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Let $G_n=2\pi n/a$. The free states $|k⟩$ and $|k-G_n⟩$ are degenerate at the Bragg point $k=G_n/2=n\pi /a$. Sincedegenerate perturbation theory restricts the Hamiltonian near this crossing toWriting $k=n\pi /a+\kappa$ and diagonalizing gives the nearly-free electron dispersion near a one-dimensional band gapThus the two continuous bands avoid crossing, and at $\kappa =0$ their separation isThis is the nearly-free electron model. The dispersion relation is the relation $E=E(k)$ between energy and Bloch wavevector.

For the specified potential,Henceand all other nonconstant Fourier coefficients vanish. The gaps are thereforeTheir centre energies are respectivelyThe extended-zone sketch consists of the free-particle parabolas shifted upward by $V_0$, with avoided crossings of these two widths at the listed Bragg points; elsewhere they meet to this order because the corresponding Fourier coefficients vanish.

For $N=1$, writeThe given evolution of the discrete scattering data givesSolving the resulting two scalar linear equations for the components of ${\psi }_1$ and substituting in the reconstruction formula givesChoosing the additive multiple of $2\pi$ so that $u\to 0$ as $X\to +\infty$ and integrating in $X$ yields the One-soliton solution of the sine-Gordon equation in light-cone coordinatesIt depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:The unique positive solution is $l=1/2$, and then

The transformations satisfy $g^s{g}^r=g^{s+r}$, $g^0$ is the identity map, and $(g^s{)}^{-1}=g^{-s}$, so they form a one-parameter group. Ifthen the chain rule givesThus $g^s$ is a Lie point symmetry. Applied to the one-soliton family, it replaces $l$ by $l^{\prime }=e^{-s}l$. Taking $e^s=2l$ gives $l^{\prime }=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce toAt fixed $x$, only $\sin (2mt)$ varies, so the Sine-Gordon breather has fundamental period

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters becomeand hence $(l^{\prime }{)}^2+(m^{\prime }{)}^2=e^{-2s}{r}^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.