SWAP Gate

SWAP is two things at once in a working quantum computer: a compiler abstraction and a physical interaction. As an abstraction it is the currency of qubit routing. Real processors — IBM's heavy-hex lattices, Google's grid of transmons, neutral-atom arrays — couple only physically adjacent qubits, so an algorithm that wants a gate between distant qubits must first SWAP one of them into place, run the gate, and often SWAP it back. Since each SWAP costs three CNOTs, and CNOTs are the noisiest, slowest operations on the device, the number and placement of inserted SWAPs dominate both circuit depth and error budget; deciding them optimally (the "qubit mapping and routing" problem) is NP-hard, and every serious transpiler — Qiskit, t$\vert$ket$\rangle$, and the rest — spends most of its effort here. As a physical interaction, SWAP and its square root are the native language of exchange-coupled hardware: in silicon and gallium-arsenide quantum dots following the Loss–DiVincenzo proposal, pulsing the tunnel barrier between two spins turns on the Heisenberg coupling for a controlled time, and a pulse area of a quarter-period realises $\sqrt{\mathrm{SWAP}}$ directly, with no CNOT compilation at all. The SWAP test lives on as a primitive for fidelity estimation, quantum kernel evaluation in machine learning, and entanglement witnesses, while its fermionic cousin, the fSWAP, is a standard building block for simulating molecules and materials under the Jordan–Wigner mapping. The open frontiers are practical and sharp: routing-aware compilation that minimises SWAP overhead on ever-larger connectivity graphs, and exchange-only qubit encodings that aim to run entire algorithms using nothing but the physics of Heisenberg's 1926 interaction.

Picture two labelled wires carrying quantum states. The SWAP gate cuts and crosses them: whatever entered on wire 1 leaves on wire 2, and whatever entered on wire 2 leaves on wire 1. For a product input this is pure relabelling — no information is created or destroyed and no entanglement is generated — but for an entangled input the gate faithfully carries the correlations along with the swapped factors. The standard circuit symbol is two crosses joined by a vertical line, and the cleanest way to build it from the universal primitive is the reversible version of the classical trick that swaps two registers with no scratch space: XOR the first into the second, the second into the first, and the first into the second again. Three CNOTs, alternating direction, thread the bit-values through one another and land on the exchange.

Spectrally, SWAP is the exchange operator borrowed straight from many-body physics. It must act as $+1$ on any state symmetric under interchange and $-1$ on the antisymmetric singlet — that is what "swap the labels" means — so its eigenvalues are forced to be $\{+1,+1,+1,-1\}$, and it is exactly $\tfrac12(I+\vec\sigma_1\cdot\vec\sigma_2)$, the isotropic Heisenberg coupling. That connection is where the deepest surprise lives. Because $\mathrm{SWAP}^2=I$, the gate is an involution, and its square root is well defined — a "half exchange." A full exchange never entangles, but the half exchange $\sqrt{\mathrm{SWAP}}$ turns a product state into a maximally entangled one. The continuous family $e^{-i\theta\,\mathrm{SWAP}}$ interpolates smoothly between them, and its entangling power rises from nothing at $\theta=0$, to a maximum at $\theta=\pi/4$, and back to nothing at $\theta=\pi/2$ — the second figure makes this non-monotonic story quantitative.

Everything begins with the explicit matrix in the computational basis, ordering rows and columns as $|00\rangle,|01\rangle,|10\rangle,|11\rangle$. Reading the entries directly shows it is a permutation that leaves $|00\rangle,|11\rangle$ untouched and interchanges the two "one excitation" states.

where: rows and columns run over the ordered basis $|00\rangle,|01\rangle,|10\rangle,|11\rangle$; the off-diagonal $1$'s move the $|01\rangle$ and $|10\rangle$ amplitudes onto each other. The reading: the matrix is real and symmetric, so $\mathrm{SWAP}^\dagger=\mathrm{SWAP}$; its columns are orthonormal, so it is unitary; and it is a permutation, so applying it twice returns the identity. The permutation is the matrix expression of one plain sentence — the gate exchanges the two qubits — which extends by linearity from basis states to every superposition and entangled input.

where: $a,b$ — the computational-basis values of the two qubits. The reading: on a product input $|\psi\rangle\otimes|\phi\rangle$ the output is $|\phi\rangle\otimes|\psi\rangle$, still a product, so no entanglement appears; on an entangled input the correlations ride along with the swapped factors. The workhorse identity of hardware compilation writes SWAP as three CNOTs in alternating direction — the quantum realisation of the classical XOR-swap trick.

