The normalized radial probability density givesConsequently the expectation value of the potential energy isHere the last equality uses $Z{e}^2/(4\pi {ϵ}_0)={\hslash }^2\beta /m_e$. Since this is an energy eigenstate, $E=⟨T⟩+⟨V⟩$, and so $\boxed{⟨T⟩=-E}$. The kinetic energy is positive because $E<0$, in agreement with the virial theorem for a Coulomb potential.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

A Stern-Gerlach experiment uses an oven and slits to produce a collimated atomic beam, conventionally of neutral silver atoms, and sends it through an inhomogeneous magnetic field onto a position-sensitive detector. The unpaired outer electron supplies the effective spin one-half magnetic dipole moment; neutral atoms avoid the ordinary Lorentz deflection of a charged beam. For an analysing direction $z$, the spatial gradient produces a force approximately $F_z={\mu }_z{\partial }_z{B}_z$. The resulting separated beam positions measure the allowed projections of angular momentum. A spin one-half beam gives two branches, rather than the continuous distribution expected from classical dipoles with arbitrary orientations. A uniform magnetic field can rotate a spin but does not itself provide this spatial separation.

To measure the rotation, first use a $z$-directed Stern-Gerlach experiment and an aperture to transmit only the $+z$ branch. Pass this prepared spin eigenstate through a region of approximately uniform field in the $x$ direction, then use a second $z$ analyser to separate and count the two populations. For a field region of length $\ell$ and beam speed $v$, the precession angle is $\theta =\lambda B\ell /v$, and the predicted fractions areVarying $B$ or the transit time gives complementary oscillations, demonstrating coherent Larmor precession. Velocity selection and a sufficiently uniform rotation field keep different particles from accumulating different phases and washing out the contrast. Detection should compare the branch fractions after allowing for losses and analyser efficiency.

Put $\Omega =\lambda B$, including its sign. The initial vector is the $+\hslash /2$ eigenstate of $S_z$; it is not an energy eigenstate of $H=\Omega S_x$. In terms of the spin eigenstates found above, $|+z⟩=(|+x⟩+|-x⟩)/\sqrt{2}$. Each energy eigenstate acquires its own unitary time evolution phase:Equivalently, $e^{-iHt/\hslash }=I\cos (\Omega t/2)-i{\sigma }_x\sin (\Omega t/2)$ follows from ${\sigma }_x^2=I$. The Born rule givesThe expectation value of the spin angular momentum is $(\hslash /2)(0,-\sin \Omega t,\cos \Omega t)$, a rotation around the $x$ axis. Its direction follows the specified sign of the Hamiltonian; changing the magnetic coupling sign reverses the precession.

In the $S_z$ eigenbasis, let a candidate eigenvector be $(a,b{)}^{\mathsf{T}}$. The eigenvalue equation is $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$, so a nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized phases givesThe two spin eigenstates are orthogonal, as required for distinct eigenvalues of a self-adjoint operator.

Write $S_i=(\hslash /2){\sigma }_i$, with ${\sigma }_i$ the Pauli matrices. Direct multiplication gives ${\sigma }_x^2={\sigma }_y^2={\sigma }_z^2=I$, and henceFor the products in the other order,Their difference gives the angular momentum commutation relationsThe first identity corresponds to the spin one-half value $s(s+1){\hslash }^2$ with $s=1/2$.

A dispersion relation specifies the allowed angular frequency $\omega (k)$ of a wavevector $k$. A single harmonic wave $A\cos (kx-\omega t)$ has constant-phase surfaces moving at the phase velocity,For two nearby components, use the identityThe slowly varying envelope moves at $\Delta \omega /\Delta k$. Taking a narrowband limit gives the group velocity,More generally, expand $\omega (k)$ around the central wavevector of a wavepacket. The linear term translates its envelope at $v_g$; the quadratic and higher terms produce group velocity dispersion and packet distortion. This interpretation needs a sufficiently narrow spectrum and weak enough absorption and distortion for an identifiable envelope to persist.

For light in a transparent material, $k=n(\omega )\omega /c$, soThe derivative is evaluated at the wavepacket's central frequency and ${\lambda }_0$ is the vacuum wavelength. In a nondispersive medium with constant $n$, the two velocities coincide and a packet can propagate without dispersion. In normal optical dispersion $dn/d{\lambda }_0<0$, the displayed model gives $v_g<v_p$.

