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A constant displacement proportional to $(1,-3)$ leaves the velocities unchanged and changes $3x+y$ by zero. It is therefore a continuous symmetry of the Lagrangian. The generalized momentum components areNoether's theorem gives the conserved quantityA direct check using the Euler-Lagrange equations isThe normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have length units and $(K_x,K_y)=(1,-3)$ dimensionless, giving $P$ the units of momentum. Alternatively, for the paper's convention with dimensionless parameter and length-valued $K_i$, use $(K_x,K_y)=(\ell ,-3\ell )$; the conserved quantity is then $\ell P$. The conserved content is identical.

The special relativity expression for momentum is $p=\gamma mv$, with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Equal speeds imply equal Lorentz factors, so the pion and proton momenta are in the ratio of their rest masses:The common dimensionless product is $\gamma v/c=p_p/(m_{p}c)=400/0.94\simeq 425.53$, giving $\gamma \simeq 425.53$ and $v/c\simeq 0.99999724$.

Write the incident and scattered photon energies as $\epsilon =hc/\lambda$ and ${\epsilon }^{\prime }=hc/{\lambda }^{\prime }$. Conservation of relativistic energy gives the recoiling electron energy $E_e=m{c}^2+\epsilon -{\epsilon }^{\prime }$. By momentum conservation, its momentum is the difference of the two photon momenta; henceInsert these into the relativistic energy-momentum relation for the electron:Expanding and cancelling terms givesSince $1/\epsilon =\lambda /(hc)$, the Compton scattering formula follows:The scale $h/(mc)$ is the electron Compton wavelength. At $\theta =0$, ${\lambda }^{\prime }=\lambda$: there is no energy or momentum transfer in this ideal forward limit, so the electron remains at rest. At $\theta =\pi$, the wavelength shift is maximal:The electron gains its largest kinetic energy, $\epsilon -{\epsilon }_{\min }^{\prime }$, and recoils forward while the photon reverses direction.

The diagram makes $x$ the remaining distance to the wall: $x$ decreases from $L$ to $0$. With the endpoint condition $m(0)=0$, integrating the supplied mass gradient givesAssume escaping sand has the same horizontal velocity as the bucket at release. There is then no horizontal recoil: the momentum flux of the outgoing sand cancels the $\dot{m}\dot{x}$ contribution to the bucket's momentum derivative. The variable-mass system equation is thereforeSet $u=-\dot{x}\ge 0$, the speed towards the wall. Since $\ddot{x}=udu/dx$, the initial condition $u(L)=0$ givesConsequently the remaining sand has kinetic energyIn terms of $z=x/L$, $K/(TL)=-z\ln z$. Its derivative is $-\ln z-1$ and its second derivative is $-1/z<0$, givingThe magnitude of the remaining sand's momentum isIt is easier to maximize its square, $p^2/(2MTL)=-z^2\ln z$. Its derivative is $-z(2\ln z+1)$, giving the unique interior maximumBoth kinetic energy and momentum tend to zero at $x\to 0^{+}$, even though the idealized speed diverges. There is no finite-mass bucket left at the endpoint, so that divergence signals the limit of the idealization.

The leaking bucket under constant tension has $x_K=L/e<L/\sqrt{e}=x_p$: the kinetic energy maximum is closer to the wall. As the bucket loses mass, its increasing speed continues to boost $K=pu/2$ after $p$ has begun to fall. Equivalently, $p\propto z\sqrt{-\ln z}$ and $K\propto z(-\ln z)$, so kinetic energy has the stronger dependence on the growing speed. The work $T(L-x)$ is shared between the remaining sand and the sand already lost; it is not all the bucket's instantaneous kinetic energy.

Take upward velocity as positive, and let $w>0$ be the rearward rocket exhaust speed. During $dt$, the ejected parcel has mass $-dm>0$ and laboratory velocity $v-w$ to first order. The momentum change of the rocket and this parcel equals the gravitational impulse:Discarding the second-order product $dmdv$ gives the vertical rocket equationWith $\dot{m}=-\alpha$ and $m=m_0-\alpha t$, integration for free flight starting from rest at $t=0$ givesThe logarithmic term is the ideal Tsiolkovsky rocket equation contribution; $gt$ is the gravity loss. Immediate upward departure requires positive initial acceleration, orAt exact equality the initial acceleration is zero, but further burning reduces the weight and permits departure immediately thereafter. If departure is allowed to be delayed while fuel burns on the pad, the condition is instead $\alpha w>m(t)g$ at some time before burnout. With final rocket dry mass $m_f$, any powered departure before burnout requires $w>g{m}_f/\alpha$; the free-flight formula then starts at the actual departure time $t_{\ell }$ with $m(t_{\ell })$, giving $v=w\ln [m(t_{\ell })/m(t)]-g(t-t_{\ell })$.

