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In first-order time-dependent perturbation theory, the resonant time factor before taking the Fermi's golden rule limit isIts central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta (\nu )$, the oscillator density of states and relevant occupations must be approximately constant over that width. Thus $t\Delta \omega \gg 1$. At the same time, the perturbative flip probability must remain much smaller than one.

Define ${\Gamma }_0=2\pi {\lambda }^{2}g(\Omega )/{\hslash }^2$, so the two flip rates are ${\Gamma }_0(n_{\Omega }+1)$ and ${\Gamma }_0{n}_{\Omega }$. A sufficient common time window for both initial spins isFor a specified initial spin, the upper condition is simply its own flip rate times $t\ll 1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.

To derive fermionic Einstein-coefficient relations, first consider an idealized emitter that can exchange one ultrarelativistic fermion with the bath, with zero chemical potential. Let ${\mathcal{D}}_{\nu }=d4\pi {\nu }^2/c^3$ be its density of states per volume and ordinary frequency, with $d$ populated helicity states; for one neutrino helicity $d=1$, whereas the corresponding photon count is two. The Fermi-Dirac distribution givesPauli blocking changes the final-state factor in downward emission from the bosonic $1+f$ to $1-f$. Define positive coefficients $B_{12},C_{21}$ and $A_{21}=C_{21}h\nu {\mathcal{D}}_{\nu }$. The upward rate per lower emitter is $B_{12}{u}_{\nu }$, and the downward rate per upper emitter isThermal detailed balance, $n_1{B}_{12}{u}_{\nu }=n_2(A_{21}-C_{21}{u}_{\nu })$ with $n_2/n_1=(g_2/g_1)e^{-h\nu /(k_{B}T)}$, yieldsFor example, substituting $e^{h\nu /(k_{B}T)}=x$ reduces the balance equation to $B_{12}=(g_2/g_1)C_{21}$ because $x{f}_{\nu }=1-f_{\nu }$.

If one insists on the original additive notation $A_{21}+B_{21}{u}_{\nu }$, its formal coefficient is negative:A negative correction means inhibited emission into occupied modes, not a negative total decay rate; $0\le f_{\nu }\le 1$ ensures positivity. The dimensions are those of part (a), and the factor of two relative to the photon density is a helicity count, independent of the statistical sign.

There is a physical qualification: ordinary electronic energy levels of the same chemical atom have the same fermion parity, and cannot interconvert by emission or absorption of a single neutrino. A realizable single-fermion emitter must change the relevant particle-number and parity sector, with matching conservation laws. For ordinary atomic levels, neutrino-pair processes would require a different calculation. A bath with nonzero chemical potential would also require the corresponding chemical-equilibrium population ratio. The calculation above is the intended single-mode statistical analogy under explicit assumptions.

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.

Interpret the box as a hard-wall cube. Its one-particle levels are ${ϵ}_{n_x{n}_y{n}_z}={\pi }^2{\hslash }^2(n_x^2+n_y^2+n_z^2)/(2m{L}^2)$ for positive integers, giving $E_0=3{\pi }^2{\hslash }^2/(2m{L}^2)$. Counting lattice points in a positive octant and differentiating the large-radius count gives the supplied continuum density of states. Expressing its prefactor through $E_0$ yieldsHere $\mathcal{A}$ denotes the dimensionless constant requested in this question, to distinguish it from the unrelated hyperfine constant of question 1. Periodic boundary conditions would have zero ground energy and would not permit this parametrization by a nonzero $E_0$.

Separate the ground-state occupation from excited particles. The finite-box ground-energy convention in Bose condensation makes the chemical potential approach the ground energy from below as that occupation becomes macroscopic. In the thermodynamic continuum approximation one shifts the energy zero to the ground state and sets the limiting chemical potential to zero in the excited-state integral. The difference between the exact shifted density and its $\sqrt{ϵ}$ bulk form gives finite-size corrections; at the transition $k_B{T}_c/E_0\sim N^{2/3}\gg 1$. ThusThe identity follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating each term. The excited states have a finite saturation capacity. Setting that capacity equal to $N$ determines condensation:Equivalently $k_B{T}_c=(2\pi {\hslash }^2/m)[n/\zeta (3/2){]}^{2/3}$, with $n=N/L^3$.

At fixed $N$, below $T_c$ the chemical potential is pinned to its limiting value andAbove $T_c$, $N^{\prime }\simeq N$ and there is no macroscopic $N_0$; the chemical potential adjusts rather than allowing the saturation expression to exceed the available particles. At fixed $T$, put $N_c=N_{\max }^{\prime }(T)$. For $N<N_c$, $N^{\prime }\simeq N$ and $N_0$ is not macroscopic; for $N>N_c$, $N^{\prime }=N_c$ saturates while $N_0=N-N_c$ grows linearly. This is particle saturation in a homogeneous Bose gas; the sharp threshold is a thermodynamic-limit approximation to finite-box occupations.