The two velocities can differ even without material dispersion: a waveguide mode with ${\omega }^2={\omega }_c^2+c^2{k}^2$ has $v_p=\omega /k>c$ and $v_g=c^{2}k/\omega <c$, with $v_p{v}_g=c^2$. The phase velocity is not a signal speed. Near strongly absorbing resonances an anomalous derivative can also make the group velocity exceed $c$ or become negative; reshaping then prevents its simple identification with information transport. Relativistic causality constrains the signal front, not the speed of an arbitrary moving phase or distorted intensity maximum.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

The convolution of two aperture functions in one dimension isWith Fourier transform convention $\hat{f}(q)=\int f(x)e^{-iqx}dx$, the convolution theorem statesThe first identity follows by substituting the convolution integral, putting $v=x-u$, and separating the resulting double integral into $\int f(u)e^{-iqu}du$ times $\int g(v)e^{-iqv}dv$. For an absolutely integrable pair the interchange is justified by absolute convergence; finite optical apertures satisfy suitable integrability assumptions.

In Fraunhofer diffraction, the observed field is the Fourier transform of the transmitted complex aperture field. For two identical slits, let $g(x)$ be one slit and use a pair of Dirac delta distributions to locate the centres:Consequently $\hat{t}(q)=2\hat{g}(q)\cos (qD/2)$. For a rectangular slit, $\hat{g}=a\mathrm{sinc}(qa/2)$, giving a single-slit diffraction envelope times the double-slit interference factor in intensity. A $\pi$ phase reversal replaces the plus between the deltas by a minus, giving $\sin (qD/2)$ instead, as in B7.

More generally, an $N$-slit array is a single-aperture function convolved with a finite comb. Its field contains the factorThis separates the narrow diffraction grating maxima from the individual-aperture envelope. The same reasoning holds in two transverse dimensions. Multiplying an existing aperture by an additional finite mask instead convolves their field transforms and can broaden features. These are operations on complex fields: intensity is the squared modulus of the final field and cannot generally be substituted into the field-level convolution theorem.

In a normal metal, the electrical resistivity can often be approximated using Matthiessen's rule,Static impurities, defects and grain boundaries scatter conduction electrons even as $T\to 0$, giving the residual resistivity. Thermal lattice phonons add a temperature-dependent contribution through electron-phonon scattering. In a clean three-dimensional acoustic model, small-angle electron-phonon transport at low temperature gives ${\rho }_{\mathrm{phonon}}\propto T^5$ at low $T$; once thermal phonon populations are approximately classical, the resistivity is approximately linear in $T$. An electron-electron contribution proportional to $T^2$ can be relevant in some low-temperature ranges.

Carbon in mild steel, together with associated disorder, strain and microstructural changes, adds scattering and generally raises the residual resistivity relative to purer iron. Thus a dilute alloy may have a similar broad phonon-dependent rise but a larger baseline. The sketch is a generic phonon-plus-disorder model, not measured iron data. Iron's magnetic scattering and phase changes, and real steel's phases and processing history, can add structure; the simple sum is approximate rather than an exact material law. The expected explanation is greater electron scattering from the added carbon and resulting disorder.

Interpret “3 mm wire” as a circular wire of diameter $3\mathrm{mm}$, and take the current density as uniform. One conduction electron per atom givesUsing $I=neA|v_d|$, the drift velocity corresponding to a current magnitude of $13\mathrm{A}$ isIt points opposite to conventional electric current. A simple distance scale over the $1/50=0.020\mathrm{s}$ cycle is $\boxed{|v_d|/f=2.71\mu \mathrm{m}}$, but this treats the magnitude as constant. For sinusoidal mains, alternating-current drift excursion gives zero net displacement after a full cycle.

If $13\mathrm{A}$ is the usual root mean square current, $v_{\mathrm{pk}}=\sqrt{2}|v_d|$ and the displacement amplitude and total absolute drift distance per cycle areIf $13\mathrm{A}$ instead denotes peak current, these become $0.431\mu \mathrm{m}$ and $1.72\mu \mathrm{m}$. The paper does not state peak versus RMS; the characteristic drift is the boxed $1.35\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-1}$ on whichever current convention is chosen. It also does not explicitly say diameter; taking $3\mathrm{mm}$ as the radius would reduce all these velocities and drift distances by four. None of these distances is the microscopic path travelled at roughly the Fermi velocity between scattering events; that path cannot be obtained from the supplied current alone.

Let $e>0$ denote the elementary charge, so an electron's charge is $-e$. In the relaxation-time Drude model with positive transport effective mass,The steady drift velocity is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }=-\mu \mathbf{E}$, defining positive electron mobility $\mu =e\tau /m^{\ast }$. The current density is opposite to the electron drift:By the definition of electrical conductivity, $\mathbf{J}=\sigma \mathbf{E}$, soIts units are $({\mathrm{m}}^{-3})(\mathrm{C})({\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1})=\mathrm{S}{\mathrm{m}}^{-1}$. The mobility definition alone establishes the result in the scalar linear-response regime; the relaxation-time equation supplies a microscopic approximation for $\mu$.

