proves the displayed exact CNOT gate identity. Both the exchange branch and the phase gate square-root convention matter; changing either requires recalculating the correction.