Singlet–Triplet and Exchange-Only Qubits

A single electron spin is the most natural qubit imaginable — two levels, up and down. Yet driving it means delivering an oscillating magnetic field to one dot among thousands, which is slow and hard to localise. Encoded spin qubits sidestep this by hiding the logical bit not in one spin's absolute direction but in the relative configuration of two or three electrons in coupled quantum dots. The control knob is then the Heisenberg exchange coupling $H=J\,\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ , an electrostatic quantity set by how far two dot wavefunctions overlap — turned up or down in nanoseconds by a gate voltage, no microwaves required. Two encodings dominate. The singlet–triplet ( $S$ – $T_0$ ) qubit takes two electrons in a double dot and works inside the $S_z=0$ sector, where the singlet $|S\rangle$ and the unpolarised triplet $|T_0\rangle$ are the logical basis: exchange $J(\varepsilon)$ splits them (one axis) and a gradient $\Delta B_z$ mixes them (the orthogonal axis), giving full single-qubit control. The exchange-only (EO) qubit uses three electrons of fixed total spin $S=\tfrac12$ ; two bonds $J_{12}$ and $J_{23}$ supply two rotation axes $120^\circ$ apart, so every gate is a pure voltage pulse and no magnetic control is needed. Both encodings live in a decoherence-free subspace: because the logical states share the same total-spin quantum numbers, a uniform field fluctuation acts as a global phase and cannot dephase them. What remains is charge noise on the detuning, which modulates $J$ — the central tension of the field.

Key takeaways

Store the qubit in the relative spin state of two or three electrons in coupled dots, and drive gates with the tunable Heisenberg exchange $J\,\mathbf{S}_i\!\cdot\!\mathbf{S}_j$ instead of global magnetic fields.
The singlet–triplet ( $S$ – $T_0$ ) qubit uses the two $S_z=0$ states of a double dot: exchange $J(\varepsilon)$ is the $\sigma_z$ axis, a field gradient $\Delta B_z$ the $\sigma_x$ axis, and together they give full SU(2).
The exchange-only qubit uses three spins in an $S=\tfrac12$ doublet; two bonds $J_{12}$ and $J_{23}$ give Bloch axes $120^\circ$ apart ( $\hat n_{12}\!\cdot\!\hat n_{23}=-\tfrac12$ ), so any gate is a short pulse sequence with no magnetic field.
Exchange is, up to a constant, the SWAP generator: $\mathbf{S}_1\!\cdot\!\mathbf{S}_2=\tfrac12\mathrm{SWAP}-\tfrac14\mathbb1$ , so a timed pulse enacts SWAP, $\sqrt{\mathrm{SWAP}}$ , or any partial swap.
Both encodings sit in a decoherence-free subspace immune to uniform magnetic noise; the residual limiter is charge noise on $\varepsilon$ , mitigated at a sweet spot where $\partial J/\partial\varepsilon=0$ .
Key idea: encode the logical qubit in the relative spin state of a small electron cluster so it occupies a decoherence-free subspace, and drive all gates with the electrically tunable Heisenberg exchange $J\,\mathbf{S}_i\!\cdot\!\mathbf{S}_j$ — replacing global magnetic-field control with local voltage control.

Where this lives today

Encoded spin qubits are now a leading solid-state platform. Singlet–triplet qubits were first shown coherently by Petta and co-workers in GaAs, where the nuclear-spin bath both supplies the gradient and caps $T_2^*$ at tens of nanoseconds; isotopically purified $^{28}$ Si strips out the nuclear spins and, with barrier-control sweet spots, has pushed single-qubit fidelities past 99%. Exchange-only qubits have been demonstrated in GaAs and silicon triple dots with all-electrical universal control and, recently, two-qubit logic — attractive because the absence of on-chip microwave lines eases the wiring of a dense 2D array. The decoherence-free subspaces are a working realisation of passive, encoding-based error suppression that complements active error correction, and Pauli-blockade readout is now fast enough that measurement is no longer the bottleneck. The open frontier is uniformity: fabricating hundreds of near-identical dots whose $J(\varepsilon)$ curves and sweet spots can all be tuned.

