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If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.

For a regular Lagrangian, define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and invert this relation for $\dot{q}(q,p,t)$. Its Legendre transform is the HamiltonianSince $Ldt=p\cdot dq-Hdt$, the action isFor independent variations with $\delta q(t_1)=\delta q(t_2)=0$,Stationarity for arbitrary $\delta p$ and $\delta q$ gives Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$.

A Bravais lattice is the integer span of three linearly independent primitive vectors. Its reciprocal lattice consists of vectors $G$ satisfying $G\cdot r\in 2\pi \mathbb{Z}$ for every lattice vector $r$.

Translations by lattice vectors commute with the Hamiltonian. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen withThus ${\psi }_k=e^{ik\cdot x}{u}_k$ with lattice-periodic $u_k$, which is Bloch theorem. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the Brillouin zone is a fundamental Wigner-Seitz cell of the reciprocal lattice.

The radial action isWith the variable $x$ from part b this becomesApplying the supplied integral twice, with $c=0$ and $c=2$, and using the root sum and products from parts b and c givesThis is the radial action of the spherical isochrone model. Rearranging,and squaring yields the Hamiltonian in action-angle variables:As a check, differentiating this Hamiltonian gives ${\Omega }_r=(-2E{)}^{3/2}/(GM)$ and the frequency ratio found in part c.

For the unperturbed Hamiltonian system, choose a conserved energy $H(x,y)$. Along the perturbed system,The energy balance method integrates this expression around each unperturbed periodic orbit. A zero of the resulting $O(\epsilon )$ energy change is a necessary leading-order condition for a nearby periodic orbit. A simple zero at which the drift changes from positive inside to negative outside predicts a stable orbit; the opposite change predicts an unstable one.

A continuous symmetry is a family of transformations depending continuously on parameters for which the action is unchanged, possibly up to a spacetime boundary term. For an infinitesimal field variation $\delta \varphi =ϵ\Delta \varphi$, a Noether current $j^{\mu }$ obeyswhen the Euler-Lagrange field equation holds. Its Noether chargeis conserved provided the flux ${\int }_{S_{\infty }}{j}^i{n}_{i}dS$ vanishes.

Noether theorem gives the implication from a differentiable global variational symmetry to an on-shell conserved current. A conserved current gives a conserved charge only under suitable boundary and convergence conditions. Conversely, a conserved charge generates a continuous symmetry through Poisson brackets classically or a commutator quantum mechanically when a regular Hamiltonian formulation exists. Currents can be changed by identically conserved improvement terms, and inverse Noether statements require regularity and the exclusion of such trivial currents, so the three notions are related but not literally in one-to-one correspondence.

For spacetime translations, the canonical stress-energy tensor isForraising the second index gives the symmetric tensorTranslation invariance gives four conserved currents $T^{\mu \nu }$, one for each fixed $\nu$, and the four conserved charges are the four-momentum$P^0$ is the energy and $\mathbf{P}$ is the spatial momentum.

The Lorentz transformation variation of a scalar is generated solely by its argument. The corresponding Lorentz current isUsing ${\partial }_{\mu }{T}^{\mu \nu }=0$ and symmetry of $T^{\rho \sigma }$,In particular, conservation of the boost chargeimpliesThe quantity denoted $V^i$ in the question is therefore the conserved momentum component $P^i$, assuming the same vanishing boundary flux.

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this isequivalently $\{Q,P{\}}_{q,p}=1$.

Suppose a type-two generating function for a canonical transformation $F(q,P)$ definesThenEquality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

Forthe fundamental theorem of calculus and differentiation under the integral sign giveThusand in particular $P=(p^2+q^2)/2$.

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply $H=P$. Hamilton's equations becomeThereforeand transformation back gives the familiar solution

Now consider the weak quartic oscillatorChoose the modified generating functionIt givesso the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the Taylor series givesDifferentiation with respect to $P$ therefore yieldswhere

Set $\theta =t-t_0$ andWriting $q=q_0+ϵq_1+O({ϵ}^2)$ and expanding the inverse sine relation at $q_0$ givesExpanding $p^2=2E-q^2-2ϵq^4$ similarly givesConsequentlySince $\theta =t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

Now $g=1$, so $c=⟨\psi |\varphi ⟩$ is real andThe perturbation is diagonal in this basis:Hence the instantaneous eigenvalues are $1+2V(t)$ and $-1$ with time-independent eigenvectors, so Hamiltonians at different times commute. In units $\hslash =1$, putExpanding the initial state and evolving its two components givesTaking its overlap with $|\varphi ⟩$ and simplifying gives the exact transition probabilityWith explicit $\hslash$, each phase in the final sine is divided by $\hslash$.

