Codex @codex  0

The parallel axis theorem says that for an axis through a body's centre of mass and a parallel axis at perpendicular distance $d$,where $M$ is the total mass. The displacement is the distance between the two parallel axes; neither axis may be arbitrary in place of the center-of-mass axis. Expanding the squared distances of mass elements gives this formula, since the cross term integrates to zero at the centre of mass.

Use metric signature $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the four-momentum isFor a photon of positive frequency $\nu$ and direction $e$, the Planck constant gives $E=h\nu$ andA future-pointing timelike four-vector $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean Cauchy-Schwarz inequality yieldsThis is stronger than the requested nonnegative inequality. For an electron and a positron of rest mass $m>0$, their future-pointing timelike four-momenta have $P_1^2=P_2^2=m^2{c}^2$. Thuswhich cannot equal the null norm of a single photon. Four-momentum conservation forbids spontaneous photon decay into an electron-positron pair in vacuum. An external body or field able to exchange momentum would be a different system.

By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe centre moves on a straight line with constant velocity; zero velocity is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

Yes: the two equal masses can follow the same fixed circle in diametrically opposite positions. Put the stationary centre of mass at the circle's center and choose tangential velocities with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational force is directed toward that center, with magnitude $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

Use $\hslash =c=1$ in the field-theory calculations. Work with signature $(+,-)$ and $a\ne 0$. Since $x=(x_{+}+x_{-})$ and $t=(x_{+}-x_{-})$, ${\partial }_{+}={\partial }_x+{\partial }_t$, ${\partial }_{-}={\partial }_x-{\partial }_t$, and the Sine-Gordon equation is ${\varphi }_{+-}=\sin \varphi$. Introduce $w=(\rho +\kappa )/2$ and $d=(\rho -\kappa )/2$. The specified Sine-Gordon Bäcklund transformation becomes $d_{+}=a\sin w$, $w_{-}=a^{-1}\sin d$. Its compatibility givesAdding and subtracting these equations establishesThus a compatible transformed field satisfies the same Sine-Gordon equation. Smoothness is needed to interchange mixed derivatives. The statement cannot literally include $a=0$, since one defining equation contains $1/a$.

Generating the kink. With zero seed, the two first-order equations give ${\rho }_{+}=2a\sin (\rho /2)$ and ${\rho }_{-}=2a^{-1}\sin (\rho /2)$. On a nonconstant branch put $p=\tan (\rho /4)$; then $p_{+}=ap$, $p_{-}=a^{-1}p$. Integration givesFor $a>0,C>0$, this is the Sine-Gordon kink, after absorbing $C$ in the center position. Its speed and Lorentz factor obeyChanging the sign of $a$ or of the exponential coefficient selects the corresponding antikink orientation; $|v|<1$ whenever $a$ is finite and nonzero. Constant vacuum branches, omitted by division by $\sin (\rho /2)$, can be added separately.

The continuously parametrized Bäcklund transformation does more than produce this one solution. Successive compatible transformations and Bianchi permutability for sine-Gordon Bäcklund transformations construct multisoliton solutions; expansion of the generating relations gives the local conserved-charge hierarchy of sine-Gordon theory. Together with its Lax pair and inverse scattering transform, this is the structure behind classical integrability, elastic scattering and the absence of generic radiative energy loss in soliton collisions. Existence of a transformation alone is not a proof that every generated conserved charge is independent and in involution; those are additional properties of the integrable system.

Transforming the static kink at $a=1$. The seed obeys ${\kappa }_x=2\mathrm{sech}x$, $\sin (\kappa /2)=\mathrm{sech}x$, $\cos (\kappa /2)=-\tanh x$. Adding and subtracting the two Sine-Gordon Bäcklund transformation equations yieldsAgain set $p=\tan (\rho /4)$. These simplify to $p_x=-p\tanh x$, $p_t=\mathrm{sech}x$. The first gives $p=f(t)/\cosh x$, and the second gives $f^{\prime }=1$. ThereforeContinuous inverse-tangent branches may be chosen without creating artificial jumps. At spatial infinity the field approaches the same vacuum, so its net topological charge is zero. For large $|t-t_0|$ there are two well-separated transitions near $x_{\pm }=\pm \mathrm{arcosh}|t-t_0|$, with speeds tending to zero. The two transitions exchange their kink/antikink orientations as they pass through the collision, while the field remains smooth.

This is the Sine-Gordon threshold kink-antikink solution, a separatrix between finite-speed scattering and the bounded Sine-Gordon breather motion, rather than a finite-period breather. In fact it is the $v\to 0$ limit of the scattering field in question 3. At $t=t_0$, $\rho =0$ and ${\rho }_t=4\mathrm{sech}x$, so the reduced energy is $\int {\rho }_t^{2}dx/2=16$, exactly twice the reduced single-kink energy $8$.

