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Put $k_Z=Z{e}^2/(4\pi {ϵ}_0)$. Gauss's law for a uniform spherical nuclear charge distribution, with the electrostatic potential matched at $R$, gives the electronic potential energySubtracting the point-nucleus Coulomb potential leaves the nonnegative perturbationThe normalized nonrelativistic hydrogen-like ion $1s$ wavefunction has $|\psi (r){|}^2=Z^3{e}^{-2Zr/a_0}/(\pi a_0^3)$. Since $ZR/a_0\ll 1$, approximate it inside the atomic nucleus by $|\psi (0){|}^2$. First-order time-independent perturbation theory then givesDefine $E_H=-Z^2\mathrm{Ry}$, as the paper specifies for the hydrogen-like ion, rather than the neutral-hydrogen value. Using $\mathrm{Ry}=e^2/(8\pi {ϵ}_0{a}_0)$ givesto leading order in $ZR/a_0$. For $A=208$, $R\simeq 7.11\mathrm{fm}$, $ZR/a_0\simeq 0.0110$, andThe finite nuclear charge radius makes the energy level less bound. The quoted frequency is a level shift divided by $h$: a line frequency shifts by $(\Delta E_u-\Delta E_l)/h$, so a decay into this shifted $1s$ level alone has the opposite sign.

The formula is a nonrelativistic, small-nuclear charge radius estimate. Keeping the exponential in the first-order integral gives $\Delta E_{1s}=\Delta E_{\mathrm{leading}}[1-5ZR/(6a_0)+O((ZR/a_0{)}^2)]$, about a one-percent correction here. More seriously, $Z{\alpha }_{\mathrm{fs}}\simeq 0.598$ for lead, so relativistic contraction is substantial. A precise physical prediction needs a finite-nucleus Dirac equation calculation; the boxed estimate is the model requested, not precision heavy-ion spectroscopy.

For a free-space electric dipole transition, the spontaneous emission rate in SI units isHere $|\mathbf{d}{|}^2$ includes the relevant dipole components; a specified single polarization requires its corresponding matrix element.

A complete scattering cycle returns the internal state to the ground state and changes the atomic momentum to $\mathbf{P}+\mathbf{p}-{\mathbf{p}}^{\prime }$. ThusFor photon-scattering recoil heating with isotropic emission of fixed momentum magnitude, $⟨{\mathbf{p}}^{\prime }⟩=0$, givingThe first term is the velocity-dependent work of absorption and can cool an atom if the photon opposes its motion. The last two are positive absorption and emission recoil energies. If the actual absorbed directions are also isotropic and independent of $\mathbf{P}$, the first term averages to zero. For approximately equal incoming and outgoing photon momenta $\hslash k$,Isotropic incident illumination alone would not guarantee isotropic absorption if Doppler-dependent excitation selected directions.

Source qualification for the trapped calculation. Set $m=M$ for the same atom. The printed Gaussian $\exp [-x^2/(2l_0{)}^2]$ has the wrong width for the specified harmonic oscillator: the printed ground and first-excited functions have squared norms $\sqrt{2}$ and $2\sqrt{2}$, respectively. The normalized oscillator states fixed by the given $l_0=\sqrt{\hslash /(m{\omega }_{\mathrm{HO}})}$ instead useAlso, one common complex phase multiplying both off-diagonal terms does not make the Hamiltonian a Hermitian operator. Use an absorption matrix element $(\hslash \Omega /2)e^{-ikx}$ and its Hermitian conjugate for emission; choosing the opposite propagation convention reverses the imaginary sideband phase but not a transition probability. The physical results below use these explicitly stated corrections. Parts (i) and (ii) also give the literal integrals of the printed functions.

Conditional motional heating and recoil averaging. With $\eta =k{l}_0/\sqrt{2}$, an oscillator eigenstate has $⟨K{⟩}_n=⟨V{⟩}_n=(n+1/2)\hslash {\omega }_{\mathrm{HO}}/2$. Therefore the absorption outcomes haveThe same statements hold for the specified emission outcomes $|e,0⟩\to |g,0⟩$ and $|e,0⟩\to |g,1⟩$. Internal excitation energy goes into, or comes from, the photon and is distinct from this external energy change. Emission into the first motional state makes the emitted photon correspondingly less energetic.

A sudden recoil from $|0⟩$ gives probabilities $P_n=e^{-{\eta }^2}{\eta }^{2n}/n!$, so $P_1\simeq {\eta }^2\ll 1$ while $P_0\simeq 1$. The rare first sideband transfers a whole trap quantum, much larger than $E_R$; the mean is neverthelessThe emission average is the same for a one-dimensional recoil of the same magnitude. Projecting isotropic three-dimensional emission onto one axis instead gives a factor $⟨{\cos }^2\theta ⟩=1/3$ for that axis.