The physical picture

Picture two electrons as a tiny compass needle whose orientation relative to itself is the qubit, not its absolute heading. A uniform field rotates both spins together and does nothing to the encoded state — which is why these qubits ignore global field noise. Only the difference between the dots moves the logical state: a gradient field, or the exchange that wants to align or anti-align the spins. Exchange is geometric — $\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ measures how parallel the spins are — and because $J$ is just the overlap of two dot wavefunctions, squeezing the electrons together cranks $J$ up steeply, a purely electrical knob. The first figure shows where $J$ comes from: a detuned double dot in which the singlet can borrow a doubly-occupied charge state but the triplet cannot.

Figure

A double quantum dot and Pauli spin blockade: the (1,1) singlet tunnels into the (0,2) charge state while the (1,1) triplet is blocked, and the exchange splitting J = 4t_c²/(U − ε) reaches about 4 μeV ≈ h × 1 GHz for t_c = 20 μeV and U − ε = 400 μeV

The exchange splitting is only one axis on the Bloch sphere, and one axis can never reach every gate. The two encodings solve this differently. The singlet–triplet qubit adds a static field gradient $\Delta B_z$ — a micromagnet or nuclear polarisation — which drives $|S\rangle\!\leftrightarrow\!|T_0\rangle$ and supplies the orthogonal axis. The exchange-only qubit adds nothing magnetic: it carries a third electron and a second bond, and three-spin geometry forces the two bond axes $120^\circ$ apart. The second figure shows both geometries.

Figure

Two ways to reach every single-qubit gate: the singlet–triplet qubit precesses at ℏΩ = √(J² + h²) about an axis set by exchange J along z and gradient h = gμ_B ΔB_z along x, while the exchange-only qubit uses two bonds J₁₂ and J₂₃ whose Bloch axes sit 120° apart, n₁₂·n₂₃ = −1/2

The core equations

The whole subject rests on one operator. Isotropic exchange between two spin- $\tfrac12$ electrons is, up to a constant, half of the two-qubit SWAP gate — which is why a timed exchange pulse is a programmable entangling operation.

H_{\mathrm{ex}} = J\,\mathbf{S}_1\cdot\mathbf{S}_2 = \frac{J}{2}\,\mathrm{SWAP} - \frac{J}{4}\,\mathbb1 \qquad (1)

where: $J$ — exchange energy (the singlet–triplet splitting), electrically tunable; $\mathbf{S}_i$ — spin- $\tfrac12$ operator of electron $i$ (in units of $\hbar$ ); $\mathrm{SWAP}$ — the operator that interchanges the two spins. Since SWAP is $+1$ on the triplet and $-1$ on the singlet, timing $J$ performs SWAP, $\sqrt{\mathrm{SWAP}}$ , or any partial swap. The singlet–triplet qubit lives inside the two-dimensional $S_z=0$ sector of the double dot, spanned by the singlet and the unpolarised triplet.

|S\rangle=\frac{|{\uparrow\downarrow}\rangle-|{\downarrow\uparrow}\rangle}{\sqrt2},\qquad |T_0\rangle=\frac{|{\uparrow\downarrow}\rangle+|{\downarrow\uparrow}\rangle}{\sqrt2} \qquad (2)

where: $|{\uparrow\downarrow}\rangle$ — spin-up in dot 1, spin-down in dot 2; $|S\rangle,|T_0\rangle$ — the logical $|0\rangle,|1\rangle$ , both with total $S_z=0$ . The two states differ only in the relative phase, so anything coupling to the total spin leaves them untouched. Two knobs act on it: exchange shifts $|S\rangle$ below $|T_0\rangle$ (a $\sigma_z$ term) and a gradient connects $|S\rangle\leftrightarrow|T_0\rangle$ (a $\sigma_x$ term); their sum is the control Hamiltonian.

H_{ST_0}=\frac{J(\varepsilon)}{2}\,\sigma_z + \frac{g\mu_B\,\Delta B_z}{2}\,\sigma_x,\qquad \hbar\Omega=\sqrt{J^2+h^2},\ \ h\equiv g\mu_B\Delta B_z \qquad (3)

where: $\sigma_z,\sigma_x$ — Pauli operators in the $\{|S\rangle,|T_0\rangle\}$ basis; $J(\varepsilon)$ — detuning-dependent exchange, the electrical ( $z$ ) axis; $\Delta B_z$ — field difference between the dots (micromagnet or nuclear gradient), the ( $x$ ) axis; $g,\mu_B$ — electron $g$ -factor and Bohr magneton; $\Omega$ — precession rate. Two non-commuting generators span SU(2), so $J$ and $\Delta B_z$ give full single-qubit control, with the qubit precessing about a tilted axis at $\sqrt{J^2+h^2}/\hbar$ . Where does $J(\varepsilon)$ come from? The $(1,1)$ singlet mixes through tunnelling with the $(0,2)$ singlet; the triplet cannot (Pauli blockade). Second-order perturbation theory yields an exchange that grows as the dots detune toward the charge transition.

