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Insert the harmonic variables into the forward map and reduce the squared trigonometric functions. This givesIndeed $p=m\dot{x}$ at this order. The second harmonic and mean displacement of a weakly cubic oscillator show asymmetry of the motion: for positive $\lambda$ the mean shifts towards negative $x$. There is no first-order frequency shift, although the waveform is no longer purely sinusoidal.

The small parameter is $|\lambda |A/(m{\omega }^2)\ll 1$. A purely cubic perturbation is unbounded in one direction, so these formulas describe small local oscillations around the stable origin, not globally confined motion at arbitrary energy. Amplitude $A$ labels the transformed orbit and must be inferred using the inverse map if physical initial position and momentum are supplied.

The coordinate $\theta$ is cyclic. Its conjugate momentum is therefore a second conserved quantity,It differs from mechanical angular momentum $m{r}^2\dot{\theta }$. The magnetic contribution is essential to the rotational Noether charge in this axisymmetric gauge.

For this neutral Coulomb pair in a uniform magnetic field, the reduced Lagrangian gives the canonical momentaInvert these relations and take the Legendre transform. The velocity-linear terms cancel against their momentum contributions, givingThere is no explicit time dependence, so $dH/dt=0$. There is no dependence on $\mathbf{R}$, so $\dot{\mathbf{P}}=0$. The distinction between kinetic momentum and magnetic pseudomomentum matters: the conserved vector is $\mathbf{P}$, not generally $M\dot{\mathbf{R}}$. The mechanical centre-of-mass momentum $M\dot{\mathbf{R}}=\mathbf{P}+e\mathbf{B}\times \mathbf{r}$ generally varies as the separation changes. A neutral pair can therefore have accelerated centre-of-mass motion in a magnetic field.

Differentiate $\mu \dot{\mathbf{r}}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$ and substitute the previous momentum equation:For a planar vector, $\mathbf{B}\times \dot{\mathbf{r}}=B(-\dot{y},\dot{x})$, which gives the two component equations with the printed signs.

For circular relative motion of a neutral Coulomb pair of separation radius $R$, $\dot{\mathbf{r}}=\omega \hat{\mathbf{z}}\times \mathbf{r}$ and $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. ThusSince $(e^{\ast }{)}^2+\mu e^2/M=e^2/4$, the two signed frequencies areOne is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal masses remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.

The actual particle paths follow by integrating the centre-of-mass velocity:Both particles move on circles about the same fixed guiding centre with a constant separation and common signed frequency, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical centre of mass also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

Separating $G=W(q)-\beta t$ gives $[W^{\prime }(q){]}^2/(2m)+m{\omega }^2{q}^2/2=\beta$. On the positive-momentum branch chooseWith $Q=\beta$ and $A=\sqrt{2\beta /(m{\omega }^2)}$, differentiation with respect to $Q$ givesSince $P=\gamma$ is constant,The displayed integral is a local positive-$p$ branch. Continuing the trajectory through turning points switches momentum branches and extends the sine solution over the complete orbit. $\beta$ is the oscillator energy, and $\gamma$ fixes its time phase.

First fix the source at the spatial origin. Fourier transform replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$ in the free term. Multiplication by a static background becomes a momentum convolution; the derivative in $A\cdot \partial G$ acts on the Green function, so it carries the internal momentum ${\mathbf{k}}^{\prime }$. ConsequentlyThe transformed point source supplies the right-hand side one. The signs follow from $A^{\mu }{\partial }_{\mu }=A^0{\partial }_t+\mathbf{A}\cdot \nabla$ in these conventions.

Source-position qualification. For a nonuniform $A(\mathbf{r})$, the general Green function is not translation invariant. In the printed basis $e^{i\mathbf{k}\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}$, its coefficient must also depend on ${\mathbf{r}}^{\prime }$, and the convolution contains an extra factor $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot {\mathbf{r}}^{\prime }}$. The displayed source equation is valid for ${\mathbf{r}}^{\prime }=0$, or with the background itself expressed in coordinates centred on that source. Equivalently, a general spatial Fourier Green function has two momentum arguments and a momentum-delta-function source. Staticity alone permits the time-difference transform, not a general position-difference-only transform.

The transformation $\psi \mapsto \psi +b$ leaves both derivatives unchanged, so the Lagrangian density and action are invariant. For infinitesimal unit field variation, the Noether current for a first-derivative scalar field is the shift current of an elastic rod,Its divergence vanishes by the equation of motion. The charge is the total longitudinal mechanical momentum to this order in the displacement field.

For $b>0$, put ${\psi }_b(z,t)=\psi (bz,bt)$. Each derivative gains a factor $b$, so ${\mathcal{L}}_b(z,t)=b^2\mathcal{L}(bz,bt)$. Changing integration variables to $(Z,T)=(bz,bt)$ contributes $b^{-2}$, proving exact invariance of the full-space action.

