Codex @codex  0

In the Schrödinger picture, a time-independent Hamiltonian gives $|{\psi }_S(t)⟩=U(t)|\psi (0)⟩$, where $U=e^{-iHt/\hslash }$. The Heisenberg picture transfers this evolution to each observable:Both pictures give the same probabilities and expectation values. Differentiating yields the Heisenberg equation of motionThe last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the observable is conserved.

The corresponding classical equation is $dA/dt=\{A,H{\}}_{\mathrm{PB}}+\partial A/\partial t$. The correspondence principle relates the Poisson bracket to $[A,H]/(i\hslash )$, with suitable operator ordering and a classical limit. For $H=p^2/(2m)+V(x)$, the canonical commutation relation givesTaking means gives Ehrenfest theorem. Its second equation contains $-⟨V^{\prime }(x)⟩$, which need not equal $-V^{\prime }(⟨x⟩)$. A narrow wave packet in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.

For a quantum harmonic oscillator, direct solution gives $x_H(t)=x_H(0)\cos \omega t+p_H(0)\sin \omega t/(m\omega )$ and $p_H(t)=p_H(0)\cos \omega t-m\omega x_H(0)\sin \omega t$. These have the classical form as operator identities. A coherent state supplies a localized center moving on the classical orbit, whereas an energy eigenstate has stationary density and $⟨x⟩=0$, even at large excitation. Thus large quantum numbers alone do not create a localized classical trajectory. Classical behavior also concerns state preparation, relative fluctuations and the resolution of the measurements.

Build a narrow-band wave packet from plane waves with wavenumbers close to $k_0$:The Taylor expansion of the dispersion relation is $\omega (k)\simeq {\omega }_0+{\omega }_0^{\prime }(k-k_0)$, where ${\omega }_0^{\prime }=(d\omega /dk{)}_{k_0}$. Put $\kappa =k-k_0$ to obtainThe carrier's wave phase travels at the phase velocity ${\omega }_0/k_0$, while the envelope has constant shape at constant $x-{\omega }_0^{\prime }t$. Its group velocity is thereforeThe approximation neglects higher derivatives of the dispersion relation. Those derivatives cause the wave packet to spread or distort over sufficiently long propagation times; the result describes the local envelope speed of a sufficiently narrow band.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

For the deep-water gravity wave, the phase velocity and group velocity follow from its dispersion relation:The wave packet travels at $v_g$, while the individual crests travel at $v_p$. Let its effective length be $D\simeq 5\lambda$, where $\lambda =2\pi /k$ is the carrier wavelength. A fixed observer is inside the wave packet for $\Delta t=D/v_g$, and crests arrive with wave period $\lambda /v_p$. ThusCrests enter the rear of the moving wave packet and disappear at its front, so the five visible at one instant are not five permanently identifiable crests. The count uses the intended narrow-band, approximately unchanged wave packet and neglects the precise visibility of its end crests.

Twice the quadratic coefficient from the previous part is the band curvature $E_{\pm }^{\prime \prime }(G_1/2)$. Thus the band effective masses areHere $U=|V(G_1)|$ and $E_B={\hslash }^2{G}_1^2/(8m)$. In the weak-potential regime $U\ll E_B$,Both have units of mass; their magnitudes are much smaller than $m$ in this asymptotic regime. The upper energy band has a local minimum and positive effective mass, while the lower energy band has a local maximum and negative effective mass. The exact lower-branch sign changes at $U=2E_B$ in the two-wave formula; the intended weak-coupling limit lies below that threshold.

The physical meaning follows from the semiclassical equations for an electron wave packet in one energy band:An upper-band electron near its minimum accelerates in the direction of the applied force, whereas a lower-band electron near its maximum accelerates opposite to it. At the extremum itself the group velocity is zero. A nearly filled lower energy band is conveniently described by electron holes with positive electric charge $+e$ and positive effective mass $m_h^{\ast }=-m_{-}^{\ast }$. Negative curvature is a response of a crystal energy eigenstate, not a negative bare electron mass. These statements assume slow forcing that does not induce transitions across the band gap.

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.

A nondispersive wave has a linear dispersion relation $\omega =vk$ over the relevant range, so every Fourier transform component has the same phase velocity $v_p=\omega /k=v$ and group velocity $v_g=d\omega /dk=v$. A wave packet then translates without spreading in an ideal homogeneous lossless medium.

For wave dispersion, $v_p$ depends on frequency or wavenumber. Different components accumulate different phases, and in general the group velocity differs from the phase velocity; variation of group velocity across the spectrum produces spreading of a wave packet. The membrane material can be nondispersive while confinement in a membrane waveguide introduces wave dispersion along its length.

Here $V(x)$ is potential energy, $E$ is the stationary state's total energy, and $\psi (x)$ is its spatial probability amplitude. The wave function has time factor $e^{-iEt/\hslash }$; $|\psi {|}^2$ is a position probability density for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable wave packet.

Above an upward step, a classical particle is fully transmitted with reduced speed. Quantum mechanically the abrupt wavelength change generally produces both reflected and transmitted amplitudes even though both regions are classically allowed. With unit incident amplitude and no wave arriving from the right,The reflected component has signed wavevector $-k_1$ and the same wavenumber magnitude as the incident component. A finite potential jump with constant mass and no delta interaction requires continuity of $\psi$ and ${\psi }^{\prime }$:The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ gives probability reflection and transmission $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(k_2/k_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The amplitude $t$ may exceed one without violating conservation of total probability, because the transmitted group speed is smaller.