J(\varepsilon)\simeq \frac{4t_c^2}{U-\varepsilon} \qquad (4)

where: $t_c$ — inter-dot tunnel coupling; $U$ — on-site Coulomb (charging) energy; $\varepsilon$ — detuning between $(1,1)$ and $(0,2)$ . At $\varepsilon=0$ this is the Hubbard result $J=4t^2/U$ ; pushing $\varepsilon$ toward $U$ makes $J$ large and fast but $\partial J/\partial\varepsilon$ large too — the origin of charge-noise sensitivity. The exchange-only qubit needs three electrons: fixing $S=\tfrac12,\,S_z=-\tfrac12$ leaves a two-dimensional logical subspace, built by coupling the pair- $(1,2)$ state to the third spin.

|0_L\rangle = |S\rangle_{12}\,|{\downarrow}\rangle_3,\qquad |1_L\rangle = \sqrt{\tfrac13}\,|T_0\rangle_{12}|{\downarrow}\rangle_3 - \sqrt{\tfrac23}\,|T_-\rangle_{12}|{\uparrow}\rangle_3 \qquad (5)

where: $|S\rangle_{12},|T_m\rangle_{12}$ — singlet/triplet of electrons 1 and 2 ( $m=+,0,-$ ); $|0_L\rangle,|1_L\rangle$ — the logical states of the three-spin doublet. Both share fixed total $(S,S_z)$ — a decoherence-free subspace — and their $(1,2)$ pair is singlet vs triplet, so an exchange pulse on that bond distinguishes them. Every gate is a timed exchange pulse: the unitary is the exponential of the exchange operator, with the rotation angle set by the pulse area.

U_{ij}(\theta)=\exp\!\big[-i\,\theta\,\mathbf{S}_i\cdot\mathbf{S}_j\big],\qquad \theta=\frac1\hbar\!\int J_{ij}\,\mathrm{d}t,\qquad \hat n_{12}\cdot\hat n_{23}=-\tfrac12 \qquad (6)

where: $\theta$ — pulse area ( $\theta=\pi$ a full SWAP, $\theta=\pi/2$ a $\sqrt{\mathrm{SWAP}}$ ); $J_{12},J_{23}$ — the two bonds of the chain; $\hat n_{ij}$ — the Bloch axis bond $ij$ rotates about. On the $S=\tfrac12$ doublet the two bonds become rotations about axes $120^\circ$ apart, and two non-parallel axes generate all of SU(2). The price of using $J$ for control is that $J$ carries noise: detuning fluctuations from charge noise translate directly into exchange fluctuations that dephase the qubit.

\frac{1}{T_2^*}\sim \frac{1}{\hbar}\,\frac{\partial J}{\partial \varepsilon}\,\sigma_\varepsilon \qquad (7)

where: $T_2^*$ — inhomogeneous (Ramsey) dephasing time; $\sigma_\varepsilon$ — RMS detuning fluctuation; $\partial J/\partial\varepsilon$ — sensitivity of exchange to detuning. Coherence is longest where the slope vanishes — the sweet spot — so barrier control is preferred. Finally, readout rides on the same blockade: a singlet enters $(0,2)$ , a triplet is forbidden, so spin becomes charge.

(1,1)\,S \to (0,2)\,S,\qquad (1,1)\,T \not\to (0,2) \qquad (8)

where: $(1,1),(0,2)$ — charge states (one per dot vs both in dot 2); $S,T$ — singlet (tunnels) vs triplet (blocked). A charge sensor sees the $(0,2)$ electron appear only for a singlet, turning spin into a measurable current.

