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The time-independent Schrodinger equation gives the even interior wavefunction $\psi =A\cos (\alpha x)$, with ${\alpha }^2=2m(U+E)/{\hslash }^2$. Outside, normalizability of a bound state requires $\psi =C{e}^{-\beta (|x|-l)}$, with ${\beta }^2=-2mE/{\hslash }^2$.

Continuity of the wavefunction and its derivative at $x=l$ gives $C=A\cos (\alpha l)$ and $-\beta C=-\alpha A\sin (\alpha l)$. A nonzero solution cannot have $\cos (\alpha l)=0$, so division givesAt the formally allowed endpoint $E=-U$, the even interior solution is constant; derivative matching would force the exterior amplitude to vanish, leaving the zero solution. Nontrivial bound states therefore have $-U<E<0$. The matching formula is the even-state condition for a finite square well.

To keep the paper's coupling convention explicit, write $g=\sqrt{\beta }$ and use physical coordinates $X=x/m$, $T=t/m$. Assume $\beta >0$ and weak coupling $g\ll 1$. The canonical real scalar field is $\phi =\varphi /g$, and the physical potential isThe symbol $g$ is the coupling usually appearing in canonical Sine-Gordon theory; the paper's $\beta$ is its square. The vacua have $g\phi =2\pi n$. A static Sine-Gordon kink has $g{\phi }_K=4\arctan e^{m(X-X_0)}$ and satisfies $\frac{1}{2}{\phi }_K^{\prime 2}=U({\phi }_K)$. Since $1-\cos (g{\phi }_K)=2{\mathrm{sech}}^2[m(X-X_0)]$, its classical mass isThe topological sector is important: this energy is measured relative to a vacuum, and the kink joins distinct vacua at the two spatial ends. Expanding around a spatially constant vacuum cannot construct this state by any finite-order perturbation in $g$.

For the general one-loop soliton mass correction, start with a canonical scalar field potential $U$ and a stable static soliton ${\phi }_K$. Write $\phi ={\phi }_K+\eta$. The term linear in $\eta$ vanishes by the classical Euler-Lagrange field equation. The quadratic fluctuation Hamiltonian isChoose a common large box and a common finite-mode regularization in quantum field theory. Expand the nonzero normal modes as $\eta ={\sum }_n{q}_n{f}_n$, with ${\mathcal{H}}_K{f}_n={\omega }_n^2{f}_n$ and normalized eigenfunctions. Each pair $(q_n,p_n)$ is a quantum harmonic oscillator contributing ground-state energy $\hslash {\omega }_n/2$. In the vacuum, replace ${\mathcal{H}}_K$ by ${\mathcal{H}}_0=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{vac}})$. Subtract the two ground-state energies and add the local counterterms evaluated on the soliton relative to the vacuum. This derives the general formulaThe prime excludes exact zero modes in field theory from oscillator quantization; their zero frequencies contribute no zero-point energy, but their role in mode counting must not be forgotten. The two sums mean a paired finite regulator, not separate subtractions of divergent answers. Equivalently the nonzero-mode term is the regulated difference of the square-root traces of the two fluctuation operators. Discrete bound states and continuum modes both contribute. This is a vacuum-subtracted soliton mass and the first term in the semiclassical soliton mass expansion.

Translation gives a zero mode in field theory because differentiating the static equation yields ${\mathcal{H}}_K{\phi }_K^{\prime }=0$. A Gaussian oscillator or an unprimed functional determinant is inappropriate along this flat direction. Replace its amplitude by the position collective coordinate $X_0(T)$ and require the residual fluctuation to obey $\int \eta {\phi }_K^{\prime }dX=0$, preventing double counting. The associated change-of-variables Jacobian supplies the zero-mode normalization. At low speed the collective-coordinate effective Lagrangian for a soliton is $-M_{\mathrm{cl}}+\frac{1}{2}{M}_{\mathrm{cl}}{\dot{X}}_0^2+\cdots$; quantizing the position gives the soliton momentum and its translational states, not an extra oscillator rest energy. More generally, every physical continuous modulus needs a collective coordinate; gauge directions require gauge fixing rather than additional physical states.

