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In LS coupling, first add the individual orbital angular momenta and spins separately, then add their totals:The Coulomb interaction between electrons must separate different atomic terms more strongly than the spin-orbit interaction mixes them. Then $L,S,J,M_J$ are useful quantum numbers, with $J=|L-S|,\dots ,L+S$. This approximation is usually better for lighter chemical atoms and well-separated atomic terms; strong configuration interaction or strong spin-orbit coupling can spoil it. The term symbol is ${}^{2S+1}{L}_J$.

The general electric-dipole selection rules, for isolated chemical atoms with definite parity and total electronic angular momentum, are a change of parity, $\Delta J=0,\pm 1$ excluding $J=0\leftrightarrow 0$, and $\Delta M_J=0,\pm 1$, with the accessible magnetic sublevels determined by the light's polarization. In pure LS coupling, the additional selection rules are $\Delta S=0$ and $\Delta L=0,\pm 1$, excluding $L=0\leftrightarrow 0$. In a one-electron description the electric dipole operator also requires $\Delta l=\pm 1$ for the active electron. These latter statements can fail as exact labels when atomic terms mix; the general parity and total-angular momentum restrictions still apply. Unresolved hyperfine structure is neglected here.

For strontium, the closed core contributes $L=S=0$. The Pauli exclusion principle makes the two equivalent $5s$ electrons a singlet state. Different radial orbitals permit both a singlet state and a triplet state. Thus the complete lists of atomic fine-structure levels areThe $5s5p$ electron configuration has odd parity; the other two electron configurations have even parity.

Measure each energy level by the wavenumber $\tilde{E}=(E-E_g)/(hc)$ in ${\mathrm{m}}^{-1}$. An upward electronic transition of vacuum wavelength $\lambda$ adds $1/\lambda$. Treat the quoted wavelengths as vacuum values to their limited accuracy; air corrections would be smaller than the precision justified here.

The strong ground-state absorption must be $5s^2{}^1{S}_0\to 5s5p{}^1{P}_1$ at $461\mathrm{nm}$. The weak ground-state electronic transition at $689\mathrm{nm}$ must reach $5s5p{}^3{P}_1$: the electric-dipole selection rule on $J$ excludes the other two triplet state energy levels, while weak spin-orbit coupling admixes singlet state character and relaxes the approximate spin selection rule.

Assume the usual ordering $E({}^3{P}_0)<E({}^3{P}_1)<E({}^3{P}_2)$, as expected from a positive effective spin-orbit coupling constant. The three nearby strong spectral lines then join these energy levels to a common $5s6s{}^3{S}_1$ energy level. Larger lower-state energy means a smaller gap and longer wavelength, so $679,688,707\mathrm{nm}$ correspond to $J=0,1,2$ respectively. The remaining $1.12\mu \mathrm{m}$ spectral line joins $5s5p{}^1{P}_1$ and $5s6s{}^1{S}_0$. Assume the latter is the upper energy level. These assignments and orderings agree with independently tabulated strontium energy levels.^[ref]

Starting with the two ground-state gaps, the construction isThe resulting energy levels, rounded to avoid suggesting spectroscopic precision, are

Both $461\mathrm{nm}$ and $1.12\mu \mathrm{m}$ exhibit the normal Zeeman effect. Each connects two singlet state energy levels, with Landé g-factor $g_J=1$ for the $J=1$ energy level and no first-order shift for $J=0$. Their frequency shifts are $0,\pm {\mu }_{B}B/h$. The weak $689\mathrm{nm}$ spectral line also has three components, but $g({}^3{P}_1)=3/2$ gives shifts $0,\pm 3{\mu }_{B}B/(2h)$; a three-component pattern alone does not establish the normal Zeeman effect. The $679\mathrm{nm}$ spectral line likewise has three components with $g({}^3{S}_1)=2$.

Let the dimensionful angular momentum operators obey $\mathbf{F}=\mathbf{I}+\mathbf{J}$ and take the interaction Hamiltonian as $H^{\prime }=A\mathbf{I}\cdot \mathbf{J}$. Sincethe energy eigenvalues at fixed $I,J$ arewhere $F=|I-J|,\dots ,I+J$. Subtraction gives the Landé interval ruleMany atomic conventions use dimensionless angular momentum operators and call the energy-valued constant itself $A$; then the right-hand side is $AF$. Each energy level has $2F+1$ magnetic sublevels with the same energy in zero field.

