= Solution
Take $\lambda\ne0$ 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 $c=\cos\alpha$, $s=\sin\alpha$, $f(x)=c\operatorname{sech}(cx)/\lambda$, so $\psi=f e^{ist}$. Initially take $0<|\alpha|<\pi/2$, avoiding the singular target-space boundary.
The canonical <Hamiltonian density> is
$$
\mathcal H_\ell=\frac{|\dot\psi|^2+|\psi_x|^2}{1-\lambda^2|\psi|^2}+|\psi|^2.
$$
Here $|\dot\psi|^2=s^2f^2$, $|\psi_x|^2=c^2f^2\tanh^2(cx)$, and $D=1-c^2\operatorname{sech}^2(cx)$. Since $s^2+c^2\tanh^2(cx)=D$, the kinetic contribution is simply $f^2$. Consequently
$$
\boxed{\mathcal M(\alpha)=2M\int f^2dx=\frac{4M}{\lambda^2}\cos\alpha.}
$$
To fix the charge sign, take the <Noether current> directly from $\delta\psi=i\varepsilon\psi$, $\delta\psi^*=-i\varepsilon\psi^*$:
$$
j^\mu=\frac{i(\psi\partial^\mu\psi^*-\psi^*\partial^\mu\psi)}{1-\lambda^2|\psi|^2},\qquad j^0=\frac{2sf^2}{D}.
$$
Thus positive internal rotation has positive charge in this convention. With $X=cx$ and the supplied integral,
$$
Q=\frac{2sc}{\lambda^2}\int_{-\infty}^{\infty}\frac{dX}{\cosh^2X-c^2},\qquad
\boxed{Q(\alpha)=\frac4{\lambda^2}\left[\operatorname{sgn}(\alpha)\frac\pi2-\alpha\right].}
$$
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 $4M/\lambda^2$, but
$$
\boxed{\lim_{\alpha\to0^\pm}Q=\pm2\pi/\lambda^2.}
$$
The displayed local density does not assign a unique endpoint charge merely by substituting $\alpha=0$.
\b[The four collective parameters.] Translation, a <Lorentz boost> of <rapidity> $b$, and the global phase symmetry produce
$$
\boxed{\psi(x,t)=\frac{c}{\lambda}\frac{e^{i\chi+i s\eta}}{\cosh(c\xi)},\quad \xi=\cosh b(x-X_0)-\sinh b\,t,\quad \eta=\cosh b\,t-\sinh b(x-X_0).}
$$
The 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>.
\b[Semiclassical quantization.] For one rest-frame internal rotation, the period in dimensionless time is $T=2\pi/|s|$. The action around the orbit is
$$
I=\int_0^Tdt\int dx(\pi_\psi\dot\psi+\pi_{\psi^*}\dot\psi^*)=\int_0^Tdt\int dx\frac{2|\dot\psi|^2}{D}=T sQ=2\pi|Q|.
$$
Equivalently this is $S+\mathcal E T$, where $\mathcal E$ is the reduced rest energy. <Bohr-Sommerfeld quantization> with $\hbar=1$ 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. Hence
$$
\boxed{Q=\pm n,\quad n=1,2,\ldots,\qquad |\alpha|=\frac\pi2-\frac{\lambda^2n}{4},\qquad \mathcal M_n=\frac{4M}{\lambda^2}\sin\frac{\lambda^2n}{4}.}
$$
On the nonsingular branch $0<n<2\pi/\lambda^2$. The elementary charge-one <mass> approaches $M$ at weak coupling. For positive $a,b$ with $(a+b)\lambda^2/4\leq\pi/2$, $\sin a'+\sin b'>\sin(a'+b')$, where $a'=a\lambda^2/4$, $b'=b\lambda^2/4$. 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 $|q_i|\geq n$ already has mass at least $\mathcal M_n$; if every $|q_i|<n$, the decreasing ratio $\mathcal M_q/q$ gives $\sum_i\mathcal M_{|q_i|}\geq(\mathcal M_n/n)\sum_i|q_i|\geq\mathcal M_n$, 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.
\b[Counting states and the endpoint.] Write $L=2\pi/\lambda^2$. If $L$ is not an integer, the regular branch gives
$$
\boxed{Q=\pm1,\ldots,\pm\lfloor L\rfloor,\qquad \text{number of charge states}=2\lfloor L\rfloor.}
$$
If $L$ 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 https://arxiv.org/pdf/hep-th/9410140[Dorey and Hollowood, section 2]. A generic classical $\lambda$ does not automatically define an integer $k$ or specify this quantum completion.
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