The Sine-Gordon kink fluctuation operator is particularly simple:while $D{D}^{†}=-{\partial }_X^2+m^2={\mathcal{H}}_0$. This supersymmetric factorization of the one-soliton potential shows stability and generates all nonzero eigenfunctions from vacuum plane waves. The sole normalizable bound state is the translational zero mode of a sine-Gordon kink, proportional to $\mathrm{sech}m(X-X_0)$; there is no positive-frequency internal bound oscillator. The continuum has $\omega (k)=\sqrt{k^2+m^2}$ and no reflection. Applying $D^{†}$ to $e^{ikX}$ gives a transmission amplitudeThis scattering phase shift changes the density of continuum modes. The high-frequency vacuum subtraction cancels the extensive vacuum contribution but still leaves a logarithmic ultraviolet divergence.

The finite part also requires consistent mode-number regularization of soliton masses. The periodic-box phase-shift quantization condition is $k_{n}L+\delta (k_n)=2\pi n$. Match $2N+1$ modes: the kink has its translation mode plus the two continuum modes at each $n=1,\dots ,N$, while the vacuum has the $k=0$ oscillator of frequency $m$ and the corresponding continuum pairs. With $\hslash =1$ and $\Lambda =2\pi N/L$, expanding $k_n-k_n^{(0)}=-\delta (k_n^{(0)})/L+o(L^{-1})$ givesIntegration by parts uses $\delta (0)=\pi$ and yieldsThe cutoff surface term for a Sine-Gordon kink tends to $-m/\pi$. It cannot be dropped merely because $\delta (\Lambda )\to 0$: $\omega (\Lambda )$ grows at the same time.

A renormalization condition must specify which mass and coupling are held fixed. For vacuum normal ordering, or cancellation of the vacuum tadpole diagram with the elementary mass fixed at $m$, the quartic interaction gives the mass countertermEvaluating this Sine-Gordon vacuum tadpole counterterm on the kink givesThe logarithmic ultraviolet divergence cancels, leaving the renormalized one-loop Sine-Gordon kink mass in the stated vacuum scheme:This example illustrates why a zero mode in field theory must be treated as a collective coordinate, why vacuum subtraction alone need not remove ultraviolet divergences, and why the finite relation between the two regulators matters. Different finite counterterms amount to different definitions of the renormalized parameters; an unregulated frequency difference without a renormalization condition is not a physical mass prediction.

In elastic diagonal factorized scattering, the Faddeev-Zamolodchikov algebra describes an exchange of two particle operators. Exchanging the pair twice must restore the original state. This is analytic unitarity:Keeping the reversed species in this identity avoids an unstated parity assumption. For the particular amplitudes derived below, crossing and analytic unitarity also imply $S_{\bar{A}A}=S_{A\bar{A}}$. Hermitian analyticity of a two-particle S-matrix states $S_{ab}(\theta {)}^{\ast }=S_{ba}(-{\theta }^{\ast })$ and, together with analytic unitarity, gives $|S_{ab}(\theta )|=1$ for real rapidity differences. Crossing symmetry analytically turns an incoming particle into an outgoing antiparticle, relating the two channels by $S_{A\bar{A}}(\theta )=S_{AA}(i\pi -\theta )$, with the reverse relation obtained by crossing again. The shift $i\pi$ follows from the sign reversal of the two-momentum $p(\theta )=(m\cosh \theta ,m\sinh \theta )$.

Put $u=\lambda /2$. The given amplitude is the unit-modulus hyperbolic scattering block $S_u(\theta )=\sinh [(\theta +iu)/2]/\sinh [(\theta -iu)/2]$. Its numerator and denominator interchange under $\theta \mapsto -\theta$, proving analytic unitarity. For real $u$ it also obeys Hermitian analyticity of a two-particle S-matrix. Applying crossing symmetry givesThis crossed amplitude also has unit modulus on the real axis. Thus both requested channel constraints hold.

