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The joint density of states counts vertical valence-to-conduction transitions at photon energy $\hslash \omega$. Optical absorption weights it by dipole or momentum matrix elements, occupation factors and optical-frequency prefactors. A Van Hove singularity occurs at a stationary transition energy ${\nabla }_{\mathbf{k}}(E_c-E_v)=0$. It can create a strong peak, edge or cusp in an absorption spectrum and thereby reveal electronic band structure. The relevant stationary quantity is the band difference, not necessarily either band separately. Forbidden optical transitions can suppress the feature; excitons, finite lifetimes and disorder can shift or broaden it. A two-dimensional saddle gives a logarithm, whereas a parabolic two-dimensional minimum gives a step. ^[ref]

Put $\delta =\hslash \omega -E_g$, $A={\hslash }^2/(2m^{\ast })>0$ and $\eta =\delta /A$. For $\delta >0$, the saddle-point optical transition contour is the two-branch hyperbolaFor $\delta <0$ it opens along y instead; at zero the limiting contour is $k_x=\pm k_y$. On either positive-energy branch,Using the opening spin factor2, one branch therefore contributesThe displayed integral in the question is this one-branch expression with ${\Lambda }_y=\pi /a$. With both hyperbola branches and the opening spin factor2, the full joint density of states is $J_{\mathrm{full}}=2J_{\mathrm{branch}}$. Equivalently, the printed coefficient describes the full two-branch contour if the spin degeneracy is1. The PDF itself has this factor-of-two mismatch; it cannot be removed by reparametrizing the complete contour.

There is also a cutoff qualification. For the strict square first Brillouin zone, $|k_x|,|k_y|\le \Lambda =\pi /a$, both coordinates must remain inside, giving ${\Lambda }_y=\sqrt{{\Lambda }^2-\eta }$ for $0<\eta <{\Lambda }^2$. Taking ${\Lambda }_y=\Lambda$ is a convenient local-saddle cutoff with the same leading singularity, not the exact intersection with that square. The quadratic dispersion is only a local band model, so the nonsingular constant depends on how the saddle region joins the actual bands.

For the source's cutoff, the logarithmic two-dimensional saddle joint density of states follows directly:Thus, in the one-branch normalization of the requested integral,The strict square cutoff has the same leading logarithm. This establishes the requested $C-\ln (\hslash \omega -E_g)$ shape, with a dimensionless logarithm after choosing the cutoff energy. Here $E_g$ is the saddle transition energy; it need not be the global minimum optical gap. The definition in the question has units of states per area per energy, despite being written as a function of frequency.

Choose the growth direction as $z$ and the well as $0<z<L$. In the envelope function approximation, write a subband state as $A^{-1/2}{e}^{i{\mathbf{k}}_{\parallel }\cdot {\mathbf{r}}_{\parallel }}{F}_n(z)u_{j0}(\mathbf{r})$, where $u_{j0}$ is the band-edge cell-periodic Bloch factor. The infinite square well hasThe envelope has the same shape for electron and hole states: changing their effective masses changes the eigenenergies, but not these infinite-well eigenfunctions. With in-plane polarization, the leading interband momentum matrix element acts on the Bloch factors, while the slow envelopes giveorthonormality therefore yields the infinite-well interband selection rule $n^{\prime }=n$. A nonzero band-level matrix element is still required; the overlap rule does not replace polarization selection rules. Finite barriers can make electron and hole envelopes different, and interface asymmetry or band mixing can relax this ideal same-index rule.

For a fixed allowed band/subband pair, parabolic dispersions give $E_c(\mathbf{k})-E_v(\mathbf{k})=E_{j,n}^{\mathrm{edge}}+{\hslash }^2{k}_{\parallel }^2/(2{\mu }_j)$. If the photon energy exceeds the edge, the contour is a circle of radius $k_{\parallel }$ and the gradient magnitude is ${\hslash }^2{k}_{\parallel }/{\mu }_j$. ThusBelow threshold the contour is absent. With the conventional $d^{2}k/(2\pi {)}^2$ normalization, the two-dimensional joint density of states per area and independent channel isWith a locally energy-independent optical matrix element, each allowed pair therefore adds a constant step to the continuum dielectric function in the approximation of the supplied equation. Different hole families need not have equal step heights: reduced masses, spin multiplicities and Bloch-state matrix elements differ. Frequency-dependent overall optical prefactors, nonparabolicity, broadening and excitonic Coulomb enhancement can modify the literal step shape outside this approximation.