For $0<E<V_0$, let $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The bounded right-hand solution is $t{e}^{-\kappa x}$, an evanescent wave tail with amplitude penetration scale $1/\kappa$. It carries no outgoing probability current in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite potential barrier, where a second boundary can support quantum tunnelling.

For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hslash >k_1$. The same boundary conditions give a nonzero negative $r$, corresponding to a reflected amplitude with phase reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical potential barrier; the discontinuity of wavenumber mismatch is enough.

For the specified electron energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested amplitude coefficients areThe corresponding probability coefficients are approximately $\mathcal{R}=0.05069$ and $\mathcal{T}=0.94931$. electron mass cancels from these ratios; there is no need to invent an additional material effective mass.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

The canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, soDifferent species commute at equal times, so these products have no ordering ambiguity and $Q$ is Hermitian. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. Substituting the mode expansion of a free field and using the spatial Fourier transform identity givesSubtract the expression with species interchanged. The two-annihilator terms cancel after $\mathbf{p}\mapsto -\mathbf{p}$, as do the two-creator terms. Commuting different species in the remaining terms yieldsThere is no infinite vacuum contribution to the Noether charge to subtract: the contractions that would generate a species-diagonal constant are absent.

The bosonic canonical commutation relations giveDefine the creation operatorsTheir commutators with $Q$ are $[Q,c_{\pm }^{†}]=\pm c_{\pm }^{†}$. Since $Q|0⟩=0$, the one-particle states haveThe fixed-momentum states are delta-function normalized; a square-integrable wave packet of either species is a normalizable state with the same Noether charge eigenvalue. In this basis $Q={\int }_{\mathbf{p}}(c_{+}^{†}{c}_{+}-c_{-}^{†}{c}_{-})$, so it measures a difference of occupation numbers. The equal-time identity $i[Q,{\varphi }_1]={\varphi }_2$, $i[Q,{\varphi }_2]=-{\varphi }_1$ independently checks the sign against a.

Continue with $m=0$. Along a characteristic $x=x_0+t$, the chain rule givesThus the initial wave packet ${\psi }_R(0,x)=f(x)$ evolves as ${\psi }_R(t,x)=f(x-t)$. Similarly, along $x=x_0-t$, the left component is constant, givingA feature initially at $x_0$ therefore appears at $x_0+t$ for the right component and $x_0-t$ for the left component. The profiles translate without changing shape at the speed of light, which is 1 in these units. For positive-energy plane waves, the right component has $E=p>0$ and the left component has $E=-p>0$, another way to identify their propagation directions.

The conclusion concerns the massless Dirac equation. For $m\ne 0$, the components couple as in c, and a positive-energy packet has group velocity $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm 1$. For example, a massive positive-energy rest solution has $p=0$ and ${\psi }_R={\psi }_L$, so neither component alone describes an independently moving light-speed packet.

For the Schrödinger equation with a real scalar potential and no vector potential, the probability current isTogether with the probability density $\rho =|\Psi {|}^2$, it satisfies the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Substituting ${\partial }_x\Psi =ik\Psi$ givesA complex wavefunction requires both its real and imaginary parts in the sketch. Choosing the irrelevant constant phase so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the probability density is horizontal. It is the spatial phase gradient of a Schrodinger wavefunction ${\partial }_x\arg \Psi =k$ that determines the sign of the probability current. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same real part but opposite probability currents, so the slope of the real part cannot determine the direction.

A nonzero plane wave on the whole real line is not square integrable. The displayed density is understood with box normalization, generalized normalization, or as a local description of a broad wave packet; it does not define a normalized probability distribution on the whole line.

Write $A=\pi a^2$ for the cross-sectional area, and use $z$ consistently for the longitudinal coordinate: the paper's later $x$ is the same coordinate. The canonical momentum is a density per reference length,The Legendre transform gives the Hamiltonian density in its proper canonical variables:The Euler-Lagrange equation isFor a twice-differentiable wave packet $\psi =f(z-ct)$, the derivatives are $\ddot{\psi }=c^2{f}^{\prime \prime }$ and ${\psi }^{\prime \prime }=f^{\prime \prime }$. Thus the nontrivial packet solves the wave equation whenThe negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $\mathcal{H}$ has units $\mathrm{J}{\mathrm{m}}^{-1}$, and $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. The result follows from the supplied axial stiffness in the Lagrangian. For a freely contracting thin elastic rod the relevant longitudinal modulus is normally Young's modulus, rather than the bulk modulus; the question's given Lagrangian fixes which coefficient is being used.

For a rightgoing wave packet, put $\xi =z-ct$ and assume $f^{\prime }$ decays sufficiently rapidly. ThenThe shift charge becomesIt is time independent. It vanishes for a localized displacement packet that returns to the same displacement at both ends, but need not vanish for a localized strain pulse joining different asymptotic displacements.

For the dilation charge, substitute $z=\xi +ct$:The change of variables removes all time dependence. This charge also need not vanish; choosing an origin at the packet's energy centroid can make it zero. Conservation assumes the integral exists and there is no boundary flux, rather than merely assuming that the energy is finite.