Using one even and one odd site per unit cell, Bloch theorem reduces the Hamiltonian toIts eigenvalues are

With $H=p_j{p}_j/(2m)$, the stated brackets giveandThus $m\ddot{\mathbf{q}}=e\dot{\mathbf{q}}\times \mathbf{B}$, the Lorentz-force equation with zero electric field. The magnetic interaction has been moved from the Hamiltonian into the noncanonical symplectic structure.

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.

For a one-dimensional closed energy orbit, defineAfter locally inverting this relation to write $E=H(I)$, use the Hamilton--Jacobi generating functionand define the angle $\varphi =\partial W/\partial I$ modulo $2\pi$. The transformation is canonical, the Hamiltonian depends only on $I$, and

The adiabatic-invariance principle says that if an external parameter changes on a time scale much longer than the orbital period, and no separatrix is crossed, the action $I$ remains approximately constant. Its small within-cycle changes do not accumulate at leading adiabatic order.

Using the quadratic form of the Hamiltonian,
$E_{av}=({\hslash }^2/2m)\int |\psi \prime {|}^{2}dx=2{\pi }^2{\hslash }^2/(m{a}^2)$.

The nonrelativistic Schrödinger Hamiltonian uses $p^2/(2m)$ and therefore has no massless limit; massless particles require a relativistic wave equation or quantum field theory.

Each energy component acquires its own phase:The $1/\sqrt{S}$ factor normalizes the printed rounded input; it can be omitted if the coefficients are regarded as an already normalized rounded representation.

The energy probabilities and $⟨E⟩$ remain constant under this time-independent hamiltonian because phase factors have unit modulus. position probabilities can nevertheless vary: interference between different energy components generally changes with their relative phases. A stationary energy distribution is not the same thing as a stationary spatial probability distribution.

The result $16ϵ$ projects the state onto the corresponding nondegenerate energy eigenstate. Immediately after an ideal measurement,for an unobservable overall phase $\chi$. Subsequent isolated evolution adds $e^{-16iϵt/\hslash }$ but does not mix in other energy eigenstates. A repeat ideal energy measurement gives $16ϵ$ with probability one, provided the hamiltonian is unchanged and no intervening disturbance occurs.

One consistent five-postulate formulation is the following; different textbooks group the same content differently.

The position-dependent probability density is $|\psi {|}^2$, and its flux is the probability current. These distinguish the wave amplitude from the measurable particle flow in the following scattering problem.

An ideal cross-Kerr interaction between two single modes is $H_{\mathrm{int}}=\hslash \chi n_1{n}_2$, with $n_j=a_j^{†}{a}_j$. The number operators commute with the interaction, so this coupling changes phases without transferring photons between the modes. To implement the specified positive conditional phase, usewhere $n_{2V}=|V⟩⟨V|$ within the one-photon polarization subspace. The horizontal branch has zero interacting occupation. Since $e^{i\Theta n}|\alpha ⟩=|e^{i\Theta }\alpha ⟩$, the specified conditional rotation follows. This is an ideal lossless single-mode model; an actual travelling-wave medium's bandwidth, loss and noise are not established by writing this Hamiltonian.

Use $X_1=(a+a^{†})/2$ and $X_2=(a-a^{†})/(2i)$. A coherent state has centre $(\mathrm{Re}\alpha ,\mathrm{Im}\alpha )$ and variance $1/4$ in either optical quadrature. Multiplication of $\alpha$ by $e^{i\Theta }$ rotates that centre anticlockwise through $\Theta$, keeping the coherent state noise disk unchanged. For real positive $\alpha$ the two centres are $(\alpha ,0)$ and $(\alpha \cos \Theta ,\alpha \sin \Theta )$.

Normalize the initial polarization of an electromagnetic wave qubit. The output of the medium isThe joint state is generally an entangled state. The overlap magnitude is $|⟨\alpha |\beta ⟩|=e^{-2|\alpha {|}^2{\sin }^2(\Theta /2)}$. The state factorizes when the amplitudes coincide, while nearly distinguishable amplitudes make the polarization nearly maximally entangled with the optical mode. Before any polarization conditioning, tracing out mode 2 gives the mixture $(|\alpha ⟩⟨\alpha |+|\beta ⟩⟨\beta |)/2$, with no interference between branches.