electron confinement raises the conduction edge and hole confinement lowers the corresponding valence edge, so their positive confinement energies add. With $L=120\mathring{\mathrm{A}}=12$ nm, the common bare-mass scale isWriting ${\eta }_e=0.067$ and ${\eta }_j=m_{h,j}^{\ast }/m_e$, the quantum-well interband thresholds areUse $E_{g,\mathrm{hh}}=E_{g,\mathrm{lh}}=1.52$ eV and $E_{g,\mathrm{so}}=1.86$ eV. The thresholds in the requested window, using the printed masses without fitting, areThe next heavy-hole and split-off thresholds are about $2.2224$ and $2.3674$ eV, outside the interval. Each threshold marks a continuum onset or shoulder, not the stronger bound-exciton peak below it. The six heavy- and light-hole onsets follow the six paired peak groups in the supplied diagram. Their first members lie roughly 20 and 12 meV below their respective onsets, as the common solution above explains.

A qualification is necessary for the split-off branch. Its p-like $j=1/2$ bloch states have nonzero $p_x$ matrix elements to the s-like conduction states, so in-plane polarization does not generally forbid them. For example, a $j=1/2$ state has a nonzero $p_x\pm i{p}_y$ orbital component with the appropriate spin, which couples to an s-like state through $p_x$. The same-envelope rule therefore allows the two split-off thresholds in the table. The first nearly coincides with the third heavy-hole edge, while the second would give an additional onset near 2.086 eV. The supplied figure and twelve peak energies appear to retain only the heavy- and light-hole series; they do not specify a split-off-series oscillator strength or identify its extra features. Omitting the split-off transitions would require an additional approximation beyond the four-band information and polarization given. Their expected ideal two-dimensional first-state peaks would be near 1.9010 and 2.0701 eV; these are predictions of the stated model, not additional listed data.

The absorptive dielectric function counts optically accessible electron-hole pairs. In the electric dipole approximation the perturbation is proportional to $\mathbf{a}\cdot \hat{\mathbf{p}}$, with $\mathbf{a}$ the light polarization and $\hat{\mathbf{p}}$ the electronic momentum operator. The squared matrix element between occupied valence and empty conduction states is the transition strength; it includes polarization and band-symmetry selection rules. The sums enumerate initial and final subbands. photon momentum is negligible on the crystal-momentum scale, so the transition is approximately vertical in the in-plane wavevector $\mathbf{k}$.

For each subband pair, $C$ is the constant-transition-energy contour in the two-dimensional Brillouin zone. Its line element is $dl$, and the reciprocal gradient weights the states satisfying energy conservation. Specifically, the level-set representation of the joint density of states isA small energy gradient means many nearly resonant states; a vanishing gradient can produce a Van Hove singularity or an edge. The displayed expression suppresses overall optical prefactors and assumes the momentum matrix element is approximately constant along each contour. More generally it stays inside that integral. Occupation factors $f_v-f_c$ are also implicit; at low temperature in an undoped well the relevant valence states are full and the conduction states empty. This is the independent-particle continuum contribution from Fermi's golden rule. Bound excitons add optical resonances below the corresponding free-pair thresholds.

The GaAs four-band model is rooted in bonding and spin-orbit coupling. Near the zone centre, the conduction edge is predominantly s-like and antibonding, while the upper valence manifold is predominantly p-like. Combining orbital angular momentum $l=1$ with spin $s=1/2$ splits the six p-derived states into a $j=3/2$ quartet and a lower $j=1/2$ doublet. The latter is the split-off valence band. Within the upper quartet, the $m_j=\pm 3/2$ and $m_j=\pm 1/2$ branches, defined relative to a propagation or quantization direction, have different curvatures and are called heavy holes and light holes. They coincide at the unstrained bulk zone centre but need not remain degenerate in a well. The conduction edge is a spin doublet. Thus “four bands” counts conduction, heavy-hole, light-hole and split-off branches, including their spin multiplicities; it does not count four individual states. Positive hole masses describe the downward valence curvature, whereas the conduction curvature is upward.