Derivation AExchange as the generator of the SWAP gate

The claim that "exchange is the SWAP generator" is what makes spin qubits programmable, so it deserves a proof. We show that the Heisenberg coupling $\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ equals, up to an additive constant, half of the two-qubit SWAP operator, so a timed exchange pulse enacts SWAP, $\sqrt{\mathrm{SWAP}}$ , or any partial swap. The route: diagonalise $\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ on the singlet–triplet basis and compare its eigenvalues with SWAP.

Step 1. Use the total-spin identity $\mathbf{S}_1\!\cdot\!\mathbf{S}_2=\tfrac12(\mathbf{S}^2-\mathbf{S}_1^2-\mathbf{S}_2^2)$ with $\mathbf{S}=\mathbf{S}_1+\mathbf{S}_2$ and each $\mathbf{S}_i^2=s(s+1)=\tfrac34$ . Why: this trades the operator product for a single Casimir $\mathbf{S}^2$ whose eigenvalues we know from angular-momentum addition.

Step 2. The singlet has $\mathbf{S}^2=0$ , giving $\mathbf{S}_1\!\cdot\!\mathbf{S}_2|S\rangle=\tfrac12(0-\tfrac34-\tfrac34)|S\rangle=-\tfrac34|S\rangle$ ; the triplet has $\mathbf{S}^2=2$ , giving $+\tfrac14|T\rangle$ . The gap is $\tfrac14-(-\tfrac34)=1$ , so $H_{\mathrm{ex}}=J\,\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ splits triplet above singlet by exactly $J$ . Why: this fixes $J$ as the measured splitting, and $J>0$ (antiferromagnetic) puts the singlet lowest.

Step 3. SWAP acts as $+1$ on the three triplets and $-1$ on the singlet. Matching eigenvalues gives the exact identity $\mathbf{S}_1\!\cdot\!\mathbf{S}_2=\tfrac12\mathrm{SWAP}-\tfrac14\mathbb1$ , hence $H_{\mathrm{ex}}=\tfrac{J}{2}\mathrm{SWAP}-\tfrac{J}{4}\mathbb1$ . Why: the constant is a harmless global phase, so up to it the exchange Hamiltonian is proportional to SWAP itself — the exchange pulse is a swap pulse.

Step 4. Exponentiating, $U(\theta)=\exp[-i\theta\,\mathbf{S}_1\!\cdot\!\mathbf{S}_2]$ with $\theta=\tfrac1\hbar\int J\,\mathrm{d}t$ gives a full SWAP (up to phase) at $\theta=\pi$ and a $\sqrt{\mathrm{SWAP}}$ at $\theta=\pi/2$ . Why: $\sqrt{\mathrm{SWAP}}$ plus single-qubit rotations is a universal gate set, so this one pulse is the architecture's entangling primitive. *Verified: singlet eigenvalue $-3/4$ , triplet $+1/4$ , splitting $1$ ; $\mathbf{S}_1\!\cdot\!\mathbf{S}_2=\tfrac12\mathrm{SWAP}-\tfrac14\mathbb1$ reproduced state-by-state (Nielsen & Chuang Ch. 7; Loss & DiVincenzo, PRA 1998).*

Go deeper · Derivation B · 1 minExchange from a detuned double dot by Schrieffer–Wolff

Derivation BExchange from a detuned double dot by Schrieffer–Wolff

The exchange used for control is not a fundamental constant but an emergent, tunable energy born from virtual charge motion; the goal is to derive its size and detuning dependence from a two-site Hubbard model. Starting from $(1,1)$ , inter-dot tunnelling mixes in the doubly-occupied $(0,2)$ and $(2,0)$ states, and second-order perturbation theory finds how far the singlet drops while the triplet stays put — the difference being $J(\varepsilon)$ .

Step 1. Only the $(1,1)$ singlet couples to a doubly-occupied state: both electrons in one dot's lowest orbital require opposite spins, so the triplet is Pauli-blocked from $(0,2)$ and $(2,0)$ and receives no second-order shift. Why: this asymmetry — singlet mixes, triplet cannot — is the origin of both the splitting and blockade readout.

Step 2. The two-electron tunnelling matrix element is $\langle(0,2)S|H_t|(1,1)S\rangle=\sqrt2\,t_c$ , with denominator $E_{(0,2)}-E_{(1,1)}=U-\varepsilon$ . Why: the denominator is what the gate voltage tunes — raising $\varepsilon$ toward $U$ strengthens the mixing.