For fine structure, the effective interaction is $a_{LS}\mathbf{L}\cdot \mathbf{S}$ with dimensionless angular momentum operators. It originates in the spin-orbit interaction, including the magnetic interaction of electron spin with orbital motion and its relativistic correction. Within an isolated LS coupling atomic term, the labels are the electron configuration, $L,S,J,M_J$, and the interval between adjacent $J$ energy levels is $a_{LS}J$. This is smaller than the electrostatic separation of atomic terms.

For magnetic-dipole hyperfine structure, the nuclear magnetic dipole moment interacts with the magnetic field generated by the electrons. Contributions include the Fermi contact interaction of penetrating $s$ electrons, orbital magnetic fields, and the dipolar field of electron spin. Projected onto one electronic $J$ energy level, their rotationally invariant magnetic-dipole part has the form $a_{IJ}\mathbf{I}\cdot \mathbf{J}$. The labels are $I,J,F,M_F$, together with the approximate electronic labels such as $L,S$. This splitting is generally smaller than fine structure, reflecting the much smaller nuclear magnetic dipole moment. An electric nuclear quadrupole moment can add a different interaction and violate the simple Landé interval rule.

At the larger electrostatic energy scale, an effective two-spin exchange interaction can also be written $a_s{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ within fixed orbitals. Its origin is the Coulomb interaction combined with the Pauli exclusion principle, rather than a large direct magnetic interaction. It separates singlet state and triplet state atomic terms; before spin-orbit coupling, useful labels include the electron configuration, $L,M_L,S,M_S$. General electrostatic atomic term separations need not obey a single Landé interval rule as $L$ varies.

A Jablonski diagram distinguishes electronic excitation from subsequent relaxation. Photon absorption takes a molecule from the electronic ground state $S_0$ to a vibrationally excited level of $S_1$ or a higher singlet state. Rapid vibrational relaxation and internal conversion usually populate the lowest vibrational levels of $S_1$. Fluorescence then emits a photon on a transition to $S_0$, often ending in one of its excited vibrational levels. Nonradiative relaxation competes with emission; intersystem crossing can also populate triplet states. Energy lost to relaxation and solvent reorganization commonly makes the fluorescence maximum occur at lower photon energy, hence longer wavelength, than the photon absorption maximum: the Stokes shift.

The intrinsic fluorescence quantum yield is the number of photons emitted by fluorescence per absorbed excitation. For a simple excited state with radiative rate $k_r$ and other decay rate $k_{nr}$,A nearby acceptor adds a nonradiative energy-transfer channel, not an emitted photon flying from donor to acceptor. The electric dipole-dipole interaction transfers the donor excitation while returning the donor toward its ground state and exciting the acceptor. The acceptor then relaxes and may fluoresce. The donor yield becomesFor the acceptor, distinguish its intrinsic yield per acceptor excitation, $Q_A=k_{r,A}/(k_{r,A}+k_{nr,A})$, from the number of acceptor photons per donor excitation. Transfer increases the supply of acceptor excitations, giving sensitized emission $E{Q}_A$ per donor excitation in the ideal model. It does not by itself change $Q_A$ if the acceptor's own decay rates remain unchanged. Environmental changes, saturation or additional quenching can change those rates, but increased acceptor intensity alone does not establish an increased intrinsic fluorescence quantum yield.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

Let $P_{12}$ exchange the complete labels of the two electrons. The spatial factors obey $P_{12}{\Phi }_{\pm }=\pm {\Phi }_{\pm }$. The singlet state spin factor is antisymmetric, while each triplet state spin factor is symmetric. Thus in all four casesas required by the Pauli exclusion principle. In particular, the $M_S=0$ triplet state remains a triplet despite having opposite measured spin projections in each product term.

For a spin-independent Hamiltonian and the same fixed spatial atomic orbitals, direct evaluation giveswhere $J_{ab}$ is the direct Coulomb integral and $K_{ab}>0$ is the exchange integral for the helium $1s,2s$ pair. ThereforeThe antisymmetric spatial wavefunction vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower energy of the maximum-spin member of this fixed electron configuration is the first of Hund's rules.

The Pauli exclusion principle requires the complete electron wavefunction to change sign on exchange of any two electrons, including both position and spin labels. Let $a=1s$ and $b=2s$ be normalized, orthogonal spatial atomic orbitals, and let $\alpha ,\beta$ denote spin up and down.

The symmetric and antisymmetric spatial functions areThe singlet state isThe three triplet states share the antisymmetric spatial factor:Every factor and complete product is normalized. Here the subscripts are $(S,M_S)$; the spatial electron configuration is $1s2s$ in each case.