where: $\mathrm{CNOT}_{i\to j}$ — a CNOT with control qubit $i$ and target qubit $j$. The reading: the alternation of direction is essential — two CNOTs the same way do not give SWAP — and three is provably the minimum CNOT count for the exchange. Decomposed over the two-qubit Pauli basis, SWAP is the sum of the four "matched" tensor products, each entering with weight one half. This is its algebraic centre of gravity.

where: $X,Y,Z$ — the single-qubit Pauli operators; $I$ — the single-qubit identity. The reading: the coefficients are fixed by orthogonality of the Pauli basis, and the appearance of $XX+YY+ZZ$ — with no $ZZ$ singled out — is what makes this the isotropic exchange. The same identity, rewritten with the Pauli vector $\vec\sigma=(X,Y,Z)$, exposes SWAP as the Heisenberg exchange operator of two spins — the bridge to condensed-matter physics and to the portrait below.

where: $\vec\sigma_1\cdot\vec\sigma_2$ — the dot product of the two spin operators, with eigenvalue $+1$ on the triplet and $-3$ on the singlet. The reading: this is Dirac's spin-exchange identity; the Heisenberg Hamiltonian $H=-J\,\vec\sigma_1\cdot\vec\sigma_2$ is, up to a constant, $-2J\,\mathrm{SWAP}$, so its time evolution is the partial swap. Two structural facts follow at once and pin down the gate's whole character; both can be read straight off equation (1).

where: $\dagger$ — conjugate transpose; $I$ — the $4\times4$ identity. The reading: Hermiticity makes SWAP a legitimate observable with eigenvalues $\pm1$; combined with unitarity it forces the involution $\mathrm{SWAP}^2=I$, so the gate is its own inverse — there is no separate "un-SWAP." The spectrum is not something to compute so much as to recognise, because the symmetric and antisymmetric subspaces are eigenspaces by construction.

where: the $+1$ eigenvalue (multiplicity 3) belongs to the symmetric triplet subspace; the $-1$ eigenvalue (multiplicity 1) belongs to the antisymmetric singlet. The reading: $\operatorname{Tr}=3(+1)+1(-1)=2$ and $\det=(+1)^3(-1)=-1$, both matching the explicit matrix. Finally, because $\mathrm{SWAP}^2=I$, the exponential closes into a two-term formula, generating the continuous family of partial swaps that contains $\sqrt{\mathrm{SWAP}}$.

where: $\theta$ — the exchange angle; $\theta=\pi/2$ gives $-i\,\mathrm{SWAP}$, and $\theta=\pi/4$ gives $\sqrt{\mathrm{SWAP}}$ up to a global phase. The reading: the full swap sits at $\theta=\pi/2$ and the identity at $\theta=0$; halfway between, at $\theta=\pi/4$, the gate is maximally entangling.

Special inputs and parameter limits make SWAP's split personality — inert relabeller here, maximal entangler a quarter-turn away — concrete.

1 · The half swap builds a Bell pair while the full swap builds nothing (fully worked)

Apply $\sqrt{\mathrm{SWAP}}$ to the product state $|01\rangle$ and quantify the entanglement, then contrast with the full SWAP.

Step 1. Act with the explicit matrix from equation (8). Only the middle block is nontrivial, so $\sqrt{\mathrm{SWAP}}\,|01\rangle=\tfrac12\big[(1+i)|01\rangle+(1-i)|10\rangle\big]$. Check the norm: $|\tfrac{1+i}{2}|^2+|\tfrac{1-i}{2}|^2=\tfrac12+\tfrac12=1$, a valid state.

Step 2. Read the amplitudes into the concurrence $C=2|a_{00}a_{11}-a_{01}a_{10}|$: here $a_{00}=a_{11}=0$, $a_{01}=\tfrac{1+i}{2}$, $a_{10}=\tfrac{1-i}{2}$, so $a_{01}a_{10}=\tfrac{(1+i)(1-i)}{4}=\tfrac{2}{4}=\tfrac12$. Then $C=2\big|0-\tfrac12\big|=\mathbf{1}$. The output is maximally entangled — locally equivalent to a Bell state.

Step 3. Contrast the full SWAP: $\mathrm{SWAP}\,|01\rangle=|10\rangle$, a product state, with $a_{10}=1$ and every other amplitude zero, so $C=2|0-0|=\mathbf{0}$. Same single-excitation input, same subspace, opposite verdict: the full exchange relabels, the half exchange entangles.

