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Let the source's top terminal have voltage phasor $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left resistors are equal, so at balance $V_L=V_0/2$. Zero electric current through the ammeter requires $V_R=V_L$, and therefore equal upper and lower right-arm electrical impedances:Write $x=\omega R_x{C}_x$. Separating real parts and imaginary parts givesThe second equation implies $1+x^2={\omega }^2{R}_x^{2}C{C}_x$. Subtract $x^2={\omega }^2{R}_x^2{C}_x^2$ to obtainA finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity byFor a detector of finite electrical impedance $Z_A$, its terminal voltage is insteadHere $Z_{\mathrm{th}}$ is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit $Z_A\to 0$ has zero terminal voltage even when the bridge is unbalanced.

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

Use periodic boundary conditions temporarily to count the allowed wavevectors. Each state occupies $(2\pi {)}^3/V$ in reciprocal space. For one acoustic polarization, the spherical shell between $k$ and $k+dk$ containsBoth positive and negative wavevector directions are already included in that sphere; no further factor of two is needed. With the linear dispersion relation $k=\omega /v_s$,A three-dimensional monatomic crystal has three acoustic branches. Under the stated common-speed approximation,This is a phonon density of states per unit angular frequency, not per unit ordinary frequency. If longitudinal and transverse speeds differ, the isotropic expression becomes $g=V{\omega }^2(v_L^{-3}+2v_T^{-3})/(2{\pi }^2)$. In the Debye model, imposing ${\int }_0^{{\omega }_D}g(\omega )d\omega =3N$ sets the cutoff; the linear approximation is not continued to arbitrarily high frequency in a finite lattice.

Hamiltonian and commutators. Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ and define $c_0=\sqrt{m\omega /(2\hslash )}$. Expanding the product of the given creation and annihilation operators,Therefore the quantum harmonic oscillator Hamiltonian operator isSimilarly,Using $[AB,C]=A[B,C]+[A,C]B$ gives

Lowering a normalized energy eigenstate. The commutator impliesFor $n>0$, nondegeneracy of this one-dimensional energy level means $\hat{a}|{\varphi }_n⟩=c_n|{\varphi }_{n-1}⟩$. Its norm is fixed by the number operator:Choosing successive eigenstate phases in the conventional way makes $c_n=\sqrt{n}$, soFor $n=0$, the same norm is zero, hence $\hat{a}|{\varphi }_0⟩=0$. A nonzero lowered ground state would have energy below the lowest allowed level, and positivity of ${\hat{a}}^{†}\hat{a}$ forbids that possibility.

Coherent-state eigenvalue. Applying the annihilation operator term by term, with the $n=0$ contribution zero, givesThusThe coherent state is normalized because $e^{-|\alpha {|}^2}{\sum }_n|\alpha {|}^{2n}/n!=1$. Its annihilation operator is not Hermitian, so the eigenvalue $\alpha$ may be complex.

Time evolution. Under the time-independent Hamiltonian operator, each energy component acquires its own phase:Since $|\alpha e^{-i\omega t}|=|\alpha |$, the normalization prefactor is unchanged. HenceThe state remains a coherent state, with its complex parameter rotating at the oscillator's angular frequency; the displayed extra factor is a global phase from the zero-point energy.

The four supporting springs act in parallel, so their total spring constant is $k=4.00\times 10^4\mathrm{N}{\mathrm{m}}^{-1}$. The driving angular frequency is $\omega =2\pi (10)=62.83{\mathrm{s}}^{-1}$. The force-driven vibration isolation estimate givesAt this mass, $f_0=\sqrt{k/m}/(2\pi )=0.894\mathrm{Hz}\ll 10\mathrm{Hz}$, and $Q=\sqrt{mk}/b\simeq 14.2$, so the assumptions are consistent. The exact forced amplitude isIt is indeed close to $1\mu \mathrm{m}$, but it is slightly above the strict requested bound. Setting the exact expression to $10^{-6}\mathrm{m}$ gives the high-mass thresholdChoose a mass strictly greater than this threshold for a strict inequality; $1.28\times 10^3\mathrm{kg}$ is a suitable rounded choice. Simply rounding the high-frequency estimate down would not guarantee the bound.

Put $S=m_A+m_B$ and use $\sin qa\simeq qa$. Expanding the square root givesThus, to leading nonzero orders,The absolute value ensures a nonnegative angular frequency for either direction. The speed of sound is the initial acoustic slope $v_s=a\sqrt{2\alpha /(m_A+m_B)}$.

The vertical axis of the original figure is angular frequency, in $10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, not ordinary frequency. Its zone-centre optical intercept is approximately ${\omega }_{+}(0)=4.8\times 10^{13}{\mathrm{s}}^{-1}$. Convert the relative atomic masses using $m_u=1.66054\times 10^{-27}\mathrm{kg}$:The long-wavelength inference of diatomic-chain parameters then givesNo factor $2\pi$ is needed because the graph already plots $\omega$.

Substituting $x=A{e}^{pt}$ into the damped harmonic oscillator equation givesFor nonnegative viscous damping, the cases areThe PDF's request to express these conditions in terms of $\gamma /m$ is dimensionally inconsistent with its definition $\gamma =b/m$: $\gamma$ already has units ${\mathrm{s}}^{-1}$. With the unnormalised equation $m\ddot{x}+b\dot{x}+kx=0$, the equivalent comparisons are $b/m\gtrless 2{\omega }_0$, or $b^2\gtrless 4mk$. Also ${\omega }_0$ is the undamped natural angular frequency, not in general the driven displacement peak.

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.

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.

For positive angular frequency, the first three branches areEach branch is even in $k$, has minimum ${\omega }_n(0)=n\pi c_m/a$, and starts with zero slope. At large $|k|$, all approach the straight asymptote $\omega =c_m|k|$. Their group velocities are $v_{g,n}=c_m^{2}k/{\omega }_n$, whereas their longitudinal phase velocities are $v_{p,n}={\omega }_n/k$ for $k\ne 0$, giving $v_{p,n}{v}_{g,n}=c_m^2$.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.

The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

Choose positive electric current in the initial discharge direction, since the source diagram does not label the initial plate polarity or a current arrow. Let $q$ be the charge remaining on the initially positive plate, so $I=-\dot{q}$. Kirchhoff's voltage law for the series RLC circuit givesThe natural angular frequency, damping coefficient and quality factor areSince $\alpha <{\omega }_0$ (equivalently $Q>1/2$), this is an underdamped RLC response. With ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}=815.82\mathrm{rad}{\mathrm{s}}^{-1}$, satisfying the initial conditions givesand differentiating givesThe electric current starts at zero with $\dot{I}(0)=V_0/L=100\mathrm{A}{\mathrm{s}}^{-1}$. Its wave period is $T_d=2\pi /{\omega }_d=7.702\mathrm{ms}$, its zeros occur every $3.851\mathrm{ms}$, and its positive and negative wave envelopes decay with time constant $30.0\mathrm{ms}$. The first maximum obeys $\tan {\omega }_{d}t={\omega }_d/\alpha$ and is $0.1151\mathrm{A}$ at $1.875\mathrm{ms}$. The electric current tends to zero as the initial capacitor energy is dissipated in the resistor.