Let $a={\varphi }_{1s}$ and $b={\varphi }_{2s}$ be orthonormal atomic orbitals. The normalized spatial factors areFor the singlet state, use the antisymmetric spin factorFor the three components of the triplet state, useEach total wavefunction obeys the Pauli principle. At coincident spatial coordinates,Consequently the triplet pair probability density vanishes at coincidence, even for its $M=0$ component. This zero is stronger than the trivial statement that any exact point has zero integrated probability in a continuum.

Keeping the same two spin states but incorrectly imposing boson exchange symmetry would require a symmetric total wavefunction. The spin-antisymmetric singlet state would then accompany ${\Phi }_{-}$, while the spin-symmetric triplet state would accompany ${\Phi }_{+}$. Thusand $\boxed{E_T^{\mathrm{hypothetical}}-E_S^{\mathrm{hypothetical}}=2K}$. The ordering reverses: the triplet would be destabilized by $2K$ relative to the singlet. A genuinely spinless boson would have no singlet/triplet spin distinction; this comparison changes exchange symmetry while retaining the question's spin degrees of freedom.

The target states here are qubit states, as appropriate to sharing singlet states of two qubits. Write $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ and choose $|{\psi }^{\perp }⟩=-{\beta }^{\ast }|0⟩+{\alpha }^{\ast }|1⟩$. These form an orthonormal basis, and the shared Bell state can be expressed asAlice knows the classical description of $|\psi ⟩$, so she can perform the projective measurement in this basis. The outcome $|{\psi }^{\perp }⟩$ has probability $1/2$ and leaves Bob in $|\psi ⟩$, up to an irrelevant overall phase. Alice sends a success bit: one for this outcome and zero for the other. Bob keeps his qubit only on success. This proves heralded remote preparation of an arbitrary qubit:The other outcome leaves $|{\psi }^{\perp }⟩$ at Bob's side; the protocol makes no claim of success in that branch.

For a block of $n$ targets, share $n$ independent singlet states. Alice measures her $j$th qubit in the basis $|{\psi }_j⟩,|{\psi }_j^{\perp }⟩$. She sends one bit indicating whether all $n$ outcomes were of the successful kind. Because the pairs are independent, all-success has probability $2^{-n}$; conditioned on it, Bob's joint state is precisely ${⨂}_{j=1}^n|{\psi }_j⟩$. Thus one-bit heralding of a remote state preparation block givesOn failure Bob discards the block, so he does not need a list of which component states failed.

For asymptotic remote state preparation by block indexing, prelabel a supply of independent candidate blocks, each containing $n$ singlet states. Alice measures blocks until she finds the first all-success block, of index $J$. She does not send a flag after each trial. Since $q=2^{-n}$ is the success probability of each block,This geometric distribution is independent of the target states. The probability of never succeeding is ${\lim }_{k\to \infty }(1-q{)}^k=0$. Alice needs to communicate only $J$; Bob retrieves the labelled qubits in that block, and thereby knows that every requested state was prepared exactly.

Here is an explicit prefix code for a rare geometric success. Set $M=2^n$ and divide $J-1=MQ+R$, where $0\le R<M$. Encode $Q$ as $Q$ ones followed by zero, then encode $R$ in exactly $n$ bits. Bob can decode this without knowing any target-state description. The message length is $L=n+Q+1$. With $r=(1-q{)}^M$,Since $(1-1/M{)}^M\le e^{-1}$,This version succeeds exactly with probability one and has the stated expected classical communication cost. It consumes $n\mathbb{E}J=n{2}^n$ singlet states on average, compatible with the unlimited entanglement assumption.

There is also a fixed-message-length version. Try at most $K=n{2}^n$ candidate blocks and send the first successful index, or a zero codeword reporting that none succeeded. Encoding the $K+1$ possibilities uses $\lceil {\log }_2(K+1)\rceil =n+{\log }_{2}n+O(1)$ bits. Its failure probability isThus fixed-length communication approaches one bit per state with success probability tending to one; variable-length communication achieves exact eventual success with expected cost approaching one bit per state. Before the index arrives, Bob's unconditional state remains independent of the target descriptions, in agreement with the no-signalling theorem.

Represent spin by $J_i=(\hslash /2){\sigma }_i$ in the basis $|\uparrow ⟩,|\downarrow ⟩$. The Pauli matrices are Hermitian and satisfy $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$, giving the general angular momentum commutation relations $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$. The particular representation has $J^2=3{\hslash }^{2}I/4$ and $J_3$ eigenvalues $\pm \hslash /2$, the features specific to spin one-half.