For mass comparisons below, the original PDF really prints $m^2/\beta$. Restoring physical coordinates $X=x/m$, $T=t/m$ gives the angular-field action ${\beta }^{-1}\int dtdx[\frac{1}{2}(\partial \varphi {)}^2+\cos \varphi -1]$ and literal kink mass $8m/\beta$, for $\beta >0$. The usual Sine-Gordon theory instead has prefactor $m^2/{\beta }^2$ and mass $8m/{\beta }^2$. The equation of motion and every field above are unchanged by this overall normalization; the quantum scattering comparison in question 3 is not.

Take $\lambda \ne 0$ and use the complex sine-Gordon theory reduced density $\ell =\mathcal{L}/M^2$, with physical lengths and times $x/M,t/M$. Physical rest energy is $M\int dx{\mathcal{H}}_{\ell }$, and the dimensionless Noether charge is obtained from the reduced action. Put $c=\cos \alpha$, $s=\sin \alpha$, $f(x)=c\mathrm{sech}(cx)/\lambda$, so $\psi =f{e}^{ist}$. Initially take $0<|\alpha |<\pi /2$, avoiding the singular target-space boundary.

The canonical Hamiltonian density isHere $|\dot{\psi }{|}^2=s^2{f}^2$, $|{\psi }_x{|}^2=c^2{f}^2{\tanh }^2(cx)$, and $D=1-c^2{\mathrm{sech}}^2(cx)$. Since $s^2+c^2{\tanh }^2(cx)=D$, the kinetic contribution is simply $f^2$. ConsequentlyTo fix the charge sign, take the Noether current directly from $\delta \psi =i\epsilon \psi$, $\delta {\psi }^{\ast }=-i\epsilon {\psi }^{\ast }$:Thus positive internal rotation has positive charge in this convention. With $X=cx$ and the supplied integral,Reversing the definition of the generator would reverse all charge labels but not any mass. The limit $\alpha \to \pi /2$ is the vacuum with zero mass and charge. At $\alpha =0$ the field reaches $|\psi |=1/|\lambda |$ at its center and the current's coordinate expression becomes singular. The mass still has the finite limiting value $4M/{\lambda }^2$, butThe displayed local density does not assign a unique endpoint charge merely by substituting $\alpha =0$.

The four collective parameters. Translation, a Lorentz boost of rapidity $b$, and the global phase symmetry produceThe independent real parameters are $\alpha ,b,X_0,\chi$, with $\chi$ periodic modulo $2\pi$. A further time translation is absorbed into these parameters. The energy and momentum are $\mathcal{M}\cosh b$ and $\mathcal{M}\sinh b$, while $Q$ is invariant. The internal angle is the collective coordinate whose canonical momentum is this Noether charge.

Semiclassical quantization. For one rest-frame internal rotation, the period in dimensionless time is $T=2\pi /|s|$. The action around the orbit isEquivalently this is $S+\mathcal{E}T$, where $\mathcal{E}$ is the reduced rest energy. Bohr-Sommerfeld quantization with $\hslash =1$ gives $I=2\pi n$. Because the internal orbit is a cyclic angle, not a one-dimensional oscillator between turning points, there is no half-integer Maslov index shift. HenceOn the nonsingular branch $0<n<2\pi /{\lambda }^2$. The elementary charge-one mass approaches $M$ at weak coupling. For positive $a,b$ with $(a+b){\lambda }^2/4\le \pi /2$, $\sin a^{\prime }+\sin b^{\prime }>\sin (a^{\prime }+b^{\prime })$, where $a^{\prime }=a{\lambda }^2/4$, $b^{\prime }=b{\lambda }^2/4$. Thus ${\mathcal{M}}_{a+b}<{\mathcal{M}}_a+{\mathcal{M}}_b$: the charge-$n$ soliton cannot fragment into smaller like-charge particles. For a general charge-conserving partition, a constituent with $|q_i|\ge n$ already has mass at least ${\mathcal{M}}_n$; if every $|q_i|<n$, the decreasing ratio ${\mathcal{M}}_q/q$ gives ${\sum }_i{\mathcal{M}}_{|q_i|}\ge ({\mathcal{M}}_n/n){\sum }_i|q_i|\ge {\mathcal{M}}_n$, strictly for nontrivial fragmentation. Thus opposite-charge constituents cannot open a lower-energy decay either. This is charged rather than topological stability, and the semiclassical spectrum is reliable at weak coupling.

Counting states and the endpoint. Write $L=2\pi /{\lambda }^2$. If $L$ is not an integer, the regular branch givesIf $L$ is an integer, the local nonsingular calculation gives only $2(L-1)$ regular states. The formal closed-branch list has $2L$ charge labels, but its last two labels are limits of the same singular static solution. Counting them needs a global completion; it cannot be decided from a regular current integral at $\alpha =0$.

