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Consider two distinguishable, noninteracting two-level systems, each with energies $0$ and $\Delta >0$. Their four microstates $(0,0),(0,1),(1,0),(1,1)$ have total energies $0,\Delta ,\Delta ,2\Delta$. Thus the three energy macrostates have state degeneracies $1,2,1$. This is a stipulated finite model, not a measured material.

Set $q=e^{-\beta \Delta }$. The four individual Boltzmann factors are $1,q,q,q^2$, so the partition function and aggregate probabilities areEach of the two middle microstates has probability $q/Z$, while their shared energy macrostate has probability $2q/Z$.

At the temperature satisfying $\Delta =k_{B}T\ln 2$, the Boltzmann factor is $q=1/2$ and the partition function is $Z=9/4$. ThereforeThe middle energy is as likely as the ground state energy because its state degeneracy is twice as large. Yet each individual middle microstate is only half as likely as the ground state microstate. The diagram places the four energies beside their individual probabilities, then shows the sums within each macrostate.

The expected value of energy follows directly by weighting the three possible energies:At $q=1/2$ this is $2\Delta /3$. An individual energy measurement still returns $0$, $\Delta$ or $2\Delta$; the expected value describes the ensemble average, rather than a fourth possible measured energy.

Take $N=20$ distinguishable, labeled particles and two boxes of equal volume. Retain only whether each particle is in the left or right box: use a bit $0$ for left and $1$ for right. Each length-20 string is one coarse microstate of this toy model. Assume the labels make independent side choices with equal probability, so the $2^{20}=1048576$ strings are equiprobable. Equal volumes motivate these weights for noninteracting particles with no preference for either box; equal volumes alone would not establish them for an arbitrary interacting gas.

Now retain only the number $k$ in the left box as the macrostate. Choosing which $k$ labels are left gives a binomial coefficient, soThe factorials count assignments to labels; the probability of a macrostate sums the probabilities of its constituent microstates. This is the binomial distribution with parameters $20$ and $1/2$. The same binomial coefficient appears in the counting of distinguishable two-level systems, although here the macrostate records occupancy rather than energy. ^[ref]

The largest binomial coefficient occurs at $k=10$, the balanced macrostate. All-left means $k=20$ and supplies just one string. ThusThere are $184756$ balanced strings for every all-left string, even though each individual string has the same probability. Exact balance has about $17.6\%$ of the total probability; equilibrium reasoning concerns the many macrostates near balance too, rather than certainty of exactly ten particles on each side. The diagram compares the counts with their natural logarithms.

Four fair, independent coins give a smaller worked example with the same binomial distribution. If $k$ is the number of heads, the five macrostates $k=0,1,2,3,4$ contain respectively $1,4,6,4,1$ microstates. Their probabilities are respectively $1/16,4/16,6/16,4/16,1/16$. In particular, two heads is one macrostate containing six ordered coin sequences, with Boltzmann entropy $k_B\ln 6$ in this counting analogy. A prescribed ordered sequence has one realization. The next diagram enumerates the same sixteen strings as left/right assignments.

The canonical ensemble is the probability distribution for a system in thermal equilibrium with a heat bath at fixed temperature $T$, with fixed particle number and external parameters. It allows energy exchange, so individual energy measurements fluctuate. With Boltzmann constant $k_B$ and inverse temperature $\beta =1/(k_{B}T)$,The heat bath is idealized as large enough that exchanges barely change its temperature; coupling must permit equilibration while contributing negligible interaction energy. A finite heat bath approximates this description only when its temperature change is negligible over the relevant exchanges. Strong coupling or a small heat bath can require a different description.

For one two-level system at positive temperature, the Boltzmann weights are $1,e^{-\beta \Delta }$, henceAt $\beta \Delta =\ln 3$, $Z_1=4/3$ and $p=1/4$. This is a probabilistic thermal mixed state. For quantum levels, its density matrix is $\rho =(1-p)|0⟩⟨0|+p|1⟩⟨1|$. The pure state $\sqrt{1-p}|0⟩+\sqrt{p}|1⟩$ has the same energy measurement probabilities, but also has quantum coherence between the energy eigenstates. Matching those probabilities does not make the states equivalent.

The partition function $Z(\beta )={\sum }_s{e}^{-\beta E_s}$ is the sum of the Boltzmann weights over individual microstates. Dividing by it makes their probabilities sum to one. When summing distinct energies instead, include the state degeneracy: $Z={\sum }_{E}g(E)e^{-\beta E}$. For two independent distinguishable two-level systems,The middle coefficient counts two microstates, exactly as in the preceding figure. For $N$ such units, $Z_N=Z_1^N$; this factorization requires additive energy and independence, and generally fails with interactions.