For the specified data, the initial mass includes the payload. The rocket fuel fraction givesThe $100\mathrm{kg}$ payload is already part of the $2500\mathrm{kg}$ final mass; it is not the entire rocket dry mass and must not be added a second time. Here $m_{0}g/\alpha =196m/s$, well below the stated rocket exhaust speed, so departure is immediate. The burnout speed isThis is the speed at burnout. Subsequent motion under the assumed uniform gravity has $v(t)=v(T)-g(t-T)$.

For a plane electromagnetic wave with linear polarization, write $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$ and $\mathbf{H}=E_0\hat{\mathbf{y}}\cos (kz-\omega t)/({\mu }_{0}c)$. The Poynting vector is instantaneous energy flux:The light intensity is its cycle average, not its peak:An incident electromagnetic wave transports longitudinal momentum $\mathcal{E}/c$ per energy $\mathcal{E}$. Perfect reflection back along the incident direction reverses that momentum, transferring $2\mathcal{E}/c$ to the mirror. Dividing by illuminated area and time gives the mean radiation pressureHere $\mathbf{S}$ denotes the incident travelling wave. The average net Poynting vector of the incident-plus-reflected standing wave is zero, although its radiation pressure is nonzero.

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.

For a sinusoidal plane electromagnetic wave in vacuum, write the maximum electric field as $E_0$. The magnetic field maximum is $B_0=E_0/c$, so the cycle-averaged Poynting vector magnitude, or light intensity, isEach photon carries energy $hf$, where $h$ is the Planck constant. Therefore the photon number flux density, namely the number per unit normal area per unit time, isMultiplying by an illuminated normal area gives the photon arrival rate. Each absorbed photon transfers momentum $hf/c$ at normal incidence, giving the radiation pressure on a perfectly absorbing surfaceThese are time-averaged quantities. At oblique angle of incidence $\theta$, the normal radiation pressure is $I{\cos }^2\theta /c$; the simple displayed expression assumes normal incidence.

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

In Compton scattering, a photon transfers energy and momentum to an initially stationary, approximately free electron. With incident and scattered wavelengths $\lambda ,{\lambda }^{\prime }$ and photon deflection angle $\theta$, momentum conservation givesCombine this with $hc/\lambda +m{c}^2=hc/{\lambda }^{\prime }+\sqrt{m^2{c}^4+p_e^2{c}^2}$ to obtainThe photon loses energy to electron recoil; the shift is independent of incident wavelength in this free-electron model and is largest at backscattering. At energies comparable with atomic binding, initial electron motion, binding and coherent scattering from the atom require additional treatment. Classical elastic scattering from a fixed charge alone does not produce this recoil wavelength shift.

In Arthur Holly Compton's spectral test, molybdenum-target X-rays scattered from graphite at $90^{\circ }$ were compared with the primary spectrum using the same slits, crystal and tube potential energy. A calcite crystal resolved the shifted line by Bragg's law; lines on both sides established angular zero. The crystal glancing angle on the plotted axis differs from the photon scattering angle. The compared curves had different intensity scales, and the paper did not directly observe the recoil electrons in this X-ray test. These controls support the wavelength comparison without converting it into a direct recoil-track measurement.^[ref]

For elastic scattering, $E_f=E_i$ and ${\mathbf{k}}_f-{\mathbf{k}}_i=\mathbf{G}$, where $\mathbf{G}$ is a reciprocal lattice vector. In the ideal static crystal, coherent scattering obeys this lattice selection rule; the crystal as a whole can take the mechanical recoil with negligible energy. crystal momentum is defined modulo $\hslash \mathbf{G}$, not as unconstrained free-particle momentum.

For observables with finite variances, the Heisenberg uncertainty principle gives $\Delta A\Delta B\ge |⟨[A,B]⟩|/2$; in particular $\Delta x\Delta p\ge \hslash /2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and Bose-Einstein statistics; half-integer-spin particles have antisymmetric full states and Fermi-Dirac statistics. The latter imply Pauli exclusion from one complete single-particle state, including spin.

Inside an infinite square well, solutions satisfying both zero boundary conditions arewith zero wavefunction outside. For the ground state, direct integration givesThe infinite-well ground-state uncertainty product follows:The ground state is not a minimum-uncertainty Gaussian. Here momentum variance is the usual kinetic-momentum second moment of the confined state; the hard-wall momentum operator has domain subtleties that do not alter this kinetic-energy result.

