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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.

For a nondegenerate two-dimensional electron gas, the two quadratic momentum terms supply $k_{B}T/2$ each by the equipartition theorem. There is no lattice restoring potential energy for a free electron. ThusOn cooling below the Fermi temperature, the Pauli exclusion principle restricts the thermally active electrons to within approximately $k_{B}T$ of the Fermi energy. The linear low-temperature heat capacity of a two-dimensional Fermi gas isThe heat capacity therefore decreases from its classical value as quantum degeneracy becomes important. The particle number of electrons and sample area are held fixed when differentiating the internal energy.

Write $U(x)=-e\varphi (x)$ for the potential energy of an electron in an electrostatic potential $\varphi$, where $e>0$. With $G=2\pi /a$, the coupling in the nearly-free electron model is the Fourier coefficientIf a symbol $V(x)$ already denotes the electron potential energy, as in part (c), it replaces $U(x)$ without another factor $-e$.

At the Brillouin zone boundary the two resonant plane waves have unperturbed energy $E_{\mathrm{ZB}}$. Their Hamiltonian operator in the two-state basis has diagonal entries $E_{\mathrm{ZB}}$ and off-diagonal entries $V_G,V_G^{\ast }$; its eigenvalues are $E_{\mathrm{ZB}}\pm |V_G|$. ThusThis is the local band gap in the two-wave nearly-free electron model.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

Regard $V(x)$ as an electron potential energy, and take $G=2\pi /a$. Each cosine supplies half its coefficient to each of its two Fourier coefficients. The nonzero coefficients areThe Fourier coefficient connecting the two plane waves at $k=\pm \pi /a$ is $V_G$, so the two-wave band gap isAt $\delta =0$ the actual periodic potential has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved Brillouin zone boundary then has zero band gap. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.

There is also a small-displacement qualification in the source: the claimed positions of the potential energy minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. ThenAt $\delta =0.01$ and $x/a=1/4-\delta =0.24$, this equals approximately $-2.802\times 10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta +O({\delta }^3)$ and $3/4+\delta +O({\delta }^3)$. This does not alter the leading Fourier coefficient or the linear opening of the band gap.

The small-slope excess length is $\int (w^{\prime }{)}^2/2$, so the first term is the work needed against uniform tension. The second is the potential energy of the downward distributed load. Their sum is total potential energy subject to the obstacle constraint.

Expand the difference exactly:The first two integrals together are nonnegative by part (b), and the square is nonnegative. ThusFor $T>0$ this also proves uniqueness: equality for two minimizers forces identical derivatives, and the fixed endpoints remove their additive constant. The convex admissible set and strictly convex energy make the variational inequality a global minimization condition.

Make the Wick rotation $t^{\prime }=-i\tau$, and write $T_E=\beta \hslash$ for the final imaginary time, where $\beta$ is inverse temperature and $\hslash$ is the reduced Planck constant. Then $d{t}^{\prime }=-id\tau$ and $dx/d{t}^{\prime }=idx/d\tau$. Thus the continued action isBoth the kinetic energy and the potential energy enter this Euclidean action with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta \hslash )/\hslash }=e^{-\beta H}$, so the Euclidean path integral isIts boundary conditions fix the two endpoint positions. They need not coincide: identifying and integrating over them would instead give the thermal trace.

The normalization of $\mathcal{D}x$ can be fixed without continuing a formal measure. Apply the Trotter product formula to $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide $T_E$ into $M$ intervals of size $ϵ=T_E/M$, and integrate the intermediate positions. The resulting time-sliced configuration-space path integral isThis also explains the upper limit $\beta \hslash$ rather than $\beta$ when the reduced Planck constant has not been set to one.