The interaction scale compares differently with confinement and transition energies:Thus the stated range makes interactions dominate the ground-state confinement energy while remaining small on the thermal scale controlling $T_c$. This is an interaction-dominated ground state with nearly ideal condensation temperature. It also gives the dilute-gas parameter $n{a}^3=N(a/L{)}^3\ll 1$. A contact model has $g=4\pi {\hslash }^{2}a/m$ and $\mu \simeq gn$; the question's $E_I$ suppresses order-one factors, and the uniform ground energy per particle is $gn/2$, rather than exactly $gn$. Weak interactions can shift $T_c$ slightly, but their dimensionless strength $a{n}^{1/3}\ll 1$ makes that a small relative effect. In a hard-wall box, boundary healing layers do not overturn bulk interaction dominance when their thickness is small compared with $L$.

At $k_{B}T\ll \mu$, thermally occupied Bogoliubov modes have $E_k\simeq \hslash c_{s}k$, with $c_s=\sqrt{\mu /m}$. There is one acoustic branch for this spinless gas. Its thermal energy determines the phonon heat capacity of a dilute Bose gas:The zero-temperature interaction energy does not itself contribute to this leading temperature derivative at fixed $N,V$. This bulk power law also requires $k_{B}T\gg \hslash c_s/L$ so that the phonon spectrum may be treated as continuous; a strictly finite box eventually has discrete-level activated behaviour.

For the noninteracting condensed Bose gas, the ideal Bose-gas heat capacity below condensation follows by integrating $ϵg(ϵ)f(ϵ)$:Its limiting value at $T_c^{-}$ is about $1.926N{k}_B$. At very low temperature the interacting phonon curve is suppressed relative to the ideal curve, has slope three on logarithmic axes, and crosses over towards particle-like $T^{3/2}$ behaviour around $k_{B}T\sim \mu$. Since $\mu \ll k_B{T}_c$, most of the interval below $T_c$ is approximately ideal at leading weak-coupling order. The figure marks a qualitative extrapolation near the transition: a zero-temperature Bogoliubov spectrum does not determine precise interacting critical heat capacity or its small transition shift. The illustrative fixed-$\mu$ interpolation is not claimed as a controlled critical-region calculation.

Warming thermally ionizes neutral donors, depleting the occupied bound ground states from which these absorption lines start. Increased scattering can also broaden the lines. Once the donors are predominantly ionized, their bound-to-bound absorption becomes weak or disappears.

The inferred binding provides the natural thermal scale:The first excitation energy similarly corresponds to about $92.5\mathrm{K}$. An order-of-magnitude estimate is therefore a temperature of order $10^2\mathrm{K}$. This is not a sharp onset: significant ionization can start below that scale because free carriers have a large translational density of states. The precise loss of line strength requires donor density and compensation, as in part (ii), rather than just equating one energy with $k_{\mathrm{B}}T$.

The density of states $D(E)$ counts available one-electron states: $D(E)dE$ is the number with energies in $[E,E+dE]$. It describes the spectrum, irrespective of whether those states contain electrons.

The density of occupied states additionally weights each state by the Fermi-Dirac distribution:Thus $Z(E)dE$ is the mean number of electrons in that energy interval, and $N_e=\int Z(E)dE$. State and electron densities must use the same normalization, including whether they are extensive or per volume and whether $D$ already includes spin degeneracy. The symbol $Z(E)$ here denotes occupied density, not a canonical partition function.

Let $D$ be the equal constant density of states per unit volume and energy in each band. In the nondegenerate semiconductor regime, the conduction-band occupation isFor bands broad compared with $k_{B}T$, integrating above the conduction edge givesThe valence-hole integral similarly gives $p=D{k}_{B}T{e}^{-(\mu -E_v)/(k_{B}T)}$. Midgap $\mu$ therefore impliesInserting the carrier densities into the transport expression yieldsOn treating the algebraic and mobility prefactor as approximately constant, the requested leading activation law isThe factor $1/2$ is essential: an intrinsic excitation creates an electron–hole pair, and charge neutrality places the chemical potential halfway across the symmetric gap. The conduction-electron population is governed by $E_c-\mu =E_g/2$, not by replacing this separation with the full gap.

There is a qualification to the proportionality as printed. A literal constant-density-of-states intrinsic semiconductor with temperature-independent mobilities retains the additional factor $T$. Indeed, using the full Fermi-Dirac distribution giveswhich reduces to the expression above when $E_g\gg k_{B}T$. Thus the pure exponential is an activation approximation, not an exact consequence of constant DOS alone. A mobility sum proportional to $T^{-1}$ would cancel the leading algebraic factor.