For spin degeneracy in filling a one-dimensional well, let $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$ and assume the spin levels are degenerate and available. Four spin-zero bosons all occupy the lowest orbital, giving $4E_1$. Four spin-$1/2$ fermions fill two spin states in each of $n=1,2$, giving $10E_1$. Spin-$3/2$ fermions have four spin projections, so all four can occupy $n=1$ in different complete states, giving $4E_1$. These fillings do not impose full spin polarization.

For a photon scattering through angle $\theta$ from a free electron initially at rest, conserve energy and momentum. Eliminating the final electron momentum by its relativistic energy relation givesor $E^{\prime }=E/[1+(E/m{c}^2)(1-\cos \theta )]$. The recoil electron gains kinetic energy $E-E^{\prime }$. The shift is zero forward, one electron Compton wavelength at $90^{\circ }$, and twice that wavelength backward.

Compton's X-ray scattering measurements compared scattered spectra at different angles and identified an angle-dependent shifted component. This supports photon momentum $p=E/c$ and inelastic recoil, beyond the unchanged-frequency prediction of classical elastic Thomson scattering. Bound electrons and coherent scattering can also produce an unshifted component; initial momentum and binding broaden or modify the ideal free-electron result. Treating an electron as initially free and stationary is therefore a kinematic approximation appropriate when the relevant photon transfer dominates binding scales.^[ref]

In a periodic crystal potential, atomic levels broaden into bands of allowed bloch states. Coupling between states of the same symmetry opens avoided crossings, including Bragg-coupled states in the nearly free-electron picture. Bonding and antibonding combinations provide a complementary tight-binding picture. A semiconductor has a filled valence band and an empty conduction band at zero temperature separated by a forbidden energy interval; thermal excitation, light and doping provide mobile electrons and holes. The gap depends on the crystal's periodic potential, bonding and symmetry, not on a classical collision barrier. In a direct gap, the relevant extrema coincide in wavevector.

For the stated parabolic edges, take $E_v(k)=-{\hslash }^2{k}^2/(2m_h^{\ast })$ and $E_c(k)=E_g+{\hslash }^2{k}^2/(2m_e^{\ast })$. Negligible photon momentum and no phonon make the transition nearly vertical in $E$ versus $k$. parabolic-band optical energy sharing divides the excess photon energy $0.32\mathrm{eV}$ in inverse proportion to the effective masses:Thus the initial and final electron energies, relative to the valence-band maximum, areTheir difference is $0.5000\mathrm{eV}$. The selected $|k|$ is about $9.33\times 10^8{\mathrm{m}}^{-1}$; either sign gives the same energy in this symmetric approximation.

A shallow donor supplies an electron in a screened impurity potential, so its localized level lies just below the conduction edge. A shallow acceptor binds a hole and has an electron-occupation level just above the valence edge; thermal ionization supplies mobile holes. The level positions are measured relative to the relevant band edge, and deep impurities need not obey this hydrogenic approximation.

For the supplied bound-carrier mass and unit effective ionic charge, dielectric screening and effective mass change the hydrogen binding energy toThis corresponds to $E_b/k_B\simeq 110\mathrm{K}$ and an effective Bohr radius about $6.35\mathrm{nm}$. The binding is shallow compared with the band gap and typical atomic binding. Trivalent denotes chemical valence, not an unscreened hydrogenic charge of three. An acceptor physically binds a hole; the question's stated $0.1m_e$ is used as its stipulated bound-carrier mass rather than silently replacing it by the separate band-edge $0.4m_e$.

A three-phonon normal process conserves crystal wavevector without a reciprocal-lattice shift, ${\mathbf{k}}_1+{\mathbf{k}}_2={\mathbf{k}}_3$. An Umklapp process instead hasThe crystal can absorb the reciprocal-lattice momentum; after reduction to the first Brillouin zone the phonon drift transporting heat is relaxed. A normal process redistributes phonons while preserving their total crystal momentum, so it is not by itself the same resistive mechanism. energy must also be conserved in either process.