virial pressure-radius curve with uniform-sphere binding energy. Put $c_s^2=k_{B}T/\mu$. Isothermality makes $\int (p/\rho )dM=c_s^2{M}_0$, so the virial theorem givesThe curve crosses zero at $r_z=B/A=G{M}_0/(5c_s^2)$, rises to its maximum atthen decreases toward zero at large radius. Since $p_0^{\prime }=(4B-3A{r}_0)/r_0^5$, a cloud displaced outward at fixed external pressure has a restoring pressure deficit only on the descending branch. Thus the homologous, isothermal radial mode is stable for $r_0>r_c$, unstable for $r_z<r_0<r_c$, and marginal at $r_c$. The negative-pressure branch is not a physical equilibrium with a positive confining pressure. This conclusion uses the stipulated gravitational-energy approximation, not the exact isothermal density profile.

critical mass from a virial pressure turning point. At the turning point, $B=3A{r}_c/4$; henceUsing the cloud's mean density $\bar{\rho }=3M_0/(4\pi r_0^3)$, the condition $r_0>4G{M}_0/(15c_s^2)$ becomesUsing the boundary density instead changes the numerical coefficient, not this scaling. The coefficient should not be identified with the exact Bonnor-Ebert sphere threshold because the assumed potential energy is different.

mass and luminosity weighting of a stellar initial mass function. Impose finite birth-mass cutoffs $m_l,m_u$. The stellar birth mass and main-sequence luminosity scale asFor a broad mass interval, the low-mass end controls the first integral and the high-mass end the second. Per logarithmic mass interval the weights are respectively $m^2\xi \propto m^{-0.35}$ and $mL\xi \propto m^{2.65}$. Finite lifetimes remove massive stars from an aging population, so the luminosity conclusion applies while those stars remain on the main sequence.

exponential closed-box gas consumption. The gross formation rate is $\Psi =ϵM_g$, but the permanent locking rate is $\alpha \Psi$. The instantaneous recycling approximation and mass conservation give ${\dot{M}}_g=-\alpha ϵM_g$, ${\dot{M}}_{\ast }=\alpha ϵM_g$. ThereforeHere $M_{\ast }$ means surviving locked stellar mass in the galactic gas fraction, not the cumulative gross mass that has passed through star formation.

locked-mass and returned-mass yield conventions. Pre-existing metals are locked at rate $\alpha Z\Psi$; the stipulated newly produced metal source is $b(1-\alpha )\Psi$. ThusFor $\alpha >0$ and initially pristine gas,The effective yield $y$ is per unit permanently locked mass; it is not the source parameter $b$. For $\alpha =0$, $M_g=M_0$ and the direct time equation gives $Z=bϵt$. For $\alpha =1$, the explicitly stipulated metal source vanishes, so this particular enrichment model gives $Z=0$. Constant yield is a trace-enrichment approximation; the formal logarithmic divergence as gas is exhausted must not be interpreted as a physical metal fraction exceeding one.

backward extrapolation of exponential closed-box star formation. Denote the present age of the universe by $t_{\mathrm{now}}$. With the last paragraph's $\alpha =1$,Adopting a present age of the universe $t_{\mathrm{now}}=13.8\mathrm{Gyr}$ for the estimate gives about $365M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$, requiring $7.3\times 10^{11}{M}_{\odot }$ of gas at that epoch in the same model. The boxed expression makes the dependence on the adopted present age explicit. This is an extreme history for a present Milky-Way-like system, although intense high-redshift star formation by itself is not forbidden.

The extrapolation forces a fixed gas reservoir to decline exponentially over many gas-depletion times. Gas accretion, outflows, evolving gas density and star-formation efficiency, delayed recycling, and galaxy assembly break that assumption. The empirical disk-averaged relation is not a universal constant-efficiency law for all gas: the original measurements combined Balmer series H-alpha, atomic gas, carbon monoxide and infrared data, with tracer calibration and dust corrections entering the inference.^[ref] Cosmological calculations explicitly support continuing cold-gas supply to rapidly forming high-redshift galaxies, providing a physical alternative to this closed-box backward extrapolation.^[ref]