Codex @codex  0

An effective one-nucleon potential energy representing the average influence of the other nucleons. The nuclear shell model treats motion in this field first, then includes residual interactions. The field experienced by a proton also includes the repulsive Coulomb potential; proton and neutron energy levels need not have identical spacings.

Filling the accessible one-species nuclear shell model states below a substantial energy gap. A closed spherical $j$ level has all $m_j$ values occupied, so their sum is zero. In the independent-particle picture, a closed core therefore contributes zero total angular momentum; a single particle outside it, or a single missing particle inside it, supplies the leading spin assignment. Residual interactions can change this simple picture.

The three-dimensional isotropic harmonic oscillator alone has one-species shell capacities $(q+1)(q+2)$ for oscillator index $q=0,1,\dots$, including spin. Summing gives closures $2,8,20,40,70,112,\dots$, rather than the complete traditional magic number sequence. Strong nuclear spin-orbit splitting can separate a high-$j$ level from its former neighbors. Realistic radial potential energies and residual interactions also matter: a chosen oscillator plus an arbitrary single constant is not a demonstrated explanation of every real nuclear gap.

Traditional nuclear shell closures can weaken along isotope chains far from stability, as in neutron-rich magnesium.^[ref]

The relevant test is the actual energy gap and the correlated nuclear quantum states, rather than membership in an immutable numerical list.

Why should adding one neutron be easy for one atomic nucleus and much less favorable for its neighbor? In the nuclear shell model, nucleons occupy discrete energy levels in a nuclear mean field. Filling all available states below a sufficiently large gap produces a nuclear shell closure. This connects nuclear separation energies with the counting of quantum states, and connects the spin of an odd atomic nucleus with its last unpaired nucleon.

The starting point is an approximate Hamiltonian operatorwhere $U_i$ is a confining potential energy, $a_i$ specifies nuclear spin-orbit splitting, and $V_{\mathrm{res}}$ contains interactions left out of the nuclear mean field. This is an effective organization of the nuclear many-body problem, rather than a derivation of the nuclear force. In a spherical approximation, addition of angular momentum combines the orbital angular momentum $l$ with spin $s=1/2$ to give $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. The Pauli exclusion principle permits one neutron in each such state and, separately, one proton. Thus a $j=7/2$ level holds eight neutrons, or eight protons; it does not hold sixteen identical neutrons. Here $n=1,2,\dots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal quantum number. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.

The linear operator $V_T(r)S_{12}$ couples nucleon spin orientations to their separation direction $\hat{\mathbf{r}}$. Here ${\mathbf{\sigma }}_i$ are Pauli matrices. A scalar radial potential energy alone omits this directional dependence.

For a bead in a stationary viscous liquid, linear drag often dominates inertia on the observed timescale. Neglecting both inertia and the random thermal force, the deterministic force balance givesHere $\zeta >0$ is the drag coefficient and $\tau$ the optical trap relaxation time: the time to reduce the deterministic displacement by a factor $e$. The bead approaches equilibrium without oscillation in this overdamped model. The approximation requires times long compared with the inertial time $m/\zeta$ and, for separated timescales, $m/\zeta \ll \zeta /k$, where $m$ is the bead's mass. For a no-slip spherical bead of radius $a$ in an unbounded liquid of dynamic viscosity $\eta$, Stokes drag law gives $\zeta =6\pi \eta a$. Nearby walls change that drag coefficient.

For example, $a=0.5\mu \mathrm{m}$, $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ and $k=0.01pN/nm=10^{-5}N/m$ give $\zeta \approx 9.42\times 10^{-9}\mathrm{N}s/m$ and $\tau \approx 0.942\mathrm{ms}$. A displacement of $100\mathrm{nm}$ gives a restoring force of magnitude $1\mathrm{pN}$, and its deterministic displacement drops to $100/e\approx 36.8\mathrm{nm}$ after one optical trap relaxation time. These are predictions for the stated model, not measured trajectories.