Set $b=1+\eta$. At fixed coordinates, $\delta \psi =\eta X$ with $X=z{\psi }^{\prime }+t\dot{\psi }$. Since the density is quadratic in derivatives,The variation is a divergence, so subtract its boundary term in the Noether current:Thus the dilatation charge of a one-dimensional elastic wave isAn overall sign change would give the same conservation law. Unlike $Q_1$, this charge has units of action, $\mathrm{J}\mathrm{s}$. Its explicit factor of $t$ is balanced by transport of the weighted momentum density; an explicitly time-dependent charge can still be conserved.

Position-dependent density. The density ${\mathcal{L}}_1$ remains independent of $\psi$ itself, so the shift Noether charge survives with its altered canonical momentum:A prescribed generic function $\rho (z)$ supplies a fixed spatial scale and breaks scale invariance. For the dilation-current candidate with the corresponding ${\mathcal{H}}_1$ and ${\Pi }_1$, the scale-breaking sources in derivative field theories giveThus $Q_2$ is generally not conserved. Constant density recovers the earlier case; specially scale-invariant profiles and suitably invariant domains would require separate treatment.

Quartic strain. The extra term $-\alpha ({\psi }^{\prime }{)}^4$ in ${\mathcal{L}}_2$ again preserves a uniform field shift. Its time momentum is unchanged, while its spatial current becomes $-AK{\psi }^{\prime }-4\alpha ({\psi }^{\prime }{)}^3$. ThereforeThe new Hamiltonian density is ${\mathcal{H}}_2={\mathcal{H}}_0+\alpha ({\psi }^{\prime }{)}^4$. Under waveform dilation, the quartic term scales with $b^4$ rather than $b^2$, andHence there is no corresponding conserved $Q_2$ for nonzero generic $\alpha$, even if the energy in its candidate expression is updated.

Field-dependent quadratic stiffness. In ${\mathcal{L}}_3$, the term $-\beta {\psi }^2({\psi }^{\prime }{)}^2$ breaks the uniform field shift:when spatial flux vanishes. Thus $Q_1$ is not generally conserved. However all terms still contain exactly two derivatives in total, with a field of zero dilation weight. The factor ${\psi }^2$ gains no extra power of $b$, so waveform dilation remains an action symmetry. Its charge isHere $\Pi =A\rho \dot{\psi }$ as before. Relative to the original charge, the energy term changes; the symmetry survives despite the nonlinear field dependence. These statements concern symmetry conservation, separately from any restrictions on parameters needed for positive elastic energy.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

A constant shift $\varphi \mapsto \varphi +ϵ$ leaves every derivative unchanged, and the density has no undifferentiated $\varphi$. Thus it is exactly invariant under this scalar-field shift symmetry. Localize the shift parameter temporarily to $ϵ(x,t)$:Integrating the last term by parts once identifies the Noether current for a constant shift with second derivatives:The Euler-Lagrange field equation is $\rho {\varphi }_{tt}+\alpha {\varphi }_{xx}+\beta {\varphi }_{xxxx}=0$, precisely the continuity equation ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. Therefore the conserved quantity, when the endpoint current vanishes or the field has suitable periodic/decaying conditions, isthe total transverse momentum. More explicitly $dQ/dt=-[\alpha {\varphi }_x+\beta {\varphi }_{xxx}{]}_{\mathrm{ends}}$. The fixed endpoints introduced later can exert transverse forces and break the global shift symmetry of the boundary problem, so vanishing endpoint current must not be inferred merely from fixed displacement.

Square the operator solutions, retaining the symmetrized cross term rather than treating $x^{\prime }(0)$ and $p^{\prime }(0)$ as commuting variables. The initial symmetrized expectation vanishes, soThe means remain zero. These are also the variances. Unless $\alpha =m\omega /\hslash$, the Gaussian state breathes: its position and momentum widths exchange periodically while the mean energy remains constant.

Split the time-evolution operator into $N$ factors with $ϵ=T/N$, and insert $N-1$ position completeness relations between them. Each factor is evaluated by inserting momentum states normalized by $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$:Multiplying the $N$ kernels and integrating the intermediate positions produces the stated time-sliced path integral, including its $N/2$ normalization power. For a free particle this Gaussian short-time kernel is exact, so the convolution identity already holds at finite $N$. Oscillatory integrals are defined with a vanishing convergence regulator, consistently fixing the square-root phase.

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

For the momentum-space propagator in a uniform force, the stated Fourier convention makes $x$ act on momentum wavefunctions as $i\hslash {\partial }_p$. HenceFor $T>0$, the characteristic has $p(s)=p^{\prime }-\alpha (s-t^{\prime })$. Along it the amplitude acquires quantum phase $-{\int }_{t^{\prime }}^{t}p(s{)}^{2}ds/(2m\hslash )$. momentum is translated without a Jacobian, givingThe cubic expression is evaluated on the delta-function support. There it equals $-i[p^{\prime 2}T-\alpha p^{\prime }{T}^2+{\alpha }^2{T}^3/3]/(2m\hslash )$, making both its sign and its continuous $\alpha \to 0$ limit clear. The jump at $T=0$ is the identity momentum kernel.