Step 3. Second-order perturbation lowers the singlet by $2t_c^2/(U-\varepsilon)$ per doubly-occupied channel; including both $(0,2)$ and $(2,0)$ near symmetry gives $J=E_{T_0}-E_S=4t_c^2/(U-\varepsilon)$ . Why: at $\varepsilon=0$ this is the Hubbard super-exchange $J=4t^2/U$ , the same physics that makes Mott insulators antiferromagnets.

Step 4. Differentiating, $\partial J/\partial\varepsilon=4t_c^2/(U-\varepsilon)^2>0$ , large in the detuning-control regime but sent to zero by tuning the barrier $t_c$ at fixed $\varepsilon$ . Why: this slope is the charge-noise lever of equation (7); killing it to first order defines the sweet spot. *Verified: reduces to $J=4t^2/U$ at $\varepsilon=0$ ; $J$ diverges as $\varepsilon\to U$ where perturbation theory breaks down (Loss & DiVincenzo, PRA 1998; Petta et al., Science 2005).*

Go deeper · Derivation C · 1 minUniversality of the exchange-only qubit from two bonds

Derivation CUniversality of the exchange-only qubit from two bonds

The exchange-only qubit makes the strongest claim — that pure Heisenberg exchange, with no magnetic field, suffices for universal single-qubit control — and the goal is to see why, and at what pulse cost. We restrict the three-spin space to the $S=\tfrac12$ doublet, project the two bonds onto it, show they become rotations about axes $120^\circ$ apart, and use the fact that two non-parallel axes generate all of SU(2).

Step 1. On the $\{|0_L\rangle,|1_L\rangle\}$ doublet, $\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ gives $-\tfrac34$ on $|0_L\rangle$ (pair-12 singlet) and $+\tfrac14$ on $|1_L\rangle$ (pair-12 triplet), i.e. $\mathbf{S}_1\!\cdot\!\mathbf{S}_2\big|_{\text{doublet}}=-\tfrac14\mathbb1-\tfrac12\,\hat n_{12}\!\cdot\!\boldsymbol\sigma$ with $\hat n_{12}=\hat z$ . Why: each bond, restricted to the code space, is a genuine Pauli rotation generator — that is what lets an exchange pulse rotate the logical qubit.

Step 2. The identity $\sum_{i<j}\mathbf{S}_i\!\cdot\!\mathbf{S}_j=\tfrac12(\mathbf{S}_{\text{tot}}^2-\sum_i\mathbf{S}_i^2)=\tfrac12(\tfrac34-\tfrac94)=-\tfrac34$ is a constant on the doublet, so the three axis vectors must sum to zero: $\hat n_{12}+\hat n_{23}+\hat n_{13}=0$ . Symmetric unit vectors summing to zero are $120^\circ$ apart, so $\hat n_{12}\!\cdot\!\hat n_{23}=-\tfrac12$ . Why: this angle is fixed by spin algebra, not device geometry, so no fine tuning gives two independent axes.

Step 3. Any $g\in\mathrm{SU}(2)$ factorises as $R_{\hat n_{12}}(\alpha)\,R_{\hat n_{23}}(\beta)\,R_{\hat n_{12}}(\gamma)$ for two non-parallel axes (an Euler decomposition), so an arbitrary single-qubit gate needs at most three, generically four, exchange pulses. Why: this is the pulse budget — a few nanosecond voltage pulses per gate, with a longer but finite sequence for CNOT.

Step 4. Because $|0_L\rangle,|1_L\rangle$ share fixed $(S,S_z)$ , a uniform field $H_B=b\,S_z^{\text{tot}}$ acts as $\propto\mathbb1$ on the code space — a global phase. Why: the decoherence-free property; collective magnetic fluctuations, which dominate in many hosts, cannot dephase the logical qubit. *Verified: axis overlap $-1/2$ from the total-spin sum rule; single-qubit gates from pure exchange, CNOT in a finite sequence — DiVincenzo, Bacon, Kempe, Burkard & Whaley (Nature 408, 339, 2000).*

Limiting cases

The exchange profile $J(\varepsilon)$ and the control geometry have several instructive regimes:

