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Use $\hslash =1$ throughout. Let $N_a=a^{†}a$, $N_b=b^{†}b$ and $N=N_a+N_b$. Operators belonging to different boson species commute. The proposed Schwinger boson representation givesConsequently the first of the spin commutation relations isThe number operator commutators $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$ and their $b$ counterparts giveThus all the spin commutation relations hold. Also $[N,S^{\pm }]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its spin, calculate the Casimir operator:The irreducible spin representation with prescribed $s$ therefore requires $N=2s$, and its normalized Fock basis correspondence isConversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer occupation numbers give precisely $m=-s,-s+1,\dots ,s$, and there are $2s+1$ such states. As a check, the spin raising operator acts aswhich is the usual spin ladder operator coefficient $\sqrt{s(s+1)-m(m+1)}$.

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

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.

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.

The $4f^7$ shell carries a large spin (its ground atomic term under Hund's rules has $L=0,S=7/2$), and the $6s$ electron penetrates the inner shells. Its nonzero probability density at the atomic nucleus supplies a potentially large Fermi contact interaction with the nuclear magnetic dipole moment. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable europium isotopes have nonzero nuclear spin, as the spectrum below also establishes.

The term symbol ${}^8{D}_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable nuclear spin, permits a wide range of total angular momentum $F$ and hence a substantial total hyperfine structure span. For $I=5/2$, $F=3,\dots ,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large hyperfine structure; the term symbol alone does not determine its constant, which also depends on electronic recoupling, radial wavefunctions and configuration interaction.

For a nucleus of spin $I>0$ and electronic $J=1/2$, the magnetic-dipole hyperfine interaction is $A\mathbf{I}\cdot \mathbf{J}$, using dimensionless angular-momentum operators. The electric-quadrupole hyperfine contribution vanishes for $J=1/2$. In zero field, $F=I\pm 1/2$ labels the levels:Their signed separation is $\Delta =A(I+1/2)$. For $A>0$, the larger-$F$ multiplet lies higher; negative $A$ reverses this order. If $I=0$ there is no such hyperfine splitting.

Let the field define $z$ and writeHere $g_I$ is the signed nuclear Zeeman coefficient expressed in Bohr-magneton units, including its nuclear-to-Bohr-magneton factor; it is not an unconverted nuclear $g$ factor. Total projection $m=m_I+m_J$ is conserved, while $F$ need not be. In the basis $|m-1/2,+1/2⟩$, $|m+1/2,-1/2⟩$, the two-state block isFor $A>0$, define $x=(g_J-g_I){\mu }_{B}B/\Delta$. Diagonalization gives the Breit-Rabi formula,For either sign of $A$, the block itself remains valid; its ordered eigenvalues are $-A/4+g_I{\mu }_{B}Bm\pm \frac{1}{2}\sqrt{{\Delta }^2+2Am(g_J-g_I){\mu }_{B}B+(g_J-g_I{)}^2{\mu }_B^2{B}^2}$.

For a stretched hyperfine Zeeman state at projection $m=\pm (I+1/2)$ there is only one state each. Their energies are exactlyAssigning two branches to each stretched projection would double-count states.

In weak fields the shift is $g_F{\mu }_{B}mB$, with $g_F=g_I\pm (g_J-g_I)/(2I+1)$ for $F=I\pm 1/2>0$; an $F=0$ state has no first-order Zeeman shift. At strong hyperfine-decoupling fields the eigenstates approach product states withThis is the hyperfine regime of the Paschen-Back effect, still assuming the field is weak enough not to invalidate the electronic $J=1/2$ manifold. States of the same $m$ mix and avoid crossing when their off-diagonal coupling is nonzero; different-$m$ levels can cross.

For hydrogen, the leading characteristic scales areOrder-one spin and nuclear magnetic factors, and state-dependent factors, are suppressed here. Gross atomic structure is Coulomb binding. Fine structure includes relativistic kinetic and spin-orbit effects, with Thomas precession affecting the spin-orbit coefficient. Hyperfine structure couples electronic motion and spin to the nuclear magnetic moment, whose inverse nuclear mass makes it much weaker. The contact spin interaction can remain in an s state where orbital spin-orbit coupling vanishes.