The usefulness goes beyond normalization. For fixed energies, derivatives of the natural logarithm of the partition function yield measurable quantities:Here $U$ is internal energy, the second expression is the energy variance, and $S$ is Gibbs entropy. For one two-level system at fixed $\Delta$, its heat capacity $C=\partial U/\partial T$ and entropy areAt $\beta \Delta =\ln 3$, these give $U/\Delta =0.25$, $C/k_B\simeq 0.2263$ and $S/k_B\simeq 0.5623$. For independent units, all three quantities add.

The figure shows why heat capacity peaks. At low temperature, the excited energy is rarely occupied; at high temperature, the probabilities are almost equal and further warming barely changes internal energy. The largest response lies between these limits, at $k_{B}T/\Delta \simeq 0.4168$, where $C/k_B\simeq 0.4392$. Meanwhile entropy rises from zero toward $k_B\ln 2$, reflecting two equally likely alternatives. This high-temperature limit belongs to the stipulated two-level system; extra accessible energy eigenstates would change it.

An oven supplies a collimated caesium beam. State-selecting magnets precede and follow two coherent microwave interaction regions; a weak static magnetic field separates sublevels. The clock transition connects $F=3,m_F=0$ and $F=4,m_F=0$. Transferred atoms reach a heated ionizing collector. A detector scan identifies the resonance before locking.^[ref]

Here $F$ labels the total angular momentum and $m_F$ its projection. The selected pair is a two-level system inside a larger set of atomic energy levels; treating every oven atom as a prepared qubit would overstate the preparation. A measured detector signal may be modeled as $S(\nu )=S_{\mathrm{bg}}+A\overline{P}(2\pi ({\nu }_0-\nu ))$, with background $S_{\mathrm{bg}}$ and gain $A>0$. Background subtraction and stable transmission are necessary before interpreting the signal as a transition probability. The velocity-averaged Ramsey line provides one explicit ideal example; its chosen dimensions are not historical specifications.

Use a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$. Here $\Gamma$ is a decay rate in inverse seconds and also the full width at half maximum of the natural linewidth expressed in angular frequency. The corresponding full width in hertz is $\Gamma /(2\pi )$; the angular half width is $\Gamma /2$. These conventions determine every factor of two and $2\pi$ below.

Let $s$ be the on-resonance saturation parameter of each beam, with total excitation weak enough that saturation and coherent coupling between beams can be neglected. The independent-beam scattering rates and net force areThe force is the photon momentum per absorption multiplied by the difference in absorption rates. With dimensionless laser detuning $d=\delta /\Gamma$ and dimensionless velocity $u=kv/\Gamma$, writeFor the example $d=-0.5$ and $u=0.25$, the positive and negative beam factors are respectively $4/13$ and $4/5$. Thus $F(0.25)=-32/65\simeq -0.4923$: the atom is moving right and the net force points left. All values in this example are dimensionless.

Subtracting the two fractions gives an especially useful sign check:For red detuning, the force opposes nonzero $v$ throughout this ideal model; its useful magnitude eventually falls at large $|u|$. Near zero, a Taylor series gives dampingAt $d=-0.5$, $F^{\prime }(0)=-2$ and $\alpha =\hslash k^{2}s$. For blue detuning the sign reverses and the small-velocity motion is amplified; at exact resonance the opposing mean forces cancel in this symmetric model. Its derivation assumes weak excitation, a closed two-level system, equal beam intensities and nonrelativistic velocities; real multilevel atoms need additional physics.

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

Use the Driven two-level Rabi model in the ordered basis $(|e⟩,|g⟩)$, with Pauli matrices satisfying ${\sigma }_z=\mathrm{diag}(1,-1)$. Let the Rabi detuning be $\delta ={\omega }_0-\omega =2\pi ({\nu }_0-\nu )$, in angular frequency, while ${\nu }_0$ and $\nu$ are ordinary frequencies in hertz. Assume two identical square pulses, equal drive oscillation phases, constant positive Rabi angular frequency $\Omega$, constant $\delta$ during pulses and flight, and no quantum decoherence or loss. The rotating-wave approximation and isolation of the chosen two-level system are required, just as for the Driven two-level Rabi model. Free flight means no transverse drive; the reference oscillator continues running.

The pulse and free-flight Hamiltonian operators in the rotating frame areThe Pauli matrix multiplication law gives $(\delta {\sigma }_z+\Omega {\sigma }_x{)}^2=q^{2}I$. Thus the exact special unitary group $\mathrm{SU}(2)$ pulse matrix for this effective Hamiltonian operator, and the flight matrix, areThe sequence is $U_p{U}_f{U}_p$. Starting in $|g⟩$, matrix multiplication and the Born rule giveThis retains the tilted rotation axis and quantum phase accumulated during each finite pulse. At resonance it gives $P_{g\to e}(0)={\sin }^2(\Omega \tau )$; with no flight it reduces to the single square-pulse response for duration $2\tau$. Both checks would fail if arbitrary ideal $\pi /2$ rotations were silently substituted at nonzero Rabi detuning.