At low temperature, large-wavevector phonons are rare, suppressing Umklapp scattering. Boundary and defect scattering then limit the mean free path; with a nearly constant boundary-limited path, ${\kappa }_{\mathrm{ph}}\simeq (C_{\mathrm{ph}}/V)v\ell /3$ grows as $T^3$. As temperature increases, heat capacity rises but Umklapp scattering shortens $\ell$, often producing a electrical conductivity maximum. At sufficiently high temperature a simple anharmonic model has $C_{\mathrm{ph}}$ nearly constant and ${\tau }_U^{-1}\propto T$, giving ${\kappa }_{\mathrm{ph}}\propto T^{-1}$. These are regime estimates, not a universal law for every material or size.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

A continuous symmetry is represented by a unitary family $U(ϵ)=e^{-iϵG/\hslash }$ with a self-adjoint conserved generator of a continuous quantum symmetry $G$. If $UH{U}^{†}=H$ for all $ϵ$, differentiating gives $[H,G]=0$. Thus, for no explicit time dependence,translation invariance conserves the corresponding momentum component; rotational invariance about one axis conserves that angular-momentum component. A central potential conserves all components and $L^2$, but the components need not be simultaneously measurable because $[L_x,L_y]=i\hslash L_z$. Time-translation invariance of a closed system gives a time-independent Hamiltonian and energy conservation.

A discrete symmetry need not have a continuously varying generator. For parity, $[H,P]=0$ lets stationary states be chosen even or odd, and preserves parity sectors. symmetry does not force every state to be an eigenstate of its conserved observable. Explicit time dependence adds $⟨\partial G/\partial t⟩$ to the conservation equation, and an external symmetry-breaking potential generally permits exchange with the environment. Degenerate eigenspaces require choosing a compatible common basis rather than assuming every eigenvector already diagonalizes every commuting operator.

Confinement to length $L$ gives a momentum scale $p\sim \hslash /L$ by the Heisenberg uncertainty principle. Using the alpha-particle mass $m_{\alpha }\simeq 6.64\times 10^{-27}\mathrm{kg}$, its speed scale is $v\sim \hslash /(m_{\alpha }L)$. Counting encounters with either wall,This is an order-of-magnitude collision-rate estimate, not a classical trajectory of a stationary quantum state. Taking $\Delta x\Delta p=\hslash /2$, a different geometric interpretation of the well size, or successive returns to the same wall changes the coefficient by factors of order unity; the robust scale is $10^{20}{\mathrm{s}}^{-1}$.

For a normalized top-hat wavefunction of width $L$ centred at zero, $\psi (x)=L^{-1/2}$ inside the interval and zero outside. Its position distribution has $⟨x⟩=0$ and $\Delta x=L/\sqrt{12}$. With Fourier convention $(2\pi \hslash {)}^{-1/2}$, the momentum amplitude isThe momentum density has long oscillatory $1/p^2$ tails. Consequently $\int p^2|\tilde{\psi }{|}^{2}dp$ diverges: the ideal discontinuous top hat has infinite r.m.s. momentum spread. Its symmetric momentum density has formal principal-value mean zero, but no finite absolute first moment either. The finite-variance form $\Delta x\Delta p\ge \hslash /2$ is therefore not saturated and cannot be evaluated using the width of the central diffraction lobe as $\Delta p$.

The first momentum zeros occur at $p=\pm 2\pi \hslash /L$, showing the inverse-width scale. Smoothing the edges sufficiently gives a finite momentum variance and restores the ordinary finite-variance uncertainty comparison; sharper edges require larger high-momentum components. It is the ideal jump, not a violation of the uncertainty principle, that causes the divergence.

Differentiating the expectation value and using ${\partial }_t|\psi ⟩=-i\hat{H}|\psi ⟩/\hslash$ and ${\partial }_t⟨\psi |=i⟨\psi |\hat{H}/\hslash$ givesThe second term means explicit operator time dependence, distinct from the state's evolution.

The commutator product rule follows by adding and subtracting $\hat{B}\hat{A}\hat{C}$:For the mathematical induction step, let $\hat{D}=[\hat{A},\hat{B}]$ and use $[\hat{A},\hat{D}]=0$. The case $n=1$ is immediate. If $[{\hat{A}}^m,\hat{B}]=m\hat{D}{\hat{A}}^{m-1}$, thenThus the claimed identity holds for every positive integer. Commutation of $\hat{B}$ with $\hat{D}$ is not needed for this particular step. This is the central-commutator differentiation rule. Termwise application to a convergent power series, on a common domain where the operations are justified, gives

For position and momentum, $[\hat{x},\hat{p}]=i\hslash \hat{I}$. Hence $[{\hat{p}}^2,\hat{x}]=-2i\hslash \hat{p}$, $[V(\hat{x}),\hat{x}]=0$ and $[V(\hat{x}),\hat{p}]=i\hslash V^{\prime }(\hat{x})$. Substitution gives Ehrenfest theorem:The mean momentum changes by the mean force. It does not follow for a general nonlinear Potential energy that $⟨V^{\prime }(\hat{x})⟩=V^{\prime }(⟨\hat{x}⟩)$: a closed classical trajectory for the mean position requires an additional condition, such as a quadratic Potential energy or a sufficiently localized wave packet.