The figure uses $q=x/\ell$, $\kappa =k/k_0$ and $s=t/(\zeta /k_0)$, with arbitrary positive reference length $\ell$ and stiffness $k_0$. Thus $U/(k_0{\ell }^2)=\kappa q^2/2$, $F/(k_0\ell )=-\kappa q$ and, from $q(0)=1$, $q(s)=e^{-\kappa s}$. At fixed drag, greater stiffness gives both a steeper potential energy well and faster relaxation. Thermal Brownian motion is omitted from these deterministic curves.

Optical tweezers hold and move a small particle using the force exerted by a tightly focused beam of light, usually from a laser. They can position transparent beads in a liquid and manipulate suitable biological cells. Attaching a molecule to a trapped bead turns the bead into a handle: a measured bead displacement, together with a calibrated spring constant, reveals the force exerted on the attachment. This makes microscopic manipulation and measurement possible without a mechanical tool touching the bead. Living biological cells were manipulated with infrared beams by Ashkin, Dziedzic and Yamane in 1987; Svoboda and Block used trapped beads to measure forces from individual molecules in 1994. ^{[ref] [ref]}.

Why can light pull a bead towards a focus? Light carries momentum. A transparent bead with a higher refractive index than the surrounding liquid bends transmitted light rays by refraction, changing their momentum; conservation of momentum requires an opposite transfer to the bead. In the geometrical optics picture, unequal illumination of a displaced bead produces a net restoring force towards the brighter region. For a particle much smaller than the wavelength, a physical optics description replaces that ray picture. The detailed force depends on particle size, refractive index, beam shape and wavelength; the qualitative attraction does not supply a universal formula for every particle.

The optical gradient force draws a suitable transparent bead towards high illumination, while the optical scattering force generally pushes along beam propagation. A stable three-dimensional optical trap needs a balance with a restoring force in each direction: transverse confinement alone does not ensure axial confinement. Strong focusing can make the backward axial gradient contribution large enough to balance the forward scattering contribution, with the equilibrium displaced from the focus. Higher refractive index alone is insufficient; absorbing particles, different index contrasts and different beam geometries can behave differently. ^[ref].

Historically, Ashkin's 1970 experiments accelerated and trapped particles using radiation pressure. The 1986 paper by Ashkin, Dziedzic, Bjorkholm and Chu demonstrated the single-beam gradient optical trap for transparent particles. These are distinct steps: the earlier work established optical manipulation, and the later work supplied the tightly focused single-beam arrangement. ^{[ref] [ref]}.

Near a stable equilibrium, a harmonic optical trap behaves like a spring; thermal calibration of an optical trap uses its Brownian motion to determine the effective spring constant. The later passage through a translating optical trap sections instead consider a moving localized potential energy, which need not remain harmonic throughout a passage.

In the Hilbert space $L^2([0,2\pi ])$, choose orthonormal functions ${\varphi }_1(x)=\cos x/\sqrt{\pi }$ and ${\varphi }_2(x)=\sin x/\sqrt{\pi }$. Their inner product is zero, but for the real potential energy $V(x)=V_0\sin 2x$,Thus ordinary orthogonality alone does not remove an off-diagonal matrix element of a spatially varying potential energy. A constant scalar potential does give zero between distinct orthonormal states. The example separates these two claims; it is not a model of localized atomic orbitals in a particular crystal.

For positive integers $m,n$, the sine functions form an orthogonal family on $(0,L)$:The same formula holds for the corresponding cosine functions when both indices are positive. Product-to-sum identities reduce the distinct-index products to cosines with nonzero integer frequencies, each with zero integral; for matching indices, the squared sine or cosine has mean $1/2$. This proves the formula and explains why separate normal modes of a uniform fixed-end string make additive contributions to its kinetic energy and potential energy. The constant cosine has squared norm $L$ rather than $L/2$.

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

Magnetic flux freezing ties a transverse fluid displacement to a bend in the magnetic field lines. For the uniform background, the transverse displacement equation isFor a sinusoidal bend, ${\partial }_z^2{\mathbf{\xi }}_{\perp }=-k_{\parallel }^2{\mathbf{\xi }}_{\perp }$. Positive pressure-anisotropy-modified magnetic tension therefore gives a restoring force, and inertia produces oscillation. In an isotropic plasma, this is precisely the positive magnetic tension $B_0^2/(4\pi )$.