For positronium, the reduced mass is $m_e/2$. Its gross levels and separation scale areThus the mean ground-state electron–Positron separation is twice the infinite-proton-mass hydrogen value: both means have the same factor $3/2$ multiplying their respective Bohr radii. Retaining proton recoil makes the ratio $2m_p/(m_p+m_e)$, very slightly below two. Separation is not the distance of either positronium constituent from its centre of mass.

Interpret the spin operators in units of $\hslash$, as required by the printed Zeeman term, so $A$ has units of energy. Two spin halves give four eigenstates. At zero field, total spin $F=0,1$ and projection $m_F$ label the singlet and triplet. At high field relative to the hyperfine scale, the individual projections $m_{s,e},m_{s,p}=\pm 1/2$ become approximately good quantum numbers. Total projection remains exact at every field, but $F^2$ does not for the two mixed states.

Choose the quantization axis along $\mathbf{B}$. The aligned-spin states $|\uparrow \uparrow ⟩$ and $|\downarrow \downarrow ⟩$ both have $E=A/4$ because their Zeeman shifts cancel. In the other product-state sector,For physical positronium $A>0$. At $B=0$ there is a triplet at $A/4$ and a singlet at $-3A/4$. At large positive field the mixed states approach $|\uparrow \downarrow ⟩$ and $|\downarrow \uparrow ⟩$, with energies $-A/4+2{\mu }_{B}B$ and $-A/4-2{\mu }_{B}B$ respectively, while the aligned pair remains degenerate at $A/4$.

The dimensionless mixing parameter is $x=4{\mu }_{B}B/A$. A crossover estimate is $x\sim 1$, rather than a sharp transition. The scale $A\sim {\alpha }^4{m}_e{c}^2$ already gives a field of a few tesla. Using the physical hyperfine interval $A/h\simeq 203.39\mathrm{GHz}$ givesThe numerical hyperfine interval is external physical input, not a number supplied by the symbolic Hamiltonian.^[ref] The high-field limit here still presumes that orbital distortion is negligible compared with Coulomb binding.

Set the ground-state energy to zero. Bose-Einstein occupation of an excited level is $[e^{(ϵ-\mu )/(k_{B}T)}-1{]}^{-1}$, with $\mu \le 0$. As particles are added, $\mu$ approaches zero and the excited-state population saturates. Separating the ground state from the density-of-states integral therefore givesThe excited-state integral must actually converge: $g\to 0$ alone is not a sufficient condition in complete generality. A power law $g\propto {ϵ}^s$ with $s>0$ suffices at low energy, as in the stated three-dimensional box.

Let $\mathcal{C}=V{m}^{3/2}/(\sqrt{2}{\pi }^2{\hslash }^3)$ denote the density-of-states prefactor, not a heat capacity. Substitution $x=ϵ/(k_{B}T)$ givesThe condensate has zero energy on this reference, so a further factor of $ϵ$ in the integral givesRelative to the classical value, the ratio is $\zeta (5/2)/[\eta \zeta (3/2)]\simeq 0.5134/\eta <0.5134$. Macroscopic occupation of the zero-energy state and the Bose distribution both lower the average energy.

At fixed volume the condensed gas has $E\propto T^{5/2}$. Unbiased random loss keeps average $E/N$ constant, henceRemoving a condensate atom removes essentially no energy and by itself leaves the excited cloud's temperature unchanged. Removing an excited atom removes its energy; during re-equilibration, some zero-energy condensate atoms join the excited population and share the remaining energy, lowering temperature. Averaging over unbiased loss gives the stated constant energy per particle, even though each individual type of loss need not preserve it separately.

Using $m\simeq 87u$ and the supplied volume,There is only one occupied internal spin state, so no factor $2F+1$ belongs in the density of states. During loss, $\eta \propto N/T^{3/2}\propto N^{2/5}$, hence $\eta /{\eta }_i=T/T_i$. Condensation disappears at ${\eta }_f=1$:As a consistency check, $N_f=N_i{\eta }_i^{-5/2}\simeq 8.15\times 10^4$ agrees with $N_c(T_f)$.