Case	Result	Meaning
Deep barrier / large negative detuning	$J(\varepsilon)\to 0$	dots decoupled; $\|S\rangle$ and $\|T_0\rangle$ degenerate, so only $\Delta B_z$ evolves the qubit (idle, or a pure $\sigma_x$ rotation)
Detuning toward $(0,2)$	$J(\varepsilon)\approx \dfrac{4t_c^2}{U-\varepsilon}$	Schrieffer–Wolff regime: $J$ grows as the singlet hybridises with the charge state — fast gates, maximal charge-noise sensitivity
Symmetric (barrier-control) sweet spot	$\left.\dfrac{\partial J}{\partial\varepsilon}\right\|_{\varepsilon^*}=0$	$J$ first-order insensitive to detuning — suppresses charge dephasing while keeping $J$ sizeable
Pure exchange, no gradient	$\Omega=J/\hbar,\ \hat n=\hat z$	one axis only — not universal alone; must add $\Delta B_z$ or a second bond
Exchange-only, balanced bonds	$J_{12}=J_{23}$ , $\hat n_{12}\!\cdot\!\hat n_{23}=-\tfrac12$	two axes $120^\circ$ apart; alternating them reaches any gate, a single bond cannot

Common trapExchange alone gives a universal single-qubit gate

It does not. In the $\{|S\rangle,|T_0\rangle\}$ basis a pure exchange pulse generates only a $\sigma_z$ rotation — one fixed axis — and a single axis can never reach an arbitrary point on the sphere. You always need a second, non-commuting control: for the singlet–triplet qubit that is the gradient $\Delta B_z$ supplying $\sigma_x$ ; for the exchange-only qubit it is a second exchange bond, whose axis sits $120^\circ$ from the first by three-spin algebra. The common error is to imagine that one $J$ -knob, pulsed cleverly, suffices for arbitrary single-qubit rotations. It never does — the impossibility is geometric, not technological. (Exchange between two spins is universal for two-qubit gates — a separate statement from single-qubit universality within one encoded qubit.)

Worked examples

1 · Exchange-gate speed from the pulse area (confidence builder)

A double dot is tuned to exchange $J/h=1\ \mathrm{GHz}$ . How long must the pulse last for a $\sqrt{\mathrm{SWAP}}$ , and for a full SWAP?

Step 1. The gate angle is the pulse area $\theta=(J/\hbar)\,\tau$ for a constant pulse of length $\tau$ ; writing $J=h f_J$ with $f_J=J/h=1\ \mathrm{GHz}$ gives $\theta=2\pi f_J\,\tau$ .

Step 2. A $\sqrt{\mathrm{SWAP}}$ needs $\theta=\pi/2$ , so $\tau=\dfrac{1}{4f_J}=\mathbf{0.25\ ns}$ ; a full SWAP needs $\theta=\pi$ , so $\tau=\dfrac{1}{2f_J}=\mathbf{0.5\ ns}$ .

Step 3. These sub-nanosecond gate times — set purely by a voltage-controlled energy scale — are the headline advantage of exchange qubits: no microwave envelope, just a fast pulse whose area you dial.

2 · Singlet–triplet precession under exchange and a gradient

An $S$ – $T_0$ qubit in silicon ( $g=2$ ) runs with $J/h=100\ \mathrm{MHz}$ and a gradient tuned so that $g\mu_B\Delta B_z/h=100\ \mathrm{MHz}$ . Find the gradient, the precession frequency, and the rotation axis.

Step 1. Invert $h=g\mu_B\Delta B_z$ . With $h=100\ \mathrm{MHz}\times h_{\mathrm{Planck}}=6.63\times10^{-26}\ \mathrm{J}$ , $g=2$ , $\mu_B=9.274\times10^{-24}\ \mathrm{J/T}$ : $\Delta B_z=\dfrac{6.63\times10^{-26}}{2\times9.274\times10^{-24}}=\mathbf{3.6\ mT}$ — a realistic micromagnet gradient.

Step 2. With $\hbar\Omega=\sqrt{J^2+h^2}$ and $J=h$ in frequency, $\Omega/2\pi=\sqrt{100^2+100^2}=\mathbf{141\ MHz}$ , a Bloch-sphere period of $\mathbf{7.1\ ns}$ .