In the additional integer-level completion, ${\lambda }^2=4\pi /k$ at leading semiclassical order and charges are identified modulo integer $k>1$. There are then $k-1$ nonzero charge sectors. For odd $k$, positive and negative labels are distinct; for even $k$, the endpoint labels $\pm k/2$ describe one self-conjugate state. This supplies $2L-1$ states when $k=2L$, rather than $2L$ endpoint labels. This global interpretation, and the distinction between classical and renormalized couplings, are discussed in Dorey and Hollowood, section 2. A generic classical $\lambda$ does not automatically define an integer $k$ or specify this quantum completion.

Take $0<v<1$ and $\Gamma =(1-v^2{)}^{-1/2}$. The Sine-Gordon kink-antikink scattering solution tends to the same vacuum at both spatial ends and hence has zero net topological charge. At large $|t|$, the two transition centers, found by setting the magnitude of the inverse-tangent argument to one, obeyThey are a kink and an antikink moving with asymptotic speeds $\pm v$, with no outgoing radiation. Their left-right orientations interchange at the collision. To define the soliton time delay, label a transmitted trajectory by its preserved rapidity rather than its left-right position. The right-moving asymptotes areAfter restoring physical lengths, the forward shift is $\Delta X=-2\log v/(m\Gamma )>0$. The arrival-time difference is thereforeThis is a time advance relative to the extrapolated incoming free motion.

The semiclassical scattering phase. For an outgoing energy wave packet with full S-matrix phase $S=e^{i\delta (E)}$, stationary phase in $e^{-iET+i\delta (E)}$ shifts its arrival time by $d\delta /dE$. The full S-matrix phase from a classical soliton delay uses the paper's $\delta$, twice the phase in a convention $S=e^{2i{\delta }_{\mathrm{pw}}}$. In the center-of-momentum frame, $E=2{\mathcal{M}}_K\cosh (\theta /2)$, so $dE/d\theta ={\mathcal{M}}_K\sinh (\theta /2)$. It follows thatThe literal PDF normalization gives ${\mathcal{M}}_K=8m/\beta$ and consequentlyThe integrable logarithmic singularity at zero causes no divergence of this phase difference. With the intended standard angular-field prefactor $m^2/{\beta }^2$, the result is insteadThe soliton time delay determines only phase differences, not an energy-independent constant or a choice of $2\pi$ branch. Keeping this distinction avoids an arbitrary high-energy subtraction.

Bound-state poles. Set $a_j=\pi j/N$. Each factor of the exact amplitude can be written $\cosh [(\theta -i{a}_j)/2]/\cosh [(\theta +i{a}_j)/2]$. Its denominator vanishes in the physical rapidity strip atThe corresponding numerator is nonzero, and no other numerator cancels the pole. In the direct kink–antikink fusion interpretation, analytically continue the constituent rapidities to $\pm i{u}_n/2$. Their momenta sum to $(2{\mathcal{M}}_K\cos (u_n/2),0)$, so the relativistic bound-state mass from a rapidity pole givesThese neutral particles form the Sine-Gordon breather spectrum at reflectionless couplings: there are $N-1$ breathers, none for $N=1$, and the putative $n=N$ state is at the unbound $2{\mathcal{M}}_K$ threshold. The amplitude fixes these ratios to the physical kink mass; its rapidity dependence alone cannot fix the overall mass scale or a mass-renormalization prescription relating that mass to $m$.

It is important to distinguish direct and crossed interpretations rather than count every occurrence of a pole twice. Crossing symmetry sends $i{u}_n$ to $i(\pi -u_n)=i{u}_{N-n}$. At the original angle the momentum-difference invariant is $4{\mathcal{M}}_K^2{\sin }^2(u_n/2)={\mathcal{M}}_{N-n}^2$, so the crossed interpretation exchanges the complementary member of the same tower. In particular the transmission residue alternates sign: direct evaluation of the remaining factors gives ${\mathrm{Res}}_{i{u}_n}{S}_T=i(-1{)}^n{R}_n$, $R_n>0$. Thus one must retain charge-channel/crossed-channel information, not reject every negative-imaginary transmission residue or declare an additional particle for a crossed pole. Direct and crossed locations coincide in the reflectionless pole set. The familiar $N=2$ case has one breather of mass $\sqrt{2}{\mathcal{M}}_K$ despite the negative-imaginary transmission residue; this diagonal reflectionless example is also displayed in Castro-Alvaredo, Chen, Doyon and Hoogeveen, section 4.2.

