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Linearizing about the phi-four kink gives the displayed one-dimensional Schrodinger operator in unit kinetic normalization. Its translational zero mode in field theory is proportional to ${\mathrm{sech}}^2(cx)$. The localized shape eigenfunction $\mathrm{sech}(cx)\tanh (cx)$ has squared frequency $3c^2$, while the continuum starts at $4c^2$. Substitution verifies both bound-state eigenfunctions. The Pöschl-Teller potential permits a short completeness argument for the bound modes. In $y=cx$ let $A_{\ell }={\partial }_y+\ell \tanh y$. Then $H_K/c^2=A_2^{†}{A}_2$, its partner is $A_2{A}_2^{†}=A_1^{†}{A}_1+3$, and $A_1{A}_1^{†}=-{\partial }_y^2+1$. The last operator has no bound states. The kernels of $A_1$ and $A_2$ give the two stated modes, while the partner spectra exclude any further normalizable bound modes.

Write $T=\tanh (ax)$ and $S={\mathrm{sech}}^2(ax)$. Usingtwo differentiations ofgiveThus it solves the time-independent Schrodinger equation in the stated Pöschl-Teller potential, with

As $x\to -\infty$ and $x\to +\infty$, respectively,There is no left-moving term, so this is a reflectionless potential:The incident and transmitted amplitudes have equal modulus becauseHence

Now set $k=i\lambda$ with $\lambda >0$. ThenAt $+\infty$ this decays for every positive $\lambda$. At $-\infty$, however, the factor tends to $\lambda -a$ while $e^{-\lambda x}$ grows. Normalizability therefore requires $\lambda =a$. In that casewhich decays at both ends. The bound-state energy is

LetOn $L^2(\mathbb{R})$, $A^{†}=-{\partial }_x+\chi \tanh (\chi x)$. The supersymmetric factorization of the one-soliton potential isIn particular, $A^{†}A$ is nonnegative, so every eigenvalue of $H$ satisfies

Equality is attained precisely when $A\psi =0$. Solving that first-order equation giveswhich is square-integrable. With unit normalization, $C=\sqrt{\chi /2}$, and

It remains to exclude another bound state. Since this Pöschl-Teller potential tends to zero at infinity, a bound-state energy must have $E<0$. If $H\psi =E\psi$, then applying $A$ togivesThe free one-dimensional Schrodinger operator has no nonzero square-integrable eigenfunction with $E<0$. Consequently $A\psi =0$, and the state must be proportional to ${\psi }_0$. Thus there is exactly one bound state, with