Step 3. The axis $\hat n=(h,0,J)/\sqrt{J^2+h^2}$ points at $\mathbf{45^\circ}$ between $z$ ( $J$ ) and $x$ ( $\Delta B_z$ ) when $J=h$ ; a quarter period ( $\approx1.8\ \mathrm{ns}$ ) is then a $\pi/2$ rotation about that tilted axis — a gate built from two static, non-commuting controls.

A note on coherence. At a detuning-sensitive point with slope $\partial J/\partial\varepsilon=0.1$ and charge noise $\sigma_\varepsilon=1\ \mu\mathrm{eV}$ , the exchange jitters by $\delta J\approx0.1\ \mu\mathrm{eV}$ , giving $T_2^*\sim\hbar/\delta J\approx\mathbf{7\ ns}$ — typical of detuning-controlled GaAs. The sweet spot ( $\partial J/\partial\varepsilon=0$ ) makes $\delta J$ only second order in $\sigma_\varepsilon$ ; with isotopic $^{28}$ Si this reaches microseconds.

Physics of the portrait

Encoded spin qubits are the invention of a small group of theorists who asked, in the late 1990s, whether the messy magnetic control of individual spins could be replaced by something an engineer could actually build at scale. The answer — exchange — runs through one figure more than any other.

Origins · 1 minDavid DiVincenzo and how this was discovered

David DiVincenzo

born 1959 · quantum-dot spin qubits and the DiVincenzo criteria

David P. DiVincenzo, long at IBM's Watson Research Center and later at RWTH Aachen and Jülich, is the theorist most responsible for turning "use electron spins as qubits" into a concrete, checkable programme. His 1998 paper with Daniel Loss (Physical Review A 57, 120) proposed confining single electrons in gated quantum dots and driving two-qubit gates with the tunable exchange $J\,\mathbf{S}_i\!\cdot\!\mathbf{S}_j$ — the founding document of the quantum-dot field. Two years later he sharpened the discipline with the DiVincenzo criteria, the checklist any physical qubit must satisfy, and in 2000 he and co-workers (Bacon, Kempe, Burkard, Whaley) proved that exchange alone, with no magnetic field, is universal if you encode a qubit in three spins — launching the exchange-only architecture. His recurring theme is the one this article turns on: prefer controls that are fast, local, and electrical, and let encoding buy back whatever universality a single knob lacks.

History

1998

— Loss and DiVincenzo propose exchange-controlled electron-spin qubits in quantum dots

2000

— DiVincenzo, Bacon, Kempe, Burkard and Whaley prove universal computation with exchange alone

2002

— Levy introduces the singlet–triplet encoding in a decoherence-free subspace

2005

— Petta and co-workers observe coherent singlet–triplet oscillations in a GaAs double dot

2012

— Maune and co-workers realise a singlet–triplet qubit in isotopically enriched silicon

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, Ch. 7 · Loss & DiVincenzo, Phys. Rev. A 57, 120 (1998) · Levy, Phys. Rev. Lett. 89, 147902 (2002) · DiVincenzo, Bacon, Kempe, Burkard & Whaley, Nature 408, 339 (2000) · Petta et al., Science 309, 2180 (2005) · Kempe, Bacon, Lidar & Whaley, Phys. Rev. A 63, 042307 (2001).

Provenance. The exchange-as-SWAP identity, the singlet/triplet eigenvalues of $\mathbf{S}_1\!\cdot\!\mathbf{S}_2$ ( $-3/4$ and $+1/4$ ), the Schrieffer–Wolff $J=4t_c^2/(U-\varepsilon)$ reducing to the Hubbard $4t^2/U$ , the doublet axis overlap $\hat n_{12}\!\cdot\!\hat n_{23}=-\tfrac12$ , and $\hbar\Omega=\sqrt{J^2+h^2}$ are standard and recomputable from Loss–DiVincenzo (PRA 1998), DiVincenzo et al. (Nature 2000), Petta et al. (Science 2005), and Nielsen & Chuang Ch. 7. A verifier should reproduce: (i) $\mathbf{S}_1\!\cdot\!\mathbf{S}_2=\tfrac12\mathrm{SWAP}-\tfrac14\mathbb1$ from the eigenvalues; (ii) $J=4t_c^2/(U-\varepsilon)$ from second-order perturbation; (iii) the $120^\circ$ axis angle from the total-spin sum rule; (iv) $\hbar\Omega=\sqrt{J^2+h^2}$ by diagonalising equation (3); (v) $\sqrt{\mathrm{SWAP}}$ in $0.25\ \mathrm{ns}$ at $J/h=1\ \mathrm{GHz}$ , and $\Delta B_z=3.6\ \mathrm{mT}$ for $100\ \mathrm{MHz}$ in silicon.