A Bloch state in a crystal experiences a periodic lattice potential. Its band dispersion relation $ϵ(k)$ encodes the lattice response, so an effective mass allows a wavepacket's acceleration under a weak applied force to be described without explicitly following the microscopic periodic forces.

In one dimension the wavepacket group velocity and semiclassical equation for crystal momentum areDifferentiating the first equation with respect to time givesDefining $F=m^{\ast }a$ yieldsThis band effective mass is controlled by band curvature, not the slope. A band minimum has positive $m^{\ast }$; a band maximum has negative electron $m^{\ast }$ and is often more conveniently described with positive-mass electron holes. In three dimensions the inverse effective mass is generally a tensor, $(m^{\ast -1}{)}_{ij}={\hslash }^{-2}{\partial }_{k_i}{\partial }_{k_j}ϵ$. The scalar formula assumes motion along a specified direction in an isolated band, weak fields and negligible interband transitions; the bare electron mass has not physically changed.

Assume sodium supplies one conduction electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

Only a shell of energy width of order $k_{B}T$ near the Fermi energy can change occupation: much lower states are blocked by the Pauli exclusion principle. The thermal-shell estimate of electronic heat capacity countsTo recover the rough coefficient requested in the paper, assign each of these active particles an excitation energy of order $3k_{B}T/2$. ThenThe differentiation matters: the active shell itself widens as temperature rises. This argument explains the small linear-in-$T$ electronic heat capacity; its shell width and average excitation energy are heuristic, so it does not establish a precise numerical prefactor.

A controlled calculation from the Fermi-Dirac distribution gives a nearby but different result. At fixed particle number, the Sommerfeld expansion yieldsThus $9/2$ is a rough estimate of the exact free-electron coefficient ${\pi }^2/2\simeq 4.935$, not an exact consequence of the stated density of states. Both require $T\ll T_F$.

The occupation probability of an individual one-electron state is the Fermi-Dirac distribution,At $T=0$, it is a step: one below $E_F$ and zero above it. Heating rounds the step over an energy range of order a few $k_{B}T$, creating empty states below and occupied states above the chemical potential. At every positive temperature, $f(\mu ,T)=1/2$.

For fixed $N,V$ in the three-dimensional free electron model, the chemical potential is determined by $N=\int g(ϵ)f(ϵ,T)dϵ$; it is not exactly constant. In the degenerate regime $T\ll T_F$, the Sommerfeld expansion givesIt is therefore reasonable in a simple low-temperature sketch to set $\mu \simeq E_F$, provided this approximation is stated. The figure includes the leading small shift. At much higher temperature the fixed-$E_F$ approximation fails and the distribution approaches the classical dilute limit. The plotted occupation is $f$, rather than the energy distribution $gf$.

Impose periodic boundary conditions on a large cubic box. The allowed wavevectors occupy a lattice of cell volume $(2\pi {)}^3/V$ in wavevector space. Including the two spin states, the number of states with magnitude at most $k$ isUsing $k=\sqrt{2m_eϵ}/\hslash$ and differentiating gives the total, spin-summed three-dimensional free-electron density of states,It counts states per unit energy; divide by $V$ for states per unit volume per unit energy. At zero temperature, the Fermi-Dirac distribution fills all states up to the Fermi energy, givingWriting $n=N/V$ givesThe energy zero is the bottom of the free-electron band. The electron Fermi temperature is $T_F=E_F/k_B$.

Use the quantum free electron model: conduction electrons move as independent particles in a three-dimensional box of volume $V$, with a uniform positive background replacing the ionic lattice potential. Neglect the detailed periodic potential and explicit electron-electron interactions; the interior one-electron dispersion relation is $ϵ={\hslash }^2{k}^2/(2m_e)$. The boundary confines the particles, and its detailed shape does not affect the bulk state count in a sufficiently large sample.

The electrons are spin-$1/2$ fermions: each spatial quantum state accommodates two opposite spin states, with occupations given by the Fermi-Dirac distribution and the Pauli exclusion principle. A fixed number $N$ of conduction electrons and thermal equilibrium are assumed. Collisions can be represented by a phenomenological momentum relaxation time for transport, but are unnecessary for the equilibrium state counting here. These assumptions distinguish the quantum model from a classical Maxwell-Boltzmann electron gas.