The ambient-temperature Fermi level is raised into the upper part of the gap, toward the conduction-band edge, relative to intrinsic silicon. The sample is an n-type semiconductor: ionized donors supply many more conduction electrons than the intrinsic population.

For a nondegenerate, substantially ionized sample with negligible acceptors, $n\simeq N_D$. With effective density of states $N_c$, the conduction-band relation givesBecause a few ppm normally gives $N_D\ll N_c$ at ambient temperature, the chemical potential remains below the conduction-band edge rather than entering it as in a strongly degenerate semiconductor. Its precise position requires the dopant density, density of states and degree of ionization; the broad qualitative shift is fixed by donor doping.

It is not necessary for $\mu$ to lie between the donor level and the conduction edge. In a substantially ionized nondegenerate sample, the chemical potential can lie below the shallow donor level, as in the illustrative diagram. Donor ionization is determined by the level's occupation relative to $\mu$; a finite-temperature Fermi level is not the highest filled one-electron level of every individual configuration.

At zero temperature, states are occupied up to $E_{\mathrm{F}}$. Write the density of states as $g(E)=C\sqrt{E}$, including any fixed volume and spin factors in $C$. ThenwhileDividing eliminates $C$:The mean is positive even at absolute zero because the Pauli exclusion principle fills a range of momenta; electrons cannot all occupy the zero-momentum state.

Potassium is electropositive and transfers electron density toward the metal. The resulting surface dipole generally lowers the metal work function at such submonolayer coverages, reducing the tunnelling barrier and its inverse decay length $q\propto \sqrt{{\Phi }_{\mathrm{effective}}}$.

At the same specified $5\mathring{\mathrm{A}}$ separation, a smaller $q$ gives a larger $\exp (-2qs)$, so the expected tunnelling current is greater than $1\mathrm{nA}$ in the barrier-dominated comparison. No numerical value follows without the modified work function and electronic density of states. A real STM image also depends on the local density of states and adsorbate corrugation; equal geometric separation does not imply equal electronic barrier height.

Assume thermal equilibrium, nondegenerate carriers, and isotropic parabolic band dispersions, with the supplied effective masses interpreted as density-of-states masses. Take one effective valley with two electron spin states for each band; any additional valley multiplicity must instead be absorbed into the masses or explicit degeneracy factors.

Let the conduction band edge be $E_c$, the valence band edge $E_v$, and $E_g=E_c-E_v$. The spin-summed conduction density of states per volume isUsing the nondegenerate limit of the Fermi-Dirac distribution and the substitution $x=(E-E_c)/(k_{B}T)$ givesThe factor $\sqrt{\pi }/2$ in the supplied integral produces the normalization. Counting missing valence band electrons similarly givesMultiplication cancels the chemical potential, establishing the semiconductor mass-action law:The result does not require an intrinsic semiconductor: semiconductor doping changes $n$ and $p$ separately while preserving this product under the stated equilibrium and nondegeneracy assumptions.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.

An intrinsic semiconductor has thermally generated conduction electrons and valence holes, with $n=p=n_i$ at equilibrium in the ideal undoped neutral material. Their abundance is controlled by its band gap, temperature and band density of states.

An extrinsic semiconductor has its mobile carrier populations substantially modified by electrically active impurities. A donor impurity can supply electrons, producing n-type material; an acceptor impurity can capture a valence electron and leave a mobile electron hole, producing p-type material. Compensation and incomplete ionization can modify these simple counts. At low temperature dopant freeze-out can occur; at sufficiently high temperature intrinsic pair production can overwhelm a fixed dopant density. Doping and the dominant transport regime are related but are not interchangeable descriptions at every temperature.

Use periodic boundaries in a box of side $L$ and include two spin states for each wavevector. One orbital state occupies $(2\pi /L{)}^d$ in wavevector space. Let $v_d$ be the volume of the unit ball in dimension $d$, namely $2,\pi ,4\pi /3$. Filling the ball of radius $k_F$ givesSince $E_F={\hslash }^2{k}_F^2/(2m)$,The density $n$ means number per length, area or volume as appropriate; it is not assigned three-dimensional units in every case.

For the same spin-summed states, the cumulative number per $d$-dimensional volume below energy $E$ is $n(E)=2v_d(2mE/{\hslash }^2{)}^{d/2}/(2\pi {)}^d$. Differentiating yields $g(E)=(d/2)n(E)/E$. Evaluating at the filling energy,In one, two and three dimensions the density of states varies respectively as $E^{-1/2}$, a constant and $E^{1/2}$. The normalization counts both signs of momentum and both spin states exactly once. A single-spin density of states or extra valley degeneracy would require a different counting convention, not an extra unexplained factor in these constants.