For resonant pulse area $\Omega \tau =\pi /2$, $|\delta |/\Omega \ll 1$ and $\tau \ll T$, the central fringe approachesIts half-height points are $|\nu -{\nu }_0|=1/(4T)$, so the central peak's full width at half maximum is $1/(2T)$ in hertz, or $\pi /T$ in angular frequency. This width assumes an effectively flat pulse envelope across that central peak, one flight time and unit contrast; it is neither a universal finite-pulse width nor a frequency uncertainty. Longer flight lets a smaller frequency difference accumulate the same observable quantum phase.

The two pulses here have the same sign and the plotted outcome is transfer to $|e⟩$. The earlier Ramsey interferometry example uses an inverse analysis pulse and reports return to its initial label. Specifying the measured outcome and pulse oscillation phases reconciles the two conventions. In a beam the lengths and speed determine $\tau$ and $T$; in the Quantronium Ramsey interference experiment electronics set the pulse times for a stationary circuit.

The ladder identity $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ proves the matrix element in the title. In time-dependent perturbation theory, a linear force excites only $n=1$ directly from $n=0$. Parity alone rules out even excited states; the ladder identity also rules out odd $n>1$. Higher-order evolution can reach other levels, so the driven oscillator is not an exact two-level system.

Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

A macrostate groups microstates sharing prescribed coarse data. The grouping depends on what is retained: here retain only total energy. For the two distinguishable two-level systems, the macrostates $E=0,\Delta ,2\Delta$ have state degeneracies $1,2,1$. Although two units are not macroscopic in size, they make this loss of information visible: knowing $E=\Delta$ does not tell us which unit is excited.

At inverse temperature $\beta \Delta =\ln 3$, each unit's excited-state probability is $p=1/4$. Their independence gives microstate probabilities $9/16,3/16,3/16,1/16$ in the order above. Adding within each macrostate givesThe total expected value of energy is therefore $\Delta (6/16)+2\Delta (1/16)=\Delta /2$, matching two units each with expected value of energy $\Delta /4$. Equal state degeneracies would give the wrong answer: macrostate probabilities combine counting with the Boltzmann factor. At very high positive temperature, all four microstates become equally likely, but these three macrostate probabilities approach $1/4,1/2,1/4$.

Maser state selection separates paths according to the Stark effect, the dependence of a molecule's energy levels on an electric field. It prepares a transmitted population by sorting. Optical pumping, by contrast, changes populations through absorption of light and subsequent transitions. Both can produce a useful population imbalance, but they are different preparation mechanisms.

A simple two-level system explains the sign of the focusing force. Ignore additional splittings and let the magnitude of the relevant projected electric dipole moment be $d$. Model the two local energy levels in an electric field of magnitude $\mathcal{E}$ byFor an ideal quadrupolar electrostatic potential $\Phi =A(x^2-y^2)$, the electric field magnitude is $\mathcal{E}=2|A|r$, where $r=\sqrt{x^2+y^2}$. Since $W_{+}$ increases with $r$, the force ${\mathbf{F}}_{+}=-\nabla W_{+}$ points inward. The lower branch has the opposite sign and is defocused. The upper branch is thus low-field seeking. This argument uses the gradient of internal energy, rather than a net electric charge on the molecule. The square-root splitting is the usual two-state description.^[ref] The idealized two-level system omits the actual rotational and hyperfine structure sublevels and their different transmission efficiencies.

A microstate is a complete microscopic specification within the chosen model. For two distinguishable, noninteracting two-level systems, label the units $A,B$ and their states $0,1$, with energies $0,\Delta$ and $\Delta >0$. The four microstates arewith total energies $0,\Delta ,\Delta ,2\Delta$. Distinguishability matters: $(0,1)$ and $(1,0)$ describe different assignments to the labeled units. Noninteraction matters too: their energies add without an extra coupling term.

For a classical particle model, a microstate can instead be a point in phase space, specifying every position and momentum. In the quantum canonical ensemble, the sum uses energy eigenstates; a general quantum state may also contain quantum coherence, which an energy label alone does not specify.

A central peak and two sidebands in the resonance fluorescence spectrum of a sufficiently strongly driven two-level system. The atom-field dressed state splitting produces sidebands near ${\omega }_L\pm \sqrt{{\Omega }^2+{\Delta }^2}$, where $\Omega$ is the resonant Rabi frequency. On resonance, changing the drive intensity changes the sideband spacing. Damping and detection conditions determine the resolved line shapes; the elastic coherent contribution must be distinguished from inelastic fluorescence.^[ref]

An experimental test drives an isolated transition with a narrow-band laser, collects fluorescence away from the intense drive beam, and scans a spectral analyzer while varying drive power. Resolving the sidebands requires spectral resolution and coupling frequency sufficient relative to the radiative linewidth. The predicted spectrum concerns transitions of the coupled system rather than simply the time dependence of its population.