The bound-state conclusion needs a coupling range. In the fundamental attractive range $0<u<\pi$, equivalently $0<\lambda <2\pi$, the denominator has a simple pole at $\theta =iu$ inside the physical rapidity strip. Its residue iswith positive imaginary coefficient. Under the usual one-particle interpretation of this direct-channel bound-state pole, it couples two charge-$+1$ particles to a new charge-$+2$ particle $B$. In the center-of-mass frame its constituents have analytically continued rapidities $\pm iu/2$, and their total two-momentum is $(2m\cos (u/2),0)$. Thus the relativistic bound-state mass from a rapidity pole givesIt is positive and less than the two-particle threshold $2m$. Without the coupling qualification the requested deduction is false: at $\lambda =0$ the amplitude is identically one after removing the apparent $0/0$ at the origin. A free massive complex scalar field has this diagonal amplitude, charge-$\pm 1$ particles and no isolated charge-$+2$ bound state. At $u=\pi$ the amplitude similarly becomes constant $-1$ and the apparent boundary pole cancels. Neither endpoint supplies the claimed particle.

For bound-state fusion of factorized S-matrices, represent $B$ as the residue of $A({\theta }_B+iu/2)A({\theta }_B-iu/2)$ at its bound-state separation. Move a third $A({\theta }_A)$ through both constituents using the Faddeev-Zamolodchikov algebra, then take the same residue. The bound-state normalization occurs on both sides and cancels. Consequently bootstrap fusion giveswhere $\theta ={\theta }_B-{\theta }_A$. This has analytic unitarity as a product of two shifted blocks. Its poles occur at $\theta =iu/2$ and $\theta =3iu/2$, modulo $2\pi i$; numerator zeroes occur at the corresponding negative positions, subject to cancellations at special couplings.

The nearer pole, $\theta =iu/2$, has residue $-2i\sin u$. It is a crossed-channel pole in diagonal factorized scattering, not a new direct-channel charge-$+3$ state. Indeed, the exchanged momentum has invariantusing $m_B=2m\cos (u/2)$. The exchanged particle is therefore the already present $A$, with the appropriate charge flow at the crossed vertex. Equivalently, crossing symmetry puts a positive-residue direct pole of $S_{B\bar{A}}$ at $i(\pi -u/2)$, corresponding to $B+\bar{A}⟶A$. This distinction avoids assigning an extra mass by applying the direct-channel formula to every pole.

The farther pole, $\theta =3iu/2$, lies in the physical rapidity strip only for $0<u<2\pi /3$, equivalently $0<\lambda <4\pi /3$. Its residue iswhich has positive imaginary coefficient in that range. It gives a charge-$+3$ bound state $C$. The relativistic bound-state mass from a rapidity pole now yieldsThe positive root in the admitted range isThis is the three-particle bound state from equal-mass fusion; in its own rest frame the constituent rapidities are $iu,0,-iu$. At $u=2\pi /3$ the farther apparent pole cancels because its numerator also vanishes; for $2\pi /3<u<\pi$ it lies outside the physical rapidity strip and does not require a new charge-$+3$ particle. Charge-conjugate partners carry charges $-2$ and $-3$ where the corresponding states exist. Fusion distinguishes an existing crossed-channel particle from a genuinely new direct-channel bound state, and the latter requires the stated smaller coupling range.