Prerequisites. DiVincenzo Criteria · Pauli exclusion and spin singlets/triplets · Heisenberg exchange interaction. Leads to. Superconducting Qubits · Dispersive Readout · Trapped-Ion Qubits.

Ask the tutor: Ask me to pulse a specific gate — say, build a Hadamard on an $S$ – $T_0$ qubit from $J$ and $\Delta B_z$ , or count the exchange pulses for a CNOT between two exchange-only qubits.

Why encode the qubit in two or three spins instead of using a single electron spin?

A single electron spin is a fine qubit, but rotating it needs an oscillating magnetic field (ESR) delivered to one dot — slow and hard to confine in a dense array. Encoding trades that for exchange, set by wavefunction overlap and so tuned by a gate voltage: fast, local, CMOS-friendly. The bonus is that the encoded states share total-spin quantum numbers, placing them in a decoherence-free subspace that ignores uniform field noise. The cost is overhead: two or three electrons per qubit, plus a second control.

What exactly does "decoherence-free" protect against, and what does it not?

It protects against collective, spatially uniform magnetic fluctuations. Because both logical states have the same total spin $S$ and projection $S_z$ , a field coupling to the total spin — the same on every electron — acts as a global phase and cannot dephase the superposition. It does not protect against anything that distinguishes the dots: a fluctuating field gradient (nuclear-spin noise in GaAs) and charge noise on $\varepsilon$ , which modulates $J$ . So the encoding removes one dominant channel; the rest are attacked by isotopic purification and sweet-spot operation.

How is the exchange coupling turned on and off electrically?

Exchange is the energy cost of the two dot wavefunctions overlapping, so changing the overlap changes $J$ . Two gate-voltage knobs do this. Detuning control tilts the double well, pushing the $(1,1)$ singlet toward the $(0,2)$ state and raising $J\simeq 4t_c^2/(U-\varepsilon)$ as $\varepsilon\to U$ . Barrier control instead lowers the tunnel barrier at fixed detuning, raising $t_c$ directly. Both are nanosecond voltage pulses on metal gates — no microwaves — and barrier control can sit at the sweet spot $\partial J/\partial\varepsilon=0$ .

Why does the exchange-only qubit need three electrons rather than two?

With two electrons you have one bond, hence one rotation axis — not universal, so you must add a magnetic gradient. Three electrons give two independent bonds, $J_{12}$ and $J_{23}$ , and three-spin algebra forces their logical axes $120^\circ$ apart ( $\hat n_{12}\!\cdot\!\hat n_{23}=-\tfrac12$ ). Two non-parallel axes generate all of SU(2), so any single-qubit gate is a short pulse sequence with no field. The third electron is the price of discarding magnetic control — a few extra pulses for a fully electrical, decoherence-free qubit.

What is the "sweet spot," and how does barrier control differ from detuning control?

The sweet spot is an operating point where the qubit's splitting is first-order insensitive to its dominant noise. Charge noise enters through $\partial J/\partial\varepsilon$ , so the sweet spot is where that slope vanishes. Detuning control sets $J$ by tilting the dots ( $\varepsilon$ ): large $J$ , but large $\partial J/\partial\varepsilon$ — noisy. Barrier control instead sets $J$ by lowering the barrier ( $t_c$ ) at fixed symmetric detuning, where $\partial J/\partial\varepsilon=0$ . Same $J$ , far weaker charge-noise coupling — worth one to two orders of magnitude in $T_2^*$ .

How does Pauli spin blockade actually read out the spin state?

Readout maps an invisible spin state onto a visible charge. From $(1,1)$ (one electron per dot), you pulse toward $(0,2)$ . A singlet completes the transition — both electrons fit in one dot's lower orbital with opposite spins — so the charge moves. A triplet cannot: two same-spin electrons in one orbital are Pauli-forbidden. A nearby charge sensor (RF single-electron transistor, quantum point contact, or gate-based dispersive readout) sees the $(0,2)$ electron appear only for a singlet — a fast, high-fidelity spin-to-charge conversion.