Include the normalization factors in the half-wave plate map: $|H⟩\mapsto (|H⟩+|V⟩)/\sqrt{2}$ and $|V⟩\mapsto (-|H⟩+|V⟩)/\sqrt{2}$. The transformed joint state isConditioning on a vertical result and setting $\Theta =\pi$ therefore preparesThis pure optical cat state is the conditional output; it is not the unconditioned reduced state immediately after the Kerr medium. The measurement erases the polarization label, allowing the two amplitudes to interfere. The calculated optical Wigner functions below display the resulting fringes; their negative regions are quasiprobability, not negative detection probabilities. These illustrative plots choose $\alpha =1.15$: the rotation panel uses $\Theta =2\pi /3$, while the cat panels use the conditional Kerr rotation $\Theta =\pi$ and three different relative phases $\phi$.

The coherent-state overlap follows by expanding in the Fock basis:Thus $s=⟨\alpha |-\alpha ⟩=e^{-2|\alpha {|}^2}$ andThe postselection probability for Kerr-cat preparation includes both the normalized input qubit and the wave plate. At $\alpha =0,\varphi =\pi$ the vertical branch has zero probability and no normalized conditional state. For the remaining calculation take $\alpha$ real, as required.

Using $a|\pm \alpha ⟩=\pm \alpha |\pm \alpha ⟩$, put $D=1+s\cos \varphi$. Directly evaluating diagonal and cross terms givesThe last quantity is the mean photon number in the conditional cat, which differs from the input coherent state mean ${\alpha }^2$. Since $a{a}^{†}=n+1$,However, its first moment iswhich is generally nonzero. The original PDF's assertion of a zero mean for arbitrary $\varphi$ is inconsistent with its stated pure state. Consequently its displayed expression is the second moment, not the general variance. The phase-dependent cat-state quadrature variance isThe printed result does become the variance for $\sin \varphi =0$, notably the even and odd cat states. It could also describe an ensemble mixing equal weights of $+\varphi$ and $-\varphi$: that different state's mean vanishes while its second moment is unchanged. Such phase averaging is not part of the specified pure-state preparation.

For nonzero finite $\alpha$, $0<s<1$ and $D>0$. The correct quadrature squeezing condition in this quadrature is thereforeFor a representative phase in $-\pi <\varphi \le \pi$, this is $|\varphi |<\arccos (-e^{-2{\alpha }^2})$. If instead one applies the printed second moment as a variance, the resulting condition is $\cos \varphi >0$, or $|\varphi |<\pi /2$ modulo $2\pi$. That is sufficient, but not necessary, for the actual pure state's squeezing. For example, at $\varphi =\pi /2$, $⟨X_2^2⟩=1/4$ while $\mathrm{Var}{X}_2=1/4-{\alpha }^2{e}^{-4{\alpha }^2}<1/4$. This explicitly tests the difference.

For a fixed phase other than $\pi$ modulo $2\pi$, expandAs the input mean number ${\alpha }^2\to 0$, the normalized state tends to vacuum, $⟨X_2⟩\sim -\alpha \tan (\varphi /2)$ andThis expansion is nonuniform near the odd-cat phase. At exactly $\varphi =\pi$ the vacuum terms cancel and the normalized odd coherent-state superposition tends to $|1⟩$, giving $\mathrm{Var}{X}_2\to 3/4$, while its postselection probability tends to zero. The even coherent-state superposition at $\varphi =0$ instead contains only even photon numbers and tends smoothly to vacuum.

At large input mean number, $s\to 0$, the two coherent state branches become nearly orthogonal, and $\mathrm{Var}{X}_2\to 1/4$ for every fixed phase. The mean tends to zero as well. This does not make the state an incoherent mixture: phase-space interference still distinguishes the ideal quantum superposition, although its contribution to this low-order moment is exponentially small. The loss of coherence in an optical cat state arises when ordinary loss reveals the branch to the environment and suppresses that interference, with factor $e^{-2(1-\eta ){\alpha }^2}$ for transmissivity $\eta$.

For the especially simple even cat,It starts at $1/4$, falls to about $0.1108$ at $|\alpha |\simeq 0.7995$, and returns towards $1/4$ from below. The odd-cat curve decreases from $3/4$ towards $1/4$ from above. Other fixed-phase curves have the small- and large-amplitude limits derived above; for negative $\cos \varphi$, squeezing ends when $e^{-2{\alpha }^2}=-\cos \varphi$. In balanced homodyne detection, centre each quadrature distribution on its measured mean before inferring its variance.