For a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$ and no additional dephasing, the natural linewidth has full width at half maximum $\Gamma$ in angular frequency, or $\Gamma /(2\pi )$ in hertz. The excited-state lifetime is $1/\Gamma$.

For two dimensionless observables on a two-level system, takeBoth are Hermitian matrices: $A$ has eigenvalues $1,0$ and $B$ has eigenvalues $1,-1$. YetFor instance, $AB|0⟩=0$ whereas $BA|0⟩=|1⟩$. In $AB$, $B$ acts first. Neither product is self-adjoint, so the product of two observables is not automatically another observable. The commutator records this dependence on order. These operator products by themselves do not specify the outcomes of consecutive measurements; that additionally requires measurement in quantum mechanics.

Optical pumping redistributes populations among internal quantum states by selective photon absorption followed by spontaneous emission. It can prepare atoms for measurement and participate in sub-Doppler cooling.

The practical aim is to make otherwise almost equally populated magnetic sublevels behave differently. Repeated absorption and emission can store a population imbalance in the long-lived ground state; a change in photon absorption or fluorescence then reveals a perturbation of that prepared distribution. This combines preparation and measurement: a weak radio wave can reorganize the populations, while abundant optical photons supply the readout.

An imbalance among ground state magnetic sublevels need not be a population inversion between optical energy levels. Optical pumping also supplies excitation in some lasers, where an appropriate population inversion matters for gain. The preparation problem here concerns the distribution within the ground state. A two-level system with only one lower quantum state cannot describe this redistribution.

A laser combines pumping, an amplifying medium and optical feedback. Stimulated emission adds photons to an occupied optical mode; net amplification requires population inversion between the relevant levels after accounting for their degeneracies. Pumping through additional levels and rapid lower-level relaxation can sustain inversion; a closed two-level system under steady incoherent resonant pumping cannot sustain it.

An optical resonator selects modes and feeds light back through the gain medium. For intensity gain coefficient $g$, internal intensity loss ${\alpha }_i$, length $L$ and mirror intensity reflectivities $R_1,R_2$, the round-trip laser threshold condition isAbove threshold, saturation limits gain and a partially transmitting mirror provides output. Typical properties are strong directionality, substantial coherence and narrow bandwidth; multiple modes and pulsed operation are possible, so perfect monochromaticity is not universal.^[ref]

Let $A_{21}$ be the spontaneous emission coefficient, and let $B_{12}\rho$ and $B_{21}\rho$ be the absorption and stimulated emission rates per atom under the radiation spectral energy density $\rho$ at resonance. Because the two levels are nondegenerate, the Einstein coefficients satisfy $B_{12}=B_{21}=B$.

The radiative balance in a nondegenerate two-level atom isThe total spontaneous and stimulated emission rates are $R_{\mathrm{sp}}=N_2{A}_{21}$ and $R_{\mathrm{st}}=N_{2}B\rho$. Dividing the balance equation by $N_{2}B\rho$ givesThis presumes nonzero drive and upper-level population, and no other population-transfer mechanism. It does not require a thermal Boltzmann distribution. In this closed driven two-level system, spontaneous decay keeps $N_1>N_2$; arbitrarily intense driving approaches equal populations, rather than producing a population inversion.

Expand the wavefunction in unperturbed energy eigenstates,The $c_j$ are interaction picture probability amplitudes, so $|c_j{|}^2$ are level populations and $|c_1{|}^2+|c_2{|}^2=1$ in this closed two-level system. A classical field ${\mathbf{E}}_0\cos \omega t$ supplies the electric-dipole interaction $-\mathbf{d}\cdot {\mathbf{E}}_0\cos \omega t$. Choosing the relative phase of the energy eigenstates makes $\Omega =-⟨1|\mathbf{d}\cdot {\mathbf{E}}_0|2⟩/\hslash$ real. This is the resonant Rabi angular frequency.

Combining the field's two exponentials with the unperturbed quantum phases produces terms oscillating at $\omega -{\omega }_0=\delta$ and at $\omega +{\omega }_0$. The rotating-wave approximation drops the latter, leaving exactly the supplied equations with coupling $\Omega /2$. It requires $|\Omega |,|\delta |\ll {\omega }_0$ and a field envelope slow enough on the carrier timescale. Spontaneous emission and dephasing must be negligible during the sequence. For a ground-state hyperfine level transition, the analogous direct drive is a magnetic dipole transition; the same equations hold with the magnetic matrix element replacing the electric one.