For a long periodic chain of length $\mathcal{L}=Na$, neighboring allowed signed wavevectors differ by $2\pi /\mathcal{L}$. Counting both signs once gives a state density per length $dk/\pi$ on the positive half-zone. Therefore the density per angular-frequency interval and per length isIt is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times 10^{-5}$ s/m and has an integrable inverse-square-root Van Hove singularity as $\omega \to {\omega }_m^{-}$. The state-count normalization isThus the scalar chain has one vibrational mode per atom, with no extra factor of two for the already counted propagation directions. If the horizontal variable were ordinary frequency $\nu$, the density would be $g_{\nu }(\nu )=2\pi g_{\omega }(2\pi \nu )$. A finite chain has discrete lines instead of an actual infinite peak; the smooth divergent curve is the large-chain continuum description.

The density of states $g(E)$ counts one-electron states per energy interval, including both electron spin orientations. It may be a total or a per-volume quantity; $U$ must have the matching normalization. The Fermi-Dirac distributionis the occupation probability of such a state. In the fixed-volume free electron model, the integration variable $E$ and the density of states do not depend on $T$. The occupation depends explicitly on $T$ and implicitly through the chemical potential $\mu (T)$ needed to preserve electron number. Thermal expansion or a changing electronic band structure would require additional effects outside this model.

At $T\ll T_F$, only states within order $k_{B}T$ of the Fermi energy change their occupation. There are order $g(E_F)k_{B}T$ such states, each carrying an excitation energy of order $k_{B}T$. Thus $U(T)-U(0)\propto g(E_F)(k_{B}T{)}^2$ and the electronic heat capacity is proportional to $g(E_F)k_B^{2}T$. Using the supplied coefficient,The shell argument fixes the scaling, not the numerical coefficient; the controlled Sommerfeld expansion fixes that coefficient at fixed electron number. Filled states far below $E_F$ cannot accept arbitrary thermal excitations because of the Pauli exclusion principle.

Let $g_v(E)$ denote the spin-summed density of states per volume. An electron moment with magnitude the Bohr magneton ${\mu }_B$ has Zeeman energies $\mp {\mu }_{B}B$. At low temperature, the two spin populations areThe common chemical potential changes only at order $B^2$, so it does not affect this linear imbalance. The magnetization is $M={\mu }_B(n_{+}-n_{-})={\mu }_B^2{g}_v(E_F)B$. Define SI volume magnetic susceptibility by $M={\chi }_{p}H$ and take the weak applied field as $B\simeq {\mu }_{0}H$. Then Pauli paramagnetism givesThe last equality uses the three-dimensional parabolic free electron model, $g_v(E_F)=3n/(2E_F)$. This susceptibility is nearly independent of temperature for $T\ll T_F$, unlike the Curie law for localized independent moments. The displayed result excludes orbital diamagnetism and interaction enhancements. If susceptibility instead denotes $\partial M/\partial B$, omit ${\mu }_0$; confusing that convention with $\partial M/\partial H$ changes the units.

In the one-band electron Hall effect model, $R_H=-1/(ne)$, with $e>0$. The measured Hall coefficient therefore givesA monovalent body-centered cubic lattice has two conduction electrons per conventional unit cell. Thus $n{a}^3=2$, andCounting two spin states within the Fermi sphere gives $k_F=(3{\pi }^{2}n{)}^{1/3}=9.0418\times 10^9{\mathrm{m}}^{-1}$. Using the specified effective mass in the dispersion,There is one mole of conduction electrons per mole of sodium. Its molar density of states at $E_F$ is $3N_A/(2E_F)$, givingUsing the volume density of states in c givesThe effective mass changes the dispersion and density of states; the electron Bohr magneton used for the spin coupling retains the bare electron mass. Replacing it by $e\hslash /(2m^{\ast })$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.

Assume thermal equilibrium, a three-dimensional nondegenerate semiconductor, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied effective masses as scalar density-of-states parameters. With conduction energy $E=E_c+\epsilon$, state counting gives the per-volume density of statesSince $E_c-\mu \gg k_{B}T$, replace the Fermi-Dirac distribution by its Boltzmann factor. Using $x=\epsilon /(k_{B}T)$ and the supplied integral,The electron hole occupation is $1-f(E)$ in the valence band. Applying the same integration below $E_v$ givesTheir product cancels the chemical potential, giving the semiconductor mass-action lawThus, for masses expressed in ordinary mass units,The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or Curie constant above. Valley degeneracy or multiple valence bands must be included in the effective density of states, or absorbed into suitable density-of-states effective masses, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.^[ref]

The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu -E_c=10k_{B}T$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq 24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq 22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.