For the final exciton comparison, define the reduced mass ${\mu }_j$ and screened exciton Rydberg $R_j^{\ast }$ byThe resulting $({\mu }_j/m_e,R_j^{\ast })$ are approximately $(0.05948,4.937\mathrm{meV})$ for heavy holes, $(0.03646,3.027\mathrm{meV})$ for light holes and $(0.04631,3.844\mathrm{meV})$ for split-off holes. In the ideal two-dimensional exciton model, the first two optically active internal states lie $4R_j^{\ast }$ and $4R_j^{\ast }/9$ below their continuum edge. Their separation is $32R_j^{\ast }/9$, independent of a common threshold offset. The corresponding three-dimensional separation would be only $3R_j^{\ast }/4$. The listed peak spacings specified for this theoretical spectrum giveHere each pair comprises the stronger lower-energy peak and its weaker nearby companion, in the order associated with the continuum edges calculated in part (c). The approximately 18-meV and 11-meV pair splittings support the two-dimensional series and are incompatible with the three-dimensional series at the supplied masses and permittivity. Equivalently, for a two-dimensional pair with energies $E_1,E_2$, the inferred edge is $E_2+(E_2-E_1)/8$; the observed pair splittings predict ground-state bindings around 20 and 12 meV, close to $4R_j^{\ast }$.

Absolute agreement should not be overstated. For example, the stated square-well parameters give the first heavy-hole edge $1.56390$ eV and the first light-hole edge $1.59161$ eV. Their ideal two-dimensional peak predictions are approximately $(1.54415,1.56170)$ eV and $(1.57950,1.59027)$ eV, respectively. The listed peaks are lower by about 1–2 meV across the families; rounded parameters and this simple effective-mass model do not reproduce every absolute energy exactly. No gap, width or mass has been silently adjusted to fit them. The peak-spacing comparison remains useful because it avoids a common edge offset. The split-off transitions predicted by the four-band data are discussed separately in part (c), rather than assigned fictitious extra peaks.

At 300 K, $k_{B}T\simeq 25.9$ meV, whereas the ideal ground-state binding energies are only about $19.7$, $12.1$ and $15.4$ meV; excited states are still more weakly bound. None of these series is expected to remain a clearly resolved, robust bound-exciton peak at room temperature in this simple estimate. Thermal ionization and phonon broadening work against their visibility. This is a scale argument, not a theorem that the exciton population or all Coulomb-enhanced absorption becomes exactly zero: actual observability also requires temperature-dependent gaps, linewidths, screening and carrier density, which are not supplied.

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

Insert the interaction-picture expansion into the Schrödinger equation. The phases multiplying the basis states already solve the unperturbed evolution, leavingHermiticity requires $H_{21}^{\prime }=\hslash ({\omega }^{\prime }{)}^{\ast }$, not $\hslash {\omega }^{\prime }$ unless the coupling is real. Since the diagonal matrix elements vanish and ${\omega }_0={\omega }_2-{\omega }_1$, expanding the cosine givesThe conjugate pair of couplings conserves $|c_1{|}^2+|c_2{|}^2$. Near resonance, the difference frequency is slow while the sum frequency is approximately $2{\omega }_0$. Rapid terms alternate sign before an appreciable slow transfer accumulates. The rotating-wave approximation retains the slow terms when $|{\omega }^{\prime }|,|\omega -{\omega }_0|\ll \omega +{\omega }_0$; resonance alone is insufficient for a strong drive. Counter-rotating corrections include small shifts of the resonance and need not remain negligible at arbitrary strength or arbitrarily long times.

For the spin preparation, use $H=-\mathbf{\mu }\cdot \mathbf{B}=-\gamma \mathbf{s}\cdot \mathbf{B}$ and $\mathbf{s}=\hslash \mathbf{\sigma }/2$. Choose $B>0$ and $\gamma >0$ to fix helicity, with $|\uparrow ⟩=(1,0{)}^T$, $|\downarrow ⟩=(0,1{)}^T$ and ${\omega }_0=\gamma B$. The initial $|\uparrow ⟩$ is the lower state. For negative $\gamma$, reverse the RF rotation sense; pulse areas use $|\gamma |$, and signed Larmor precession determines the corresponding phase adjustment.