In the stated order of two-particle basis vectors, the operator isIts eigenvalue-one vectors are $|\uparrow \uparrow ⟩$, $|\downarrow \downarrow ⟩$ and $(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$, all symmetric under interchange. The eigenvalue-minus-three vector is $(|\downarrow \uparrow ⟩-|\uparrow \downarrow ⟩)/\sqrt{2}$, antisymmetric under interchange. These are the triplet state and singlet state, respectively.

Identical spin-half particles are fermions. With no other degrees of freedom to carry antisymmetry, only the singlet is allowed. Indeed, for total angular momentum $J=(\hslash /2)({\sigma }^{(A)}+{\sigma }^{(B)})$,This gives $2{\hslash }^2$ on the triplet and zero on the singlet, corresponding to total spin one and zero, as required by addition of angular momentum.

Both particles can occupy $l=0$ with opposite spins, giving a unique singlet state at $E=0$, quantum degeneracy 1. At the next distinct energy ${ϵ}_R$, one orbital is $l=0$ and the other is $l=+1$ or $-1$. For each orbital choice there is one symmetric spatial state paired with the singlet state, and one antisymmetric spatial state paired with three triplet state spin projections. HenceSwapping particle labels adds no independent state after the required exchange symmetry is imposed.

The lowest singlet state isFor the lowest triplet states take $\lambda =\pm 1$ and $m_s=-1,0,1$:orthogonality of $u_0$ and $u_{\lambda }$ verifies wavefunction normalization of the antisymmetric spatial factor. At this same energy there are also two excited singlet states, obtained by replacing its minus sign with a plus sign and the spin factor by $|0,0⟩$. Together they reproduce the eight states counted in part a.

At coincidence ${\varphi }_1={\varphi }_2$, the triplet state spatial factor vanishes, while the lowest singlet state has finite, uniform pair probability. Consequently a weak short-range repulsion raises the singlet state energy more strongly: antisymmetry suppresses close encounters in the triplet state. In the contact-interaction limit the triplet state contribution is exactly zero, whereas the singlet state shift is positive. A finite-range interaction gives a nonzero but suppressed triplet state shift. This explains a relative energy increase; it does not by itself establish a crossing or justify using unchanged noninteracting wavefunctions for arbitrarily strong repulsion.

For independent angular momenta ${\mathbf{J}}_1,{\mathbf{J}}_2$, set $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Since different subsystems commute, $\mathbf{J}$ satisfies the same angular-momentum algebra. The product basis $|j_1{m}_1⟩|j_2{m}_2⟩$ diagonalizes $J_1^2,J_2^2,J_{1z},J_{2z}$; the coupled basis diagonalizes $J_1^2,J_2^2,J^2,J_z$. Its labels obeyThe Clebsch-Gordan coefficients are the unitary change of basis between them. Start from the maximum-weight state $j=j_1+j_2,m=j$, apply $J_{-}=J_{1-}+J_{2-}$ with wavefunction normalization, and construct the remaining multiplets by orthogonality.

For two spin-one-half systems, the symmetric triplet state is $|\uparrow \uparrow ⟩$, $(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, $|\downarrow \downarrow ⟩$, with $j=1$; the antisymmetric singlet state $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ has $j=0$. The dimension identity ${\sum }_j(2j+1)=(2j_1+1)(2j_2+1)$ checks completeness. Coupling orbital and intrinsic spin similarly gives total atomic angular momentum and organizes spin-orbit energy splittings. Exchange symmetry is an additional restriction when the constituents are identical.

Addition of two spin-one-half angular momenta gives total spin $S=0$ or $S=1$. The normalized singlet state and triplet states areThey have $S^2$ eigenvalues $0$ and $2{\hslash }^2$, and $S_z=M\hslash$. The singlet state is antisymmetric under a particle exchange operator, whereas all three triplets have symmetric spin-exchange parity. For two identical electrons the full state must be antisymmetric: a singlet state pairs with a spatial state with symmetric exchange parity, a triplet state with an antisymmetric spatial state. Thus two electrons in the same atomic orbital must form a singlet state.

The identity ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=(S^2-S_1^2-S_2^2)/2$ gives $-3{\hslash }^2/4$ in the singlet state and $+{\hslash }^2/4$ in the triplet state. Hence an exchange interaction distinguishes their energies, with ordering determined by its coupling sign. The singlet state has perfect opposite-spin correlations along any common measurement axis and is entangled. The $M=0$ triplet state is also entangled, while the $M=\pm 1$ triplet states are product states along the chosen axis.

