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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.