A circularly rotating magnetic drive uses two transverse fields in quadrature during $0<t<\tau$:With $U_0(t)=e^{+i{\omega }_{0}t{\sigma }_z/2}$, their interaction Hamiltonian is exactly constant,The minus sign in the second RF component selects co-rotation for the specified positive gyromagnetic ratio. Exponentiating this Pauli matrix givesThe laboratory phase of a resonant spin pulse includes free precession. Returning to the laboratory frame and removing a common phase,Unlike a linearly polarized RF field, this exactly co-rotating field has no counter-rotating component. A single linearly polarized field $B_{x,\mathrm{peak}}\cos ({\omega }_{0}t+\phi )$ gives the same slow rotation in the weak-drive rotating-wave approximation with $\Omega =|\gamma |B_{x,\mathrm{peak}}/2$. For the circular field above, $\Omega =|\gamma |B_1$. Keeping these amplitude conventions distinct avoids a factor-of-two pulse-duration error.

For a single harmonic pulse detuned by $\Delta ={\omega }_0-\omega$, the effective rotating-frame field tilts out of the transverse plane. Its rotation rate becomes $\sqrt{{\Omega }^2+{\Delta }^2}$, and its maximum population transfer is ${\Omega }^2/({\Omega }^2+{\Delta }^2)$. Exact inversion to $-z$ is impossible with one fixed nonzero-detuning pulse. Half transfer, needed for a transverse eigenstate, gives the equatorial reachability of a detuned spin pulse criterion $|\Delta |\le \Omega$. With that condition, changing the duration and RF phase can still prepare $+x$ or $+y$ at the chosen laboratory stopping time. The mere statement $\omega \approx {\omega }_0$ does not ensure this condition, because the drive can be much weaker than either frequency. If $|\Delta |>\Omega$, neither transverse target is reached by that single pulse. These restrictions concern fixed-frequency, fixed-amplitude single-pulse control; retuning, chirping or phase-controlled composite pulses change the available controls and can remove the restriction. No dissipative spin relaxation is included in this coherent two-state model.

Hermiticity supplies the reverse matrix elements, also equal to the real number $t$. In an orthonormal basis for the degenerate eigenspace, the perturbation iswhere ${\mathbf{J}}_3$ is the all-ones matrix. Its symmetric vector $(1,1,1)/\sqrt{3}$ has eigenvalue $2t$. Every vector whose components sum to zero has eigenvalue $-t$; that subspace has dimension two. For example, $(1,-1,0)/\sqrt{2}$ and $(1,1,-2)/\sqrt{6}$ form an orthonormal basis for it.

The first-order energies are thereforeThe trace of the shifts is $2t-t-t=0$, agreeing with the zero diagonal. The ordering depends on the sign of $t$; the formula does not assume that the symmetric state is the lowest. Diagonalizing the perturbation inside the degenerate space is essential: the original diagonal entries alone would incorrectly predict no splitting.

Stimulated emission adds a photon to an already occupied radiation mode; spontaneous emission can occur when that mode initially contains no photons. In the electric-dipole approximation the atom-field interaction is $H_{\mathrm{int}}=-\mathbf{d}\cdot \mathbf{E}$. Quantizing the radiation field makes the emission matrix element into a mode with occupation $n$ proportional toThis photon occupation enhancement of emission makes the emission rate into that mode proportional to $n+1$: the part proportional to $n$ is stimulated and the remaining part is spontaneous. Absorption is proportional to $|⟨n-1|a|n⟩{|}^2=n$. The photon energy is the atomic energy difference, up to any specified recoil or line broadening.

The stimulated contribution preferentially populates the stimulating mode, including its propagation direction and polarization; it provides optical gain. Spontaneous photons populate the available modes according to the transition's dipole matrix elements and radiation environment. They are not universally isotropic. An excited bare atom with the field vacuum is not an eigenstate of the fully coupled Hamiltonian, which resolves the apparent paradox of a nominally stationary excited state radiating.^[ref]

For lower and upper levels 1 and 2, with radiation spectral energy density $u_{\nu }$ per unit ordinary frequency, the rates areThermal detailed balance with the Planck spectrum givesThe statistical weights $g_i$ here are level degeneracies, not Landé g-factors. A definition of $B$ multiplying intensity or angular-frequency density would change the normalization factors.

For a specified free-space electric-dipole transition, summing radiation modes givesAppropriate sums over final magnetic substates and averages over initial substates are needed for an unresolved degenerate level. The radiative lifetime is $1/A_{21}$ only when this is the sole decay channel. In a laser, net stimulated gain requires $N_2/g_2>N_1/g_1$, as well as sufficient gain to overcome cavity loss; population inversion alone does not specify the lasing threshold.