Indistinguishable particles have the same intrinsic properties and cannot be assigned physically observable permanent labels. Exchanging their mathematical coordinates leaves all physical predictions unchanged. In ordinary three-dimensional quantum mechanics the total state, including spin, obeysThe Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions. Exchange does not mean physically transporting two particles along some prescribed trajectories; it is the symmetry of the many-particle state under permutation of its arguments.

For orthonormal one-particle states $u,v$, normalized two-particle spatial combinations are $[u(1)v(2)\pm v(1)u(2)]/\sqrt{2}$. They illustrate coherent exchange amplitudes. If both fermions have the same complete one-particle state, the antisymmetric combination vanishes. This is the Pauli exclusion principle: a fermionic mode has occupation zero or one. Bosonic modes can have arbitrarily large occupation. A many-fermion product basis is antisymmetrized by a Slater determinant.

It is the full spin-spatial state that must have the appropriate symmetry. For two electrons, an antisymmetric spin singlet state can accompany a symmetric spatial orbital, so opposite-spin electrons can occupy the same spatial orbital. A symmetric spin triplet instead needs an antisymmetric spatial state. Thus exclusion concerns identical complete states, not an unconditional prohibition on sharing a position or spatial orbital.

Counting allowed occupations gives Bose-Einstein statistics and Fermi-Dirac statistics. For an ideal gas in thermal equilibrium,Bosonic accumulation permits Bose-Einstein condensation; exclusion gives a filled Fermi sea and degeneracy pressure even at zero temperature. In the dilute classical limit the occupations are small and both expressions approach the Maxwell-Boltzmann distribution. Exchange effects become important when wavepackets overlap appreciably, often characterized by number density times the cube of the thermal wavelength.

Composite particles follow their total exchange statistics while remaining intact. The composite statistics of neutral atoms depend on electrons as well as nucleons, as in the two lithium isotopes above. Symmetrization can produce bunching or antibunching in joint detection probabilities even in the absence of an interaction potential; it is not an additional classical force.

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.

For a double quantum dot, write $(n_L,n_R)$ for occupation. Tune a bias window allowing the cycle $(0,1)\to (1,1)\to (0,2)\to (0,1)$. Two electrons in the same lowest orbital must form a singlet state by the Pauli exclusion principle. The $(0,2)$ triplet requires an excited orbital, with an energy cost ${\Delta }_{ST}$ above the accessible window. Approximately spin-conserving tunnelling transfers $S(1,1)$ into $S(0,2)$, but cannot transfer $T(1,1)$ into that singlet. Once a triplet loads, transport stops until spin relaxation or mixing occurs: this is Pauli spin blockade. The reverse bias can avoid trapping in this sequence, producing rectification. Coulomb charging sets the allowed occupancies, but the extra blockade is a spin-selection restriction, not merely an energetically forbidden charge state.

magnetically driven spin-blockade lifting uses electron paramagnetic resonance. A transverse oscillating magnetic field drives spin transitions atFlipping one spin in $|\uparrow \uparrow ⟩$ gives $|\downarrow \uparrow ⟩=(|T_0⟩-|S⟩)/\sqrt{2}$, which has a singlet component and can resume charge transfer. A perfectly uniform simultaneous rotation of two identical spins preserves their triplet character; effective relative rotation needs unequal local resonance conditions or another singlet–triplet mixing mechanism. In a real GaAs double dot, inhomogeneous nuclear fields can supply that difference. Hyperfine mixing, spin-orbit effects, cotunnelling and ordinary relaxation also produce leakage and must be distinguished from driven resonance.

A demonstrated device used surface Ti/Au gates to define two dots in a GaAs/AlGaAs electron gas $90\mathrm{nm}$ below the surface. An insulated on-chip gold coplanar stripline supplied the oscillating magnetic field. With the dots biased in spin blockade, RF excitation produced current peaks at the two field polarities satisfying resonance. Peak positions varied linearly with frequency, giving $|g^{\ast }|=0.35\pm 0.01$. RF applied to a side gate instead gave negligible response, testing magnetic rather than electric excitation. Gate pulses and RF bursts further produced coherent spin-rotation oscillations in the charge readout.^[ref]