Both this process and evaporative cooling of atoms require particle loss and re-equilibration. Ordinary evaporation preferentially removes energetic atoms, reducing $E/N$, and can cool an uncondensed classical gas. Here loss is energy independent, $E/N$ stays constant, and cooling relies on the condensed gas's unusual equation of state; random loss of a classical ideal gas would leave its temperature unchanged. Exact noninteraction would prevent collisional rethermalization: the imposed equilibrium assumption can be understood as weak thermalizing interactions whose equation-of-state corrections are neglected.

An electric dipole transition has amplitude proportional to $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$. The electric dipole operator is odd under parity, so the initial and final states must have opposite parity. As a rank-one spherical tensor it also requires $\Delta J=0,\pm 1$, excluding $J=0\leftrightarrow 0$, and $\Delta m_J=0,\pm 1$. These follow from angular-momentum addition and the Wigner-Eckart theorem; vanishing Clebsch-Gordan coefficients can impose further restrictions. Relative to the quantization axis, $\pi$ polarization drives $\Delta m=0$, while ${\sigma }^{\pm }$ drive $\Delta m=\pm 1$.

For a single active electron, parity and the orbital triangle rule combine to give $\Delta l=\pm 1$. In LS coupling, the spin-independent electric dipole operator gives $\Delta S=0$ and $\Delta L=0,\pm 1$, excluding $L=0\leftrightarrow 0$; total parity must still change. Spin-orbit mixing can weakly relax the approximate spin rule. For resolved hyperfine states the corresponding triangle rule is $\Delta F=0,\pm 1$, excluding $F=0\leftrightarrow 0$, with $\Delta m_F=0,\pm 1$. These labels require the appropriate coupling regime. There is no universal E1 rule demanding a particular change in principal quantum number. An E1-forbidden line may occur through another multipole or a multiphoton process.

Take the field along $z$ and write $x={\mu }_{B}B$, with Bohr magneton ${\mu }_B$. The printed Hamiltonian uses dimensionless angular-momentum operators, with eigenvalues $L(L+1)$ and $m_L$; physical angular momenta carry an additional $\hslash$. Thus $\beta$ and $x$ are energies. Assume the usual positive spin-orbit interaction coefficient $\beta >0$. Reversing its sign reverses the zero-field ordering but not the state count.

An electron has $S=1/2$. addition of angular momentum givesThe coupled angular-momentum state count can be checked directly. The product basis contains $(2L+1)\times 2=4L+2$ states. The coupled basis containsCoupling therefore reorganizes the states rather than creating or losing them. There is no hyperfine multiplicity because $I=0$.

Use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ to obtainThe two spin-orbit interaction energies are $\beta L/2$ and $-\beta (L+1)/2$. In a weak field, project the Zeeman effect onto a fixed $J$ multiplet:This is the first-order energy formula, not a statement that $H_Z$ has no matrix elements between different $J$ values. For a common $m=m_J$ and a conventional choice of relative phase, its complete coupled-basis block isThe inter-multiplet Zeeman coupling terms give second-order weak-field corrections because the two spin-orbit interaction multiplets are already separated.

In the strong-field basis the Zeeman part is diagonal, $E_Z=x(m_L+2m_S)$. WriteIts diagonal matrix elements are $\beta m_L{m}_S$. Its off-diagonal matrix elements connect $(m_L,+1/2)$ to $(m_L+1,-1/2)$ with coefficientand the reverse spin change has $m_L+1$ replaced by $m_L-1$ in the corresponding ladder coefficient. These terms preserve $m_J$ but mix the two product states. To first order in $\beta /x$, the energy correction is simply $\beta m_L{m}_S$.

For the P state, $L=1$. Measured relative to its gross-structure centroid, the six low-field levels areThus the upper quartet starts at $\beta /2$ with slopes $+2,+2/3,-2/3,-2$ in a plot against $x$; the lower doublet starts at $-\beta$ with slopes $\pm 1/3$. Their zero-field separation is $3\beta /2$.

At high field all six product-state energies, retaining the term linear in $\beta$, areThe two zero-slope levels have different $m_J$ and no mutual matrix element, so their first-order degeneracy does not require an additional degenerate perturbation. At finite field they split at order ${\beta }^2/x$.