Fermi's golden rule converts a weak transition matrix element and the available density of final states into a transition rate. For a static perturbation $W$ and a continuum,When the matrix element is nearly constant on the energy shell, this becomes $(2\pi /\hslash )|W_{fi}{|}^2\rho (E_i)$. The energy delta function is to be integrated, rather than interpreted as a finite rate to one exactly resonant discrete level. The observation time must resolve a smooth final-state continuum while remaining short enough that first-order depletion is small. A periodically driven perturbation instead gives the corresponding absorption and emission energy shells.

For elastic scattering from a weak localized potential $W(\mathbf{r})$, use box-normalized free states $⟨\mathbf{r}|\mathbf{k}⟩={\mathcal{V}}^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$. ThenThe number of final states is $\mathcal{V}{k}_f^2\mathrm{d}{k}_f\mathrm{d}\Omega /(2\pi {)}^3$. With $E_f={\hslash }^2{k}_f^2/(2m)$, the density per energy and solid angle is $\mathcal{V}m{k}_f/[(2\pi {)}^3{\hslash }^2]$. Dividing the resulting golden-rule rate by the incident single-particle flux $j_i=\hslash k_i/(m\mathcal{V})$ givesThis is the golden-rule derivation of the Born cross-section. For the elastic problem $k_f=k_i$, so the first Born scattering amplitude and cross-section areThe box volume cancels, as a physical cross-section must. The minus sign is fixed by the outgoing-wave scattering convention; the golden-rule rate alone determines only the modulus.

For example, $W(r)=g{e}^{-\kappa r}/r$ has $\tilde{W}=4\pi g/(q^2+{\kappa }^2)$, hence $f_B=-2mg/[{\hslash }^2(q^2+{\kappa }^2)]$, with $q=2k\sin (\theta /2)$. Replacing the exact scattering state by the incident plane wave is the essential Born approximation. It can fail for strong scattering, low-energy resonances and situations requiring repeated scattering; a small perturbation and a suitable continuum are separate conditions.^[ref]

Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

A spherical tensor operator of rank $k$ transforms under rotation like an angular momentum $k$. The Wigner-Eckart theorem writes its matrix elements asThe reduced matrix element contains the radial and other dynamical information; the Clebsch-Gordan coefficients determine the dependence on the projections. Ratios within a fixed pair of multiplets therefore follow without a radial integral. The angular selection rules include $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$.

For the position vector, choose the rank-one componentsLet $R=⟨{\lambda }^{\prime },3/2\Vert T^{(1)}\Vert \lambda ,1/2⟩$. The stretched Clebsch-Gordan coefficients and one lowering step giveFor example, apply the total angular momentum ladder operator to the stretched state. The result has components with coefficients $1/\sqrt{3}$ and $\sqrt{2/3}$, corresponding to lowering the initial angular momentum and the vector component, respectively.

Only the $q=+1$ component of $x$ contributes to $M_1$. The $q=0$ operator $z$ cannot connect the projections in $M_2$, so that matrix element vanishes. With the convention above,Thus the requested relative magnitudes and their squared values areThese ratios presume a common nonzero reduced matrix element. The position operator has odd parity, so an electric dipole transition requires opposite-parity multiplets; for one electron the electric-dipole selection rule also gives $\Delta \ell =\pm 1$. The displayed $j,m$ labels alone do not establish that condition. If the common reduced matrix element vanishes, all three matrix elements vanish and no nonzero ratio can be inferred.

The Zeeman effect comes from $-\mathbf{\mu }\cdot \mathbf{B}$. For an electron, the orbital and spin magnetic moments are approximately $-{\mu }_B\mathbf{L}/\hslash$ and $-2{\mu }_B\mathbf{S}/\hslash$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the spin-orbit interaction: the electron's motion through the atomic electric field produces a magnetic interaction in its rest frame. The relativistic correction, including Thomas precession, and the radial expectation value are incorporated into $K$. This effective Hamiltonian retains the linear Zeeman splitting and spin-orbit coupling, while omitting diamagnetic and hyperfine structure corrections.

Define the angular momentum ladder operators by $L_{\pm }=L_x\pm i{L}_y$ and similarly for $S_{\pm }$. Direct expansion givesso adding $L_z{S}_z$ gives the required scalar product. The off-diagonal terms change $m_{\ell }$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good quantum number at every magnetic field.