A sudden unobserved kick initially raises kinetic energy by $E_R$ and leaves the position distribution, hence potential energy, unchanged. It creates a coherent superposition of motional states. The equal kinetic/potential increments above concern energy-eigenstate-conditioned outcomes, or time-averaged/dephased motion; these are different measurement conditions. There is no contradiction.

With $\Gamma >{\omega }_{\mathrm{HO}}$, motional sidebands are not cleanly resolved, and a short-lived excited atom has little time to move before decay. In the stronger sudden limit $\Gamma \gg {\omega }_{\mathrm{HO}}$ the recoil-kick picture is direct. For only the stated inequality, finite linewidth weights can change numerical factors; it does not fix an exact continuous-wave scattering rate. The average heating remains of recoil order per absorption or emission, and of order $2E_R$ when both absorption and emission kicks are counted at full magnitude. For the axial component of isotropic three-dimensional emission, the emission contribution instead carries the factor $1/3$ noted above. Strong confinement suppresses the chance of changing $n$, but does not make its averaged recoil energy disappear. This agrees with free-space recoil heating while replacing small continuous energy changes by rare quantized motional changes.

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.

For an ideal weakly driven two-level atom, opposing red-detuned beams give velocity-dependent absorption and a friction force near zero velocity. Random photon recoils also cause momentum diffusion. Their balance gives the minimum Doppler temperatureat detuning $\Delta =-\Gamma /2$ in the low-saturation model, where $\Gamma$ is the spontaneous population-decay rate, equivalently the natural full linewidth expressed as an angular frequency. This limit assumes the atom has no useful ground-state internal structure.

Real atoms possess magnetic sublevels. In a suitable multilevel transition, position-dependent polarization gives different optical couplings and different AC Stark energy shifts for those sublevels. Counterpropagating beams with orthogonal linear polarizations have nearly uniform total intensity but a periodically varying polarization. Thus spatially varying light shifts can exist even without an intensity standing wave.

In Sisyphus cooling, an atom on one ground-state branch climbs a light-shift hill and loses kinetic energy. Near the upper part of the hill, optical pumping tends to transfer population to a lower branch. spontaneous emission makes this transfer dissipative and randomizes the emitted photon direction. The atom can then climb another hill. Finite pumping time is essential: instantaneous local equilibration would remove the delayed internal-state response that produces friction. The original multilevel analysis relates these optical-pumping cycles to the spatially displaced light shifts.^[ref]

For large detuning compared with the natural linewidth, useful scales areup to angular-coupling and Rabi-frequency convention factors. Here $\Omega$ is the optical Rabi angular frequency. The slow-atom capture range is of order $kv\lesssim {\Gamma }_{\mathrm{op}}$. Typical energies in the Sisyphus regime track the light-shift scale rather than only $\hslash \Gamma$, so reducing intensity can lower temperature below the Doppler limit. Capture becomes narrower, however, and recoil heating, imperfect polarization, magnetic fields and scattered light eventually matter. Define recoil energy $E_r=(\hslash k{)}^2/(2m)$; this is an important scale, not a claim that every sub-Doppler method has the same universal final temperature.^[ref]

The experimental test must establish a narrow velocity distribution independently of fluorescence brightness. William D. Phillips's account describes moving a time-of-flight probe above and below the cloud, deliberately heating the atoms to check a previously absent signal, and release-and-recapture comparisons that included gravity and finite beam geometry. These tests separated very cold atoms from an insensitive detector or a faulty thermometer.^[ref] In a modern implementation, field cancellation and polarization checks test the mechanism, while independent expansion thermometry checks the inferred temperature. sub-Doppler cooling exploits internal-state dynamics omitted from the two-level Doppler model; it does not contradict that model within its assumptions.

Take $v>0$ and the displayed potential energy to be the position-dependent mechanical free energy $G(x)$ of a bound motor. On the descending ramp $G^{\prime }=-G_0/x_1$, and on the ascending ramp $G^{\prime }=G_0/(P-x_1)$. Thus the force conditional on being bound isIt is periodic, piecewise constant and undefined at idealized sharp corners; the corners have zero measure in averages. An unbound motor has zero force.

A cohort transported through position $x$ obeys $d/dt=vd/dx$. Its bound probability therefore satisfiesDefine $B_0=k_1/(k_1+k_2)$, $\alpha =(k_1+k_2)/v$, $\beta =k_2/v$, $a=e^{-\alpha x_1}$ and $q=e^{-\beta (P-x_1)}$. The exact periodic steady state includes a small carry-over $b_{\mathrm{in}}=B(0)=B(P)$:These expressions follow by solving each first-order ordinary differential equation, requiring continuity at $x_1$, and imposing periodicity.