Write $T_p=p^2/(2m)$ and $V=m{\omega }^2{x}^2/2$. Dropping commutators of order ${\beta }^2$ permits $e^{-\beta H}\simeq e^{-\beta V}{e}^{-\beta T_p}$. With $⟨x|p⟩=e^{ipx/\hslash }/\sqrt{2\pi \hslash }$, the momentum Gaussian givesIntegrating the diagonal yields $Z\simeq 1/(\beta \hslash \omega )$, so the leading high-temperature density kernel isThis unsymmetrized product has an apparent off-diagonal asymmetry at the omitted order; reversing the product puts $x^{\prime }$ in the potential factor. A manifestly Hermitian approximation uses $e^{-\beta V/2}{e}^{-\beta T_p}{e}^{-\beta V/2}$, replacing $\beta m{\omega }^2{x}^2/2$ by $\beta m{\omega }^2(x^2+x^{\prime 2})/4$ with the same leading normalization. The condition is $\beta \hslash \omega \ll 1$; the Gaussian off-diagonal width is of order $\hslash \sqrt{\beta /m}$.

First trace the state explicitly specified in (c). orthogonality in the $y$ oscillator removes terms with different occupation numbers:For a thermal oscillator of frequency $\omega$, the ratio of successive Boltzmann weights is $e^{-\hslash \omega /(k_{B}T)}$. Consequently the conditional answer for that two-mode squeezed state isAt $\theta =0$ this is interpreted by the limit $T_{\mathrm{sq}}=0$. The reduced state is mixed for nonzero squeezing although the joint state is pure; its effective thermal statistics arise from entanglement, not from heating by a bath.

The ground-state identification must still be qualified. For the true coordinate-coupled ground state, direct normal-mode averages giveTheir ratio is $m^2{\omega }_{+}{\omega }_{-}$, rather than $m^2{\omega }^2$. Therefore this reduced state is not thermal with respect to the original bare frequency $\omega$ for $k\ne 0$. It does have an exact thermal representation for an effective oscillator of mass $m$ and frequencyThe corresponding occupation basis is that of $\nu$, not that of the original $a$. This follows by matching the Gaussian variances to $⟨x^2⟩=\hslash \coth [\hslash \nu /(2k_{B}T)]/(2m\nu )$ and its momentum counterpart. A direct reduced-Gaussian eigenfunction calculation gives the same effective frequency and geometric eigenvalues.^[ref]

Normalize the ring modes as ${\phi }_j(x)=L^{-1/2}{e}^{i{k}_{j}x}$, with distinct $k_j=2\pi {\ell }_j/L$ and integers ${\ell }_j$. Inserting $\hat{\psi }(x)={\phi }_1(x)a_1+{\phi }_2(x)a_2$ into the contact interaction givesOrthogonality makes $I_{ijkl}=1/L$ when $k_i+k_j=k_k+k_l$, and zero otherwise: this is momentum conservation in the projected Hamiltonian operator. There is one all-mode-1 term, one all-mode-2 term, and four mixed terms with one particle in each mode. A pair conversion between distinct ring modes would require $2k_1=2k_2$ and therefore vanishes.

Using $a_j^{†2}{a}_j^2=n_j(n_j-1)$ and $a_1^{†}{a}_2^{†}{a}_1{a}_2=n_1{n}_2$ yields the contact interaction between two ring modes,The factor four comes from two independent choices of ordering for the creation operators and annihilation operators. This is the two-mode projected model; a full ring contact interaction can also scatter particles into other modes while conserving total momentum.

Put $\alpha =\Delta \tau /\hslash$ and $\beta =E{x}_0\tau /\hslash$. The Pauli matrix multiplication law gives the pulse-and-precession spin one-half mapBecause $C^2+|\mathbf{v}{|}^2=1$, a suitable angle and unit vector for the rotation areIf the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The Bloch vector $\mathbf{r}=⟨\mathbf{\sigma }⟩$ rotates through $\theta$ about $\mathbf{n}$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give quasienergies $\pm \hslash \theta /(2\tau )$ modulo $2\pi \hslash /\tau$.

In the two-orbital subspace the position operator is $X=x_0{\sigma }_x$. The Heisenberg equation of motion between pulses gives $P/m=(i/\hslash )[H_0,X]=(2\Delta x_0/\hslash ){\sigma }_y$, henceThus two components of the Bloch vector directly describe the mean position and momentum; its third gives the orbital population imbalance. The electric kick rotates about the position axis, leaving $⟨X⟩$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s⟩$, one complete period giveswhich exhibits the induced motion explicitly. These are projected two-orbital observables.