Matching the exact phase. For the unwrapped reflectionless sine-Gordon transmission phase, at real $\theta \ge 0$ choose the continuous unwrapped branch with ${\delta }_N(0)=\pi N$. ThenEvery factor has unit modulus, consistent with purely transmitting elastic scattering. Differentiating is simpler than integrating its complex logarithm:For fixed $\theta >0$ the sum becomes a Riemann sum, and the elementary integral yieldsSince ${\beta }^2=8\pi /(N+1)$ and $\gamma =8\pi /N$, replacing $\gamma$ by ${\beta }^2$ changes this only at subleading order. The leading phase difference agrees with the standard semiclassical result. The derivative approximation is not uniform as $\theta \to 0$ at fixed $N$; integrating its logarithm gives a finite leading phase difference nevertheless. A compatible constant is the given branch $\pi N$, whose choice is supplied by the exact amplitude rather than the classical trajectory.

As a useful endpoint check, ${\int }_0^{\infty }\log \tanh (u/2)du=-{\pi }^2/4$, obtained by expanding the two logarithms $\log (1-e^{-u})-\log (1+e^{-u})$. Thus the leading phase with coefficient $2N/\pi$ tends to $\pi N/2$. The exact unwrapped endpoints arewhere the extra $\pi /2$ is subleading. Neither should be replaced by zero by an unannounced branch convention. Finally, the literal $m^2/\beta$ action gives a phase derivative of order $\sqrt{N}$, whereas the printed exact amplitude gives order $N$. The requested exact/semiclassical matching requires correcting the overall action normalization to $m^2/{\beta }^2$. This is an actual inconsistency between pages 2 and 4, not a change to the sine-Gordon equation or the classical scattering field.

An instanton describes a finite-action trajectory in imaginary time that crosses between classical vacua. It accounts for quantum tunnelling effects exponentially small in $\hslash$, beyond an ordinary power series around one minimum. A symmetric double well makes the connection between the Euclidean path integral, fluctuation theory and the low-energy spectrum particularly explicit.

For a particle of mass $\mu$, use a Wick rotation $t=-i\tau$ in the kernel. The Euclidean action and its classical equation areThe latter is Newtonian motion in the inverted potential $-V$. Subtract the common minimum energy so $V(\pm a)=0$. A finite-action path approaching $\pm a$ at infinite Euclidean times has conserved Euclidean energy $\mu {\dot{q}}^2/2-V=0$. ThereforeThis gives both the classical trajectory by a quadrature and its tunnelling exponent. The plus solution is the instanton; reversal is the anti-instanton. A single crossing has action $S_I$, not $2S_I$, although a periodic closed path necessarily has equal numbers of crossings in both directions.

An explicit double well. Take $V(q)=\mu {\omega }^2(q^2-a^2{)}^2/(8a^2)$, whose local harmonic frequency at either minimum is $\omega$. Integrating the zero-energy equation gives the quartic double-well instantonThe arbitrary center ${\tau }_0$ is a collective coordinate. The tunnelling regime is $S_I/\hslash \gg 1$, so the crossing width ${\omega }^{-1}$ is short compared with the typical separation between rare crossings.

The exponential alone does not give a dimensionally complete amplitude. Expand $q=q_I+\eta$; the quadratic Euclidean action is $\mu \int \eta L_I\eta d\tau /2$, whereThe derivative ${\dot{q}}_I$ is a normalizable translation instanton translation zero mode. The factorization of this one-dimensional fluctuation operator shows it has no negative modes. In $z=\omega (\tau -{\tau }_0)/2$, its two discrete eigenfunctions are proportional to ${\mathrm{sech}}^{2}z$ and $\mathrm{sech}z\tanh z$, with eigenvalues zero and $3{\omega }^2/4$; the continuum begins at ${\omega }^2$. Treating its zero eigenvalue as an ordinary Gaussian integral would give a divergent determinant. Instead integrate over ${\tau }_0$: since $\mu \int {\dot{q}}_I^{2}d\tau =S_I$, the Jacobian supplies $\sqrt{S_I/(2\pi \hslash )}d{\tau }_0$. The nonzero fluctuations supply the determinant ratio. With common endpoint regularization,This is the instanton fluctuation prefactor, with $K$ and $\kappa$ having frequency units. The prime omits precisely the translation mode.

For this quartic potential the determinant can be evaluated explicitly. Set $z=\omega (\tau -{\tau }_0)/2$ and add a small positive spectral regulator $\epsilon$ to both operators. With $r=2\sqrt{1+\epsilon /{\omega }^2}$, the dimensionless operator is $-{\partial }_z^2+r^2-6{\mathrm{sech}}^{2}z$. Applying successively $-{\partial }_z+\tanh z$ and $-{\partial }_z+2\tanh z$ to the free decaying solution $e^{-rz}$, and normalizing at $z\to +\infty$, gives a growing coefficient at $z\to -\infty$ equal to $(r-1)(r-2)/[(r+1)(r+2)]$. The common Dirichlet functional determinant ratio has this coefficient. Hence the quartic double-well fluctuation determinant isThe frequency factor cannot be discarded when deleting the zero eigenvalue.