The stipulated $v\ll k_2(P-x_1)$ implies $q\ll 1$, so the incoming bound fraction is exponentially small. Dropping it gives the simpler intended formsThe occupancy rises towards $B_0$ on the binding interval, then decays over length $v/k_2$ outside it. The mean force per member of the full ensemble at a given position is $B(x)F_{\mathrm{bound}}(x)$, distinct from the force per bound motor. The right panels of c show both the occupancy and the conditional force.

Integration by parts and the zero endpoint values turn the Rayleigh quotient into a positive kinetic energy integral:This is the parabolic variational trial for a particle in a box. The potential energy vanishes inside the box.

Wavefunction normalization requires $\alpha >0$. For the normalized probability weight proportional to $e^{-\alpha x^2}$, the Gaussian integral and its derivatives giveSince ${\tilde{\Psi }}^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\tilde{\Psi }$, the kinetic energy and potential energy expectations areThusThe second derivative is positive, and the energy diverges at both ends of $\alpha >0$, so this is the global minimum within the Gaussian function family. The best Gaussian function upper bound isThis is an upper bound, not the exact quartic ground energy.

Take $\hat{V}\psi =V(x)\psi$ with real $V$, and $\hat{T}=-({\hslash }^2/2m)d^2/d{x}^2$. The operator domain matters: use sufficiently smooth wavefunctions for which the following boundary terms vanish at both infinities. A real multiplication operator givesTwo applications of integration by parts giveThus the potential energy and kinetic energy operators obey the stated Hermitian operator identity on this common domain. For $D=d/dx$, one application of integration by parts instead givesThe derivative is anti-Hermitian, rather than Hermitian; the minus sign cannot generally be removed. Multiplication by $-i\hslash$ makes the corresponding momentum operator Hermitian. These integrations establish the requested symmetry identities; full self-adjointness of an unbounded operator additionally requires equality of its domain with its adjoint's domain.

Quantum tunnelling is penetration of, or transmission through, a finite potential barrier into a classically forbidden region, where the total energy is less than the potential energy. The wavefunction generally remains nonzero there, so the Born rule assigns a nonzero probability of detection. This does not require an increase in total energy; it is the classical relation between a definite position and a definite kinetic energy that cannot be imposed on the quantum state. For a bound quantum harmonic oscillator, the relevant quantity is the probability of finding the particle beyond its classical turning points; it is not an escaping transmitted beam.

Let $u=r-r_e$ measure displacement from the equilibrium bond length and put the minimum of potential energy at zero. The harmonic model and the Morse potential areThe harmonic approximation is a symmetric parabola and has no dissociation limit. The Morse potential has a steep compressed side and a shallower extended side, approaching $D_e$ as $r\to \infty$. Matching the curvatures at the minimum gives $k=2D_e{\alpha }^2$. With $\tilde{\omega }$ denoting a spectroscopic wavenumber rather than an angular frequency, the energy levels areHere $x_e$ is the dimensionless spectroscopic anharmonicity constant. The harmonic spacing is constant, while the Morse oscillator spacing is $E_{v+1}-E_v=hc\tilde{\omega }[1-2x_e(v+1)]$ and decreases as the levels approach dissociation. Only finitely many Morse oscillator levels are bound. The exactly solvable potential was introduced to model this anharmonic behaviour.^[ref]

For the harmonic model put $q=\sqrt{\mu \omega /\hslash }u$, with $\omega =2\pi c\tilde{\omega }$, and $N={\pi }^{-1/4}$. The first three normalized reduced-coordinate wavefunctions areTheir numbers of nodes are zero, one and two, and their parities are even, odd and even. The $v=2$ nodes are at $q=\pm 1/\sqrt{2}$. The diagrams below use analytic curves and arbitrary illustrative scales, with both potentials given the same local curvature.

Take the dipoles to point along $+z$, and let their surface number density be $n_s$. An annulus of radius $\rho$ and width $d\rho$ contains $n_{s}2\pi \rho d\rho$ dipoles. The electron's electrostatic potential energy is thereforeFor $z\ne 0$, the integral without $2\pi$ equals $z/|z|$, givingThus the jump's magnitude isFor the chosen outward dipole orientation the electron-energy change from the negative to positive side is negative. The electrostatic voltage jump has the opposite sign. This distinguishes the requested magnitude from the sign of the work function change.

Each face excludes polymer centers out to distance $a/2$. At face contact, the combined overlap strip has width $a$ and length $L$. Ignoring end corrections of order $a^2$, the two-dimensional Asakura-Oosawa model therefore givesThe final expression retains only the leading term in $d/\ell$. Its dimensions are correct because $\rho$ is a number per area. The zero of this effective potential energy is the well-separated, nonoverlapping-exclusion-layer configuration.