Particles moving predominantly along bent magnetic field lines contribute a transverse momentum flux. Their larger parallel pressure opposes the restoring curvature force supplied by the magnetic field and perpendicular pressure. Once $p_{\parallel 0}-p_{\perp 0}>B_0^2/(4\pi )$, the net force points along the original displacement: a larger bend produces a larger outward force, giving positive feedback. Equivalently, the bending contribution to the quadratic potential energy is $\frac{1}{2}{\mathcal{T}}_0|{\partial }_z{\mathbf{\xi }}_{\perp }{|}^2$ and has become negative.

In this static, homogeneous fluid model the unstable roots areWith $e^{-i\omega t}$ the root $+i\gamma$ grows. Its real angular frequency is zero, so the unstable pattern grows without propagation in the equilibrium rest frame; its partner decays. Advection by a background flow would change that frame-dependent statement. The growth increasing without bound as $|k_{\parallel }|$ increases is a limitation of this long-wavelength model: finite gyroradius and other kinetic corrections cannot be neglected at sufficiently short scales.

Put $r^2={\varphi }_i{\varphi }_i$. Completing the square in the scalar potential givesSince $\lambda >0$, its global minima form the vacuum manifoldHere $\mu$ denotes the positive square root of the parameter ${\mu }^2$. The origin is a local maximum, not a ground state. Choose $⟨\mathbf{\varphi }⟩=(v,0,0)$. Its stabilizer subgroup is the SO(2) subgroup of rotations in the $(2,3)$ plane, so the stated SO(3) global symmetry in field theory is spontaneously broken to SO(2).

Write ${\varphi }_1=v+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The kinetic terms stay canonically normalized, while the scalar mass matrix isEquivalently, the quadratic scalar potential is ${\mu }^2{h}^2$. Therefore the particle masses areThe two angular Goldstone bosons match $\dim SO(3)-\dim SO(2)=3-1=2$. The radial displacement changes $r$ and costs potential energy, giving the massive Higgs mode. These are tree-level masses; the Goldstone theorem protects the angular zero-mass modes in the exact global theory.

The WKB approximation, also called JWKB, describes a short local de Broglie wavelength in a slowly varying potential energy. For $p(x)=\sqrt{2m[E-V(x)]}$ in a classically allowed region, its two local waves areThe phase follows the classical momentum; its amplitude varies to conserve probability current. In a forbidden region, putting $\kappa (x)=\sqrt{2m[V(x)-E]}$ gives $\psi \simeq {\kappa }^{-1/2}[C{e}^{{\int }^x\kappa /\hslash }+D{e}^{-{\int }^x\kappa /\hslash }]$, with boundary conditions selecting the physical combination.

The WKB validity condition is $\hslash |p^{\prime }|/p^2\ll 1$, with the analogous condition on $\kappa$. It fails at a turning point where $E=V$. For a simple turning point, locally linearizing V yields an Airy function; matching fixes the relative amplitudes and a $\pi /4$ phase. Two such turning points give the two-turning-point WKB quantization conditionFor an opaque barrier, the WKB tunnelling exponent instead givesThe factor two distinguishes probability from amplitude. Matching determines the prefactor; the exponential alone is not accurate near the barrier top. These formulas connect bound state energy eigenvalues and quantum tunnelling to classical actions. ^[ref]

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

Put $\mathcal{T}=\beta \hslash >0$, the interval of imaginary time, and assume $m,\omega >0$. For a variation $x_0+ϵ\eta$ with $\eta (0)=\eta (\mathcal{T})=0$, the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is $A\cosh (\omega \tau )+B\sinh (\omega \tau )$. Applying the two endpoint conditions gives $A=x$ and $B=-x\tanh (\omega \mathcal{T}/2)$. Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since ${\dot{x}}_0(\mathcal{T})=\omega x\tanh (\omega \mathcal{T}/2)$ and ${\dot{x}}_0(0)=-\omega x\tanh (\omega \mathcal{T}/2)$,The positive quadratic form $m{\int }_0^{\mathcal{T}}({\dot{\eta }}^2+{\omega }^2{\eta }^2)d\tau$ is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.

Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).