An exact interpolation makes the requested whole-field sketch unambiguous. The stretched levels areFor $m=\pm 1/2$, use product states $|m-1/2,+1/2⟩$ and $|m+1/2,-1/2⟩$. Their Hamiltonian block issoThese energies reduce to the quartet and doublet at zero field and to the six strong-field expressions above. Within a fixed $m_J$, the nonzero off-diagonal element prevents a crossing of the two branches. Levels in different $m_J$ sectors can cross without hybridizing. For the S state, $L=0$ and the spin-orbit interaction term vanishes, leavingthroughout this Hamiltonian's range of validity.

For electric dipole transitions at low field, $\Delta J=0,\pm 1$ and $\Delta m_J=0,\pm 1$ apply, with the usual exclusion of $J=0\to 0$. Each P state has opposite parity to the S state. The $P_{1/2}\to S_{1/2}$ transition has four allowed sublevel pairs. For $P_{3/2}\to S_{1/2}$, the two stretched upper states each have one destination and the other two each have two, giving six. Hence there are 10 allowed state-to-state channels at weak nonzero field. Relative to the two zero-field transition energies, their Zeeman shifts are

In the decoupled high-field limit, an electric dipole acts on orbital coordinates, so $\Delta m_S=0$. Each of the six P product states connects to the S state with the same spin and $m_L^{\prime }=0$. Thus there are 6 allowed channels in the limiting product-state description. Their photon energies areThis Paschen-Back transition-channel count is not the count of distinct frequencies: to first order in $\beta$ the two $m_L=0$ channels coincide, giving 5 distinct frequencies, with shifts $0,\pm (x+\beta /2),\pm (x-\beta /2)$. If the residual spin-orbit interaction splitting is unresolved there are the three usual orbital Zeeman groups.

The apparent loss of four channels is a change in eigenstate mixing, not a change in Hilbert-space dimension. At finite large field, four weak spin-forbidden Paschen-Back lines arise from small opposite-spin admixtures, with intensities vanishing as $(\beta /x{)}^2$. A generic finite-field calculation therefore retains ten nonzero channels; six carry finite strength in the complete-decoupling limit. Observation direction and polarization select which of these channels a detector actually receives.

evaporative cooling removes atoms selectively from the high-energy tail of a trapped gas. The remaining sample has less energy per atom; elastic collisions then redistribute energy and restore a thermal distribution at lower temperature. Random atom loss would generally remove average energy and would not provide the same cooling mechanism. Repeated energy selection and rethermalization, rather than one removal event, is the essential cycle.

Using the evaporation truncation parameter $\eta$, let the escape threshold be ${ϵ}_{\mathrm{cut}}=\eta k_{B}T$. As the gas cools, a fixed threshold is populated by fewer atoms, so evaporation stalls. In forced evaporative cooling the threshold is lowered during the sequence. In a magnetic trap, a radiofrequency evaporation knife flips trapped weak-field-seeking atoms into untrapped spin states where the Zeeman splitting matches the photon energy. Energetic atoms reach that resonance surface; sweeping the radiofrequency lowers the energy knife. In an optical dipole trap, lowering the trap depth offers a different implementation, although it usually also weakens confinement.

A simple evaporative cooling energy balance shows when atom loss is useful. For a classical gas in a three-dimensional harmonic trap of fixed frequencies, equipartition gives $E=3N{k}_{B}T$, counting both kinetic and potential energy. Suppose escaping atoms remove average energy ${ϵ}_{\mathrm{ev}}=(\eta +\kappa )k_{B}T$, with $\kappa$ a distribution-dependent excess above the threshold. For $dN<0$,Thus preferential removal cools when the removed energy exceeds $3k_{B}T$. More stringently, cooling must raise the phase-space density of an atomic gas, not merely lower the temperature:For this harmonic-trap phase-space density, with fixed geometric-mean trap frequency $\bar{\omega }$, $d\ln \mathcal{D}/d\ln N=1-3\alpha$. Therefore phase-space density rises during atom loss if $\alpha >1/3$. The numerical factors here belong to the stated harmonic-trap model, not every evaporative apparatus.

There must be enough elastic collisions to repopulate the tail: ${\Gamma }_{\mathrm{el}}\sim n\sigma \bar{v}$ should exceed damaging loss and heating rates, and the ramp should allow rethermalization. Increasing density can improve elastic collisions but also enhances inelastic and three-body losses. Very rapid removal can leave a nonequilibrium cloud; very slow removal exposes it to background-gas losses. Lowering an optical trap can reduce density and collision rate, so a falling temperature is not by itself evidence of efficient evaporation.