Treat the marble as a frictionless point particle; no size, rolling constraint or friction law is supplied. Neglect its vertical kinetic energy as requested, but retain the gravitational potential energy $mgz=m{\omega }^2(x_1^2+y_1^2)/2$.

For completeness keep the general constant cup rotation rate $\nu =\Omega +\varphi$. In cup components the center's velocity is $\mathbf{w}=R\Omega (\sin \varphi t,\cos \varphi t)$. With $\mathsf{J}(x,y)=(-y,x)$, the horizontal velocity norm is $|\dot{\mathbf{q}}+\nu \mathsf{J}\mathbf{q}+\mathbf{w}|$. ThusIn the nonrotating translating case $\varphi =-\Omega$, this reduces toThe time-only constant $m{R}^2{\Omega }^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent LagrangianIt differs by a total derivative and a time-only term, so gives the same trajectories but different expressions for the canonical momenta. The small-vertical-velocity approximation is typically consistent at small surface slope ${\omega }^2|\mathbf{q}|/g$; it need not remain valid during unlimited resonant growth.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.

Write $w=\sqrt{1+2k}$. The initial conditions on the normal mode coordinates are $u(0)=v(0)=0$ and $\dot{u}(0)=\dot{v}(0)=k$. Solving the two harmonic oscillator equations and recovering the original angles gives the exact solution within the harmonic approximation:Their angular velocities are ${\dot{\theta }}_1=k(\cos t+\cos wt)/2$ and ${\dot{\theta }}_2=k(\cos t-\cos wt)/2$, which verify all four initial conditions.

To expose the beat, define $\Delta =w-1$ and $\Omega =(w+1)/2$. Replacing $1/w$ by $1$ changes each angle by $O(k^2)$ uniformly in time. The sum-to-product identities then giveThe displayed angular velocities are exact in the linear system. The absolute and signed beat envelopes have leading amplitudes $k|\cos (\Delta t/2)|$ and $k|\sin (\Delta t/2)|$. Because $\Delta =k+O(k^2)$, at $\tau =\pi /k$ we have $\Delta \tau /2=\pi /2+O(k)$ and $\Omega \tau =\tau +\pi /2+O(k)$. HenceThe first simple pendulum has negligible oscillation amplitude. Its kinetic energy is $T_1(\tau )=O(k^4)$, while the second has $T_2(\tau )=k^2{\cos }^2\tau /2+O(k^3)$. Thus, to the retained order, all instantaneous kinetic energy belongs to the second simple pendulum. The fast phase can make that kinetic energy small: the whole initial energy $k^2/2$ is transferred as the second simple pendulum's combined kinetic energy and gravitational potential energy, sinceThis is energy transfer after an impulse in two weakly coupled oscillators. It does not require $\cos \tau =1$.

At $2\tau$, $\Delta (2\tau )/2=\pi +O(k)$, so the beat has returned the oscillation to the first simple pendulum:The energy localization returns, although the fast oscillation phase need not equal its initial phase. The exact first minimum of the leading beat envelope is at $\pi /\Delta$, whose leading value is $\pi /k$. Keeping the slow phase $\Delta t$ is essential: expanding $\sin (wt)$ at fixed $t$ and then using $t=O(1/k)$ would lose the energy exchange. Although angles are retained only to order $k$, their leading energies are of order $k^2$.

Put $y=e^{-r/\xi }$. Differentiating the Morse potential givesThe finite-separation minimum has $y=B/(2A)$. ThusThe positive bond-dissociation energy is $D=B^2/(4A)=0.250\mathrm{eV}$, measured relative to the separated-atom value $V(\infty )=0$. This distinguishes the well's negative potential energy from the positive energy required to break the bond.

For the displacement $x=r-a$, the Taylor expansion isConsequentlyThe restoring force is $-\kappa x$ to leading order, identifying the bond's Hooke's law spring constant.

There are two chemical atoms in the primitive unit cell, hence six displacement coordinates and six phonon branches at each wavevector. Three rigid translations cost no restoring potential energy in the long-wavelength limit. ConsequentlyFor propagation along $[100]$, one acoustic phonon has longitudinal polarization along $x$ and two have transverse polarization along $y,z$. The cubic crystal symmetry makes those two transverse directions equivalent. In terms of the elastic stiffness tensor and the mass density,Thus there are two distinct acoustic velocities in general, one longitudinal and a doubly degenerate transverse value. This degeneracy of normal modes concerns the two transverse polarizations; it does not imply equality with the longitudinal speed of sound.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.