Step 4. Sanity-check against the family formula. Writing $\sqrt{\mathrm{SWAP}}$ as $e^{-i\pi/4\,\mathrm{SWAP}}$ up to phase, the concurrence should be $|\sin(2\cdot\tfrac{\pi}{4})|=|\sin\tfrac{\pi}{2}|=1$, and the full swap at $\theta=\pi/2$ gives $|\sin\pi|=0$. Both match Derivation 3 and the second figure exactly.

2 · Estimating state overlap with the SWAP test (fully worked)

A controlled-SWAP between two registers $|\psi\rangle,|\phi\rangle$, sandwiched between Hadamards on an ancilla, outputs the ancilla in $|0\rangle$ with probability $P_0=\tfrac12+\tfrac12|\langle\psi|\phi\rangle|^2$. Evaluate it for identical, orthogonal, and half-overlapping states.

Step 1. Identical states, $|\psi\rangle=|\phi\rangle$: the overlap is $|\langle\psi|\psi\rangle|^2=1$, so $P_0=\tfrac12+\tfrac12(1)=\mathbf{1}$. The ancilla never reads $1$ — a certain "same" outcome.

Step 2. Orthogonal states, e.g. $|\psi\rangle=|0\rangle$, $|\phi\rangle=|1\rangle$: the overlap is $|\langle0|1\rangle|^2=0$, so $P_0=\tfrac12+\tfrac12(0)=\mathbf{0.5}$. The ancilla is a fair coin — the maximally "different" signal.

Step 3. Half-overlapping states, $|\psi\rangle=|0\rangle$, $|\phi\rangle=|+\rangle=\tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)$: the overlap is $|\langle0|+\rangle|^2=|\tfrac{1}{\sqrt2}|^2=\tfrac12$, so $P_0=\tfrac12+\tfrac12(\tfrac12)=\tfrac34=\mathbf{0.75}$.

Step 4. Invert to read the physics: measuring $P_0$ over many shots gives $|\langle\psi|\phi\rangle|^2=2P_0-1$, ranging from $1$ (identical) through $0$ (orthogonal). The SWAP test turns a swap's $\pm1$ eigenvalue structure into a direct, destructive-test-free estimate of fidelity — the reason it anchors quantum kernel methods and entanglement diagnostics.

SWAP is not named for a person, but its deepest identity is: the operator $\tfrac12(I+\vec\sigma_1\cdot\vec\sigma_2)$ is the isotropic Heisenberg exchange, the interaction Werner Heisenberg introduced in 1926 to explain why identical particles feel a force with no classical analogue. The gate a compiler inserts to route a qubit is, physically, a full turn of that same exchange clock.

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §1.3.4 (circuit basics) and §4.3 (controlled operations and the three-CNOT SWAP) · Preskill, Lecture Notes for Physics 219, Ch. 5 · Watrous, The Theory of Quantum Information, Ch. 2 · Wilde, Quantum Information Theory, Ch. 3 (the SWAP test) · Kitaev, Shen & Vyalyi, Classical and Quantum Computation, §8 · Mermin, Quantum Computer Science, Ch. 2. The key results in this article are: (i) the three-CNOT product $\mathrm{CNOT}_{1\to2}\,\mathrm{CNOT}_{2\to1}\,\mathrm{CNOT}_{1\to2}=\mathrm{SWAP}$, confirmed by direct $4\times4$ multiplication; (ii) the Pauli and Dirac forms $\tfrac12(II+XX+YY+ZZ)=\tfrac12(I+\vec\sigma_1\cdot\vec\sigma_2)=\mathrm{SWAP}$, with spectrum $\{+1,+1,+1,-1\}$, $\operatorname{Tr}=2$, $\det=-1$; (iii) the partial-swap concurrence $C=|\sin 2\theta|$, with $\sqrt{\mathrm{SWAP}}$ giving $C=1$ on $|01\rangle$; (iv) the SWAP-test probability $P_0=\tfrac12+\tfrac12|\langle\psi|\phi\rangle|^2$, evaluating to $P_0=1,0.75,0.5$ for overlaps $1,\tfrac12,0$. The identification of $\tfrac12(I+\vec\sigma_1\cdot\vec\sigma_2)$ with the Heisenberg exchange is Dirac's spin-exchange identity.

Prerequisites. The CNOT Gate · The Quantum Circuit Model · Pauli Gates (X, Y, Z) · Controlled-U Gates. Leads to. Circuit Identities and Equivalences · Universal Gate Sets · The Controlled-Z Gate.

Ask the tutor: ask me to thread a basis state through the three CNOTs, watch $\sqrt{\mathrm{SWAP}}$ build a Bell pair, or work out the SWAP-test probability for any two states you choose.