The dilute-gas condensation experiments made this tradeoff concrete. Eric A. Cornell and Carl Wieman's account explains how selective escape and contraction of a trapped cloud raise phase-space density, while the need for rapid two-body equilibration and slow three-body loss restricts usable density. It also describes expansion imaging as an experimental diagnostic.^[ref] Useful controls compare different knife ramps and hold times, measure both atom number and expansion temperature, and check trap-frequency calibration. Pure adiabatic expansion can cool without increasing phase-space density. Successful evaporation pays an atom-number cost to increase the remaining gas's phase-space occupancy.

A two-photon transition transfers population through two interactions with the electric-dipole operator. For absorption of two photons,The intermediate atomic states enter a second-order amplitude rather than necessarily being populated. Suppressing field-normalization factors, its structure isBoth time orderings contribute coherently. Close to a real intermediate resonance, linewidths and intermediate-state population must be treated explicitly; an off-resonant virtual-state approximation then ceases to be adequate. Well away from saturation the absorption rate scales as $I_1{I}_2$, or as $I^2$ for a single applied frequency.

Each electric-dipole interaction reverses parity, so two such interactions connect states of the same parity. The product operator contains angular tensor ranks $0,1,2$, subject to triangle and polarization selection rules. Under the degenerate two-photon selection rule, identical equal-frequency photons have no antisymmetric rank-one contribution cancels; ranks zero and two remain. This permits a $J=0\to 0$ transition and excludes $J=0\to 1$ in that degenerate E1-E1 case. It does not universally forbid every $\Delta J=1$ transition, because a rank-two tensor can also connect some such pairs. For a single orbital electron, the corresponding orbital changes are $\Delta L=0,\pm 2$, with spin unchanged in the nonrelativistic electric-dipole approximation.

In Doppler-free two-photon absorption, counterpropagating equal-frequency photons give first-order Doppler cancellation when one photon is absorbed from each beam:The laser frequency is half the transition frequency. Same-beam two-photon events can still produce a broader Doppler-shifted background; residual beam-angle errors, unequal frequencies, AC Stark shifts and finite transit time limit precision. Thus a narrow two-photon resonance requires control of beam geometry and intensity, not just high laser power.

A stimulated Raman transition is the absorption-emission version: the atom absorbs from one beam and emits into the other, transferring between two long-lived states through an off-resonant upper state. It instead obeysCounterpropagation now enhances momentum transfer rather than cancelling it. With single-photon Hamiltonian matrix elements $\hslash {\Omega }_i/2$ and a common large angular detuning $\Delta$, eliminating the excited state givesThe two-photon detuning must include differential AC Stark and recoil shifts. Large detuning suppresses spontaneous scattering but also reduces coherent coupling at fixed optical power.

The coherent Raman route is useful for atomic beam splitters and mirrors. Mark Kasevich and Steven Chu demonstrated a laser-cooled sodium matter-wave interferometer using stimulated Raman transitions; its controlled internal-state changes accompanied separated external wave packets.^[ref] Testing this interpretation requires scanning the two-photon detuning, controlling relative laser phase and checking output populations against pulse timing. A two-photon process is defined by its coherent two-interaction pathway; absorption-absorption and absorption-emission have different resonance and recoil conditions.

Assume dilute nondegenerate electrons and electron holes, parabolic three-dimensional band edges with the stated scalar band effective masses, spin multiplicity two, one effective valley for each edge, and thermal equilibrium. Take the unchanged host ground state to have unit activity and neglect bound excitons. The effective density of states values areFor a Fermi level $\mu$ between the band edges,The factors of $\mu$ cancel on multiplication. With $E_g=E_c-E_v$, the semiconductor mass-action law isThis density-form equilibrium constant has units ${\mathrm{m}}^{-6}$. A dimensionless thermodynamic constant uses a specified reference number density $n^{\circ }$, giving $K=K_n/(n^{\circ }{)}^2$. The dot representing the host ground state is not a freely translating gas species requiring a further gas partition function. For an intrinsic semiconductor, charge neutrality gives $n_e=n_h=n_i=\sqrt{K_n}$.

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 $2J+1$.

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.