The delta function in the artificial constant potential collapses the convolution. Put $b=e\gamma$ and $K^2=k^2+m^2$. ThenInverse Fourier transform gives the stated real-frequency integral with exponential $e^{-i\omega (t-t^{\prime })}$. For $K^2>b^2$, the poles have opposite real parts and common imaginary part $b$. For $0<K^2<b^2$, both are imaginary, at $i[b\pm \sqrt{b^2-K^2}]$. They remain in the upper half-plane if $b>0$, or the lower half-plane if $b<0$. At equality they merge into a second-order pole at $ib$. For $K^2=0$ one pole reaches the real axis, requiring a prescription.

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.

For the supplied Lagrangian, variation with respect to ${\varphi }^{\ast }$ usesThus $\square \varphi +2ie{A}^{\mu }{\partial }_{\mu }\varphi +ie({\partial }_{\mu }{A}^{\mu })\varphi +m^2\varphi =f$. In the Lorenz gauge,For real external fields and driving, the conjugate equation has $-2ie{A}^{\mu }{\partial }_{\mu }$ acting on ${\varphi }^{\ast }$. The supplied model has no $e^2{A}^2|\varphi {|}^2$ term; one must not insert such a term from a different minimally coupled Lagrangian.

The local gauge symmetry contains the constant-phase subgroup. For constant $\alpha$, $\delta A_{\mu }=0$, and Noether's theorem applied to the matter fields givesThus ${\partial }_{\mu }{J}^{\mu }=0$ on the matter field equations. More explicitly, allowing $\alpha (x)$ in the matter variation while holding $A$ fixed gives $\delta \mathcal{L}=J^{\mu }{\partial }_{\mu }\alpha$. On shell it also equals ${\partial }_{\mu }(\alpha J^{\mu })$, so equality for arbitrary $\alpha$ implies conservation. In the full gauge transformation, varying $A$ cancels the $J^{\mu }{\partial }_{\mu }\alpha$ term; this is the related Noether second-theorem identity, not a new independent charge for every gauge function.

Under the combined local transformation,The derivative of the phase cancels the added gauge potential. The conjugate derivative gains the opposite phase, so their product and $|\varphi {|}^2$ are invariant. AlsoEvery term of the interacting Lagrangian is therefore invariant.

The derivatives needed for varying $A_{\nu }$ areTherefore the electromagnetic Euler-Lagrange equation isAntisymmetry of $F$ and commutation of derivatives imply ${\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, hence ${\partial }_{\nu }{J}^{\nu }=0$. In ordinary derivatives, $J^{\mu }=j^{\mu }-2q^2{A}^{\mu }|\varphi {|}^2$; the interaction contribution must not be omitted.

An infinitesimal constant phase change has $\delta \varphi =-iq\alpha \varphi$ and $\delta {\varphi }^{\ast }=iq\alpha {\varphi }^{\ast }$. Both the derivative product and the mass term are invariant. Noether's theorem gives, per unit $\alpha$,Thus the global phase symmetry supplies the conserved current in part a.

Use metric signature $(+,-,-,-)$ and units $\hslash =c=1$. Varying ${\varphi }^{\ast }$ and $\varphi$ independently givesIn the divergence of the stated current, the products of first derivatives cancel, leavingThe two identical mass terms cancel on the field equations.

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.

The general action identity isTreating $q,Q,t$ as independent and comparing coefficients givesChoose a complete integral of the Hamilton-Jacobi equation,Then $\mathcal{H}=0$, and Hamilton's equations become ${\dot{Q}}_i={\dot{P}}_i=0$. Completeness and local invertibility are needed: a solution with too few independent parameters cannot generate an arbitrary trajectory.

When the Hamiltonians agree, exactness of the one-form givesThe action argument of part b then proves canonicity. Equivalently, exterior differentiation preserves the symplectic two-form. For a time-dependent transformation, the exact differential must supply the full action identity; a fixed-time check alone would not establish that the new Hamiltonian equals the old one without a time-dependent correction.

The assumed relation makes the old and new actions differ by the endpoint term $G(t_f)-G(t_i)$. Fixed endpoint values of both coordinate sets make its variation zero. Therefore stationarity of the old action is equivalent to stationarity ofVarying independent $Q_i,P_i$ and integrating $P_i\delta {\dot{Q}}_i$ by parts givesBoth coefficients vanish: the new variables satisfy Hamilton's equations, so a locally invertible transformation with this action identity is canonical.

For a regular Lagrangian define $p_i=\partial L/\partial {\dot{q}}_i$ and $H={\sum }_i{p}_i{\dot{q}}_i-L$. Its differential isThe velocity differentials cancel. Using $\partial L/\partial q_i={\dot{p}}_i$ and comparing independent coefficients givesThe last equality concerns explicit partial derivatives in their respective natural variables; it does not assert that total time derivatives are always negatives.

Substitution of the proposed sine and cosine displacements into either linearized equation gives $m{\omega }^2+qB\omega -k=0$. ThusThese are signed angular frequencies, not two positive frequency magnitudes. To first order, with $\epsilon \ll {\rho }_0$,The orbit is therefore a circle of radius $\epsilon$ about $({\rho }_0,0)$ at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf{z}}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.

Direct variation of the Lagrangian givesFor a static configuration, $k\rho =qE\cos \theta$ and $qE\rho \sin \theta =0$. Since $qE>0$ and the radius is positive,The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_{\rho }=0$ but $p_{\theta }=qB{\rho }_0^2/2$, rather than zero canonical angular momentum.

The canonical momenta areTaking the Legendre transform givesIts four Hamilton equations, in a polar chart with $\rho >0$, areCanonical angular momentum includes the vector-potential contribution and is not simply mechanical angular momentum.

Choose electric scalar potential $\Phi =-Ex$ and the symmetric gauge $\mathbf{A}=B(-y,x,0)/2$, whose curl is $B\hat{\mathbf{z}}$. Minimal coupling gives $L=m{\dot{\mathbf{r}}}^2/2-k{\mathbf{r}}^2/2+q\mathbf{A}\cdot \dot{\mathbf{r}}-q\Phi$. Since $x=\rho \cos \theta$ and $x\dot{y}-y\dot{x}={\rho }^2\dot{\theta }$,The magnetic term is velocity dependent; it is not an ordinary position-only potential.

The new kinetic matrix is diagonal, givingFor the first mode, $\varphi =0$: the line between the centres stays vertical, both centres translate together, and $\dot{\psi }=2\dot{\theta }$. For the second, $\theta =0$: the outer hoop stays still while the inner hoop rolls within it, with $\dot{\psi }=-\dot{\varphi }$. These coordinate motions are independent only at the retained quadratic order.

Let $\psi$ be the small hoop's absolute spin angle with the same angular sign convention. Equating the two contact-point velocities cancels the large hoop centre's translation. The remaining tangential equality isThis rolling constraint inside a moving circular track adds small-hoop rotational energy $M{a}^2(2\dot{\theta }-\dot{\varphi }{)}^2/2$. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, soStatic friction enforces rolling; ideal no-slip contact dissipates no energy.

The Euler-Lagrange equations from the quadratic Lagrangian areFor a normal mode proportional to $e^{i\omega t}$, let $\lambda ={\omega }^{2}a/g$. Nonzero amplitudes requireHenceThe corresponding amplitude ratios are $\Phi /\Theta =2\lambda /(1-\lambda )$. The lower-frequency mode has $\Phi /\Theta =2(\sqrt{2}-1)$, with the angles in phase; the higher-frequency mode has $\Phi /\Theta =-2(1+\sqrt{2})$, with opposite signs. The small hoop's spin remains zero in both modes.