An ideal isolated resonant spin has Rabi oscillation probability $P_{\mathrm{flip}}(t)={\sin }^2({\Omega }_{R}t/2)$. For a linearly oscillating field of peak amplitude $B_{\mathrm{ac}}$, the rotating-wave convention is ${\Omega }_R=|g^{\ast }|{\mu }_B{B}_{\mathrm{ac}}/(2\hslash )$. Repeated initialization, isolated manipulation and blockade-sensitive readout can therefore map a damped oscillatory singlet yield into an average current. Nuclear fluctuations, readout rates and dephasing change the envelope and contrast; an ideal sine squared is not a reconstruction of experimental data. Compare RF on/off, resonance tracking with frequency, drive-amplitude dependence and detuning controls. Current enhancement alone does not prove coherent rotations; the pulse-duration oscillations provide the further time-domain test.

An electrostatically defined quantum dot uses depletion gates to confine a small island laterally in a two-dimensional electron gas. Two narrow barriers allow weak tunnelling to source and drain; a plunger gate tunes the island's electrochemical levels, and additional gates can form a double quantum dot. Adjusting the gates resolves successive charge states and eventually reaches the single-electron regime. An electrical dot is defined by confinement and tunnel coupling, not simply by the drawn metal area.

In a constant interaction model,Here $Q_g$ is gate-induced charge and $C_{\Sigma }$ total island capacitance. Successive additions cost $E_{\mathrm{add}}=e^2/C_{\Sigma }+{ϵ}_{N+1}-{ϵ}_N$, with spin, exchange and correlations modifying the orbital term. Coulomb blockade appears when no allowed charge transition lies in the source–drain energy window. Gate sweeps give charge steps and transport peaks; finite-bias Coulomb diamonds constrain charging and orbital energies under a stated bias convention. Resolved single-electron operation requires thermal and tunnel broadening smaller than the relevant addition energies.

Quantum-dot charge sensing avoids relying on a measurable current through a nearly isolated dot. A nearby quantum point contact biased on a sensitive conductance slope, or a single-electron transistor, responds capacitively when one electron enters or leaves the dot. A time trace of sensor current can therefore distinguish $N$ from $N-1$ and resolve individual tunnelling events. The sensor measures charge; a spin-dependent step is needed to measure spin.

For one electron, Zeeman splitting gives a lower and a higher spin level separated by $E_Z=|g^{\ast }|{\mu }_B|B|$. In energy-selective spin readout, initialize an empty dot and pulse both levels below the reservoir chemical potential to load one electron. Then move to a readout configurationAn electron in the higher state can tunnel out, temporarily leaving the dot empty, after which a lower-state electron can refill it. An electron initially in the lower state remains trapped. The charge sensor records a pulse for the first outcome and no escape pulse for the second. This is spin-to-charge conversion; it measures the spin projection along the field through a state-dependent tunnelling probability. A primary single-shot experiment uses this conversion with a quantum-point-contact charge detector.^[ref]

The readout window must be long enough to detect escape but short compared with excited-spin relaxation:Detector bandwidth must also resolve the empty-state dwell time before refilling. Thermal activation, relaxation before escape, missed fast events and sensor backaction cause errors. Since the excited electron can leave and be replaced, this simple sequence does not preserve an arbitrary initial spin state.

For coupled dots, label charge occupation by $(n_L,n_R)$. Gate-controlled detuning can make $(1,1)$ and $(0,2)$ accessible. Two electrons in the lowest orbital of one dot must form a singlet state; a triplet state requires a different orbital or another higher-energy configuration. Choose a measurement window where the $(0,2)$ singlet is available but the $(0,2)$ triplet is not. Thenunder spin-conserving tunnelling. The second process is forbidden by spin symmetry and the Pauli exclusion principle, producing Pauli spin blockade. Charge sensing distinguishes transfer from a blocked $(1,1)$ state. This measures singlet versus triplet content; without a reference spin or additional control it is not a measurement of each electron's absolute spin direction separately.

The exchange interaction in a double quantum dot provides control of these states. In an effective spin description,For a symmetric weak-tunnelling two-site model with onsite charging cost $U$, $J\simeq 4t_c^2/U$. Gates change tunnel coupling $t_c$ and detuning, allowing preparation and evolution followed by a charge-conversion measurement. Spin mixing from local magnetic-field differences or spin-orbit effects can lift blockade, so the sensing interval must be chosen before unwanted relaxation obscures the initial state. Single-dot readout uses spin-dependent escape; double-dot readout can use spin-dependent charge transfer. In both cases the measurable signal is a resolved charge change.