The Rayleigh-Taylor instability occurs when a denser fluid is supported against an effective acceleration by a lighter fluid. Let density ${\rho }_h$ occupy the upper half-space, density ${\rho }_l<{\rho }_h$ the lower half-space and gravity point downward. A displaced heavy-fluid finger lowers gravitational potential energy as it sinks; neighboring light fluid rises. For incompressible inviscid half-spaces without surface tension, a disturbance $\eta \propto e^{\sigma t+ikx}$ hasThe density ratio in this expression is the Atwood number. With the density ordering reversed, ${\sigma }^2<0$ and the interface supports stable gravity-wave oscillations in this ideal model.

Surface tension suppresses sufficiently short wavelengths, and viscosity changes the growth dynamics. A magnetic field parallel to the interface can resist perturbations that bend its lines; interchange directions that do not bend the field need separate consideration. In astrophysics, effective deceleration can make dense ejecta or a swept-up shell supported by a lower-density hot interior unstable. The effective acceleration and density ordering must be evaluated in that moving interface's frame; simply seeing a dense shell does not fix the stability sign.

For a steady inviscid barotropic fluid in a conservative gravitational potential, define $h(\rho )=\int dp/\rho$. Dotting the Euler equation with the velocity givesThe Bernoulli function is therefore constant along a streamline. It balances kinetic energy, pressure work or enthalpy, and gravitational potential energy per unit mass. Different streamlines can have different constants; for an irrotational flow, the identity $\nabla H=\mathbf{u}\times (\nabla \times \mathbf{u})$ makes the constant common throughout a connected region.

For an incompressible fluid, $h=p/\rho$ up to a constant, giving $u^2/2+p/\rho +\varphi =C$. For a constant-entropy perfect-gas streamline, $h=\gamma p/[(\gamma -1)\rho ]$. Together with mass conservation, this explains conversion of enthalpy into kinetic energy in a nozzle and the need to include gravity in an escaping wind. Viscous losses, heating and entropy production require their corresponding energy balance. Across an adiabatic shock the total stagnation enthalpy can still be conserved, but the upstream isentropic pressure-density law cannot simply be continued through the entropy jump.

Write $k=e^2/(4\pi {ϵ}_0)$ and retain the nuclear Coulomb potential. In spherical coordinates, the total potential energy iswhen $\beta =-\alpha /2$; $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. Treat the added term using quadratic anisotropic perturbation of hydrogen and degenerate perturbation theory in the $n=2$ hydrogen energy eigenbasis. It preserves parity and $m$, so it cannot mix $2s$ with $2p$, or distinct $2p$ magnetic sublevels.

For a hydrogen atom, the hydrogen radial second moment givesThe angular expectation values of ${\cos }^2\theta$ are $1/3$ for an $s$ orbital, $3/5$ for $p_0$, and $1/5$ for $p_{\pm 1}$, obtained by integrating the squared spherical harmonics. Therefore the five lowest spatial quantum states, counting statistical weight, have first-order energy levelsEach spatial quantum state has two electron spin states when spin-orbit coupling and the magnetic field are omitted. Thus these are five spatial states but only three distinct first-order energies; five distinct energies would be an incorrect reading. The accidental $2s,2p_{\pm 1}$ equality can be lifted at higher order. Quantitatively the added shifts must also be small compared with neighboring principal-shell gaps; $\alpha \ll k/a_0^3$ is an asymptotic condition, not a guarantee for arbitrary large $n$.

For general $\beta$, the $1s$ expectation value isIt does not cancel every order. At $\beta =-2\alpha$, $V_{\mathrm{ext}}=-2\alpha r^2{P}_2(\cos \theta )$ couples $1s$ to $d$ orbitals. For example its angular matrix element with $Y_{20}$ is $1/\sqrt{5}$, and the $3d$ radial matrix element is nonzero. The second-order nondegenerate perturbation theory sum has negative denominators, so this is a concrete nonzero higher-order contribution.

There is also a global limitation: for either negative $\beta$ above, $V(0,0,z)=-k/|z|+\beta z^2\to -\infty$. A localized wavepacket translated far along $z$ has fixed kinetic energy but arbitrarily negative potential energy, proving there is no true lowest energy level. The results describe local perturbative resonance poles, or a potential completed to a confining form far from the atomic nucleus. No globally exact ground-state cancellation exists for the potential as printed.