Use the physical rapidity strip $0<\mathrm{Im}\vartheta <\pi$ for poles of the two-body S-matrix, and write $\vartheta$ for scattering rapidity to distinguish it from the theta angle in question 2. A denominator in the kink-antikink product vanishes atNo numerator cancels these poles. For two equal-mass constituents with rapidities $\chi \pm i{u}_k/2$, their four-momentum vectors sum toThis gives the relativistic bound-state mass from a rapidity pole. The ordered breather spectrum isIt increases strictly with $k$. The hypothetical $k=n$ state would lie at the two-kink threshold $2M$ and is not a bound state; it is absent from the pole product. At these couplings $M=mn/\pi$ and ${\beta }^2=8\pi /(n+1)$. In particular $m_1\to m$ at weak coupling $n\to \infty$. This is the Sine-Gordon breather spectrum at reflectionless couplings. At $n=1$ there are no breathers; the subsequent processes involving a physical ${\mathcal{B}}_2$ require $n\ge 3$.

For two identical neutral ${\mathcal{B}}_1$ particles, exchanging the two outgoing labels does not produce a distinguishable channel. In one spatial dimension the elastic final momenta are the incoming pair, up to interchange. Thus there is one scalar identical-particle amplitude, rather than separately observable transmission and reflection amplitudes.

Put $a=\pi /(2n)$, so the basic amplitude uses $\sin (2a)$. Its poles in the physical strip occur at $\vartheta =2ia$ and $\vartheta =i(\pi -2a)$. The first is the direct bound-state pole. Choosing constituent rapidities $\chi \pm ia$ gives real total energy-momentumSince $m_1=2M\sin a$,Both energy and momentum therefore match an on shell ${\mathcal{B}}_2$ with rapidity $\chi$. The complementary pole is its crossed-channel partner. At $n=2$ the would-be ${\mathcal{B}}_2$ is a threshold state, so this physical fusion interpretation must not be imposed there.

The bound-state fusion of factorized S-matrices treats a bound particle as its on shell constituents with analytically continued rapidities. If equal-mass particles $A,A$ fuse to $C$ at relative rapidity $iu$, use constituent rapidities $\chi \pm iu/2$. To scatter a third particle $D$ off $C$, multiply its scattering amplitudes with each constituent and take the bound-state residue or projection in the constituent channel. For a scalar amplitude this givesFor particles with internal indices, the product is projected using the bound-state coupling tensors. The heuristic reason is factorized scattering: conserved higher charges prevent particle production and fix the rapidity data, so the third particle scatters through the constituents by successive two-body processes. Consistency of different orders is the Yang-Baxter equation. Without integrability, an independent three-body interaction would invalidate this simple bootstrap product.

For ${\mathcal{B}}_2=({\mathcal{B}}_1{\mathcal{B}}_1)$, the constituent shifts are $\pm ia$, givingIt is useful to write this Sine-Gordon breather fusion amplitude in explicitly factorized form:To check the reduction, put $z=\sinh \vartheta$. Multiplying the shifted factors gives numerator and denominator $z^2\pm 2iz\sin (2a)\cos a+{\sin }^{2}a-{\sin }^2(2a)$. Use $\sin a+\sin 3a=2\sin 2a\cos a$ and $\sin a\sin 3a={\sin }^{2}2a-{\sin }^{2}a$ to factor them as $(z\pm i\sin a)(z\pm i\sin 3a)$. The product tends to one at large positive real rapidity, fixing its overall phase in this bootstrap convention.

For $n\ge 3$, the nearest pole to the real axis is $\vartheta =ia$. It is simple and comes from the first factor. In the crossed, or t-channel, the momentum carried between the external particles is their difference. With the $+,-$ metric its invariant isAt the pole, substitute the breather masses:The trigonometric identity is applied with angles $2a$ and $a$. Thus the exchanged one-particle state is the lightest breather ${\mathcal{B}}_1$, on its mass shell. This is crossed-channel lightest-breather exchange. The external momenta at a bound-state pole are analytically continued; on shell here means the invariant mass relation and conservation of the continued energy-momentum, not a pole at real physical rapidity. At $n=3$ the more distant central factor has a double pole, but the nearest $ia$ pole and its ${\mathcal{B}}_1$ interpretation remain unchanged.

Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector $Q=0$ and in the one-kink topological sector $Q=1$. In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals ${\Psi }_Q$. Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with $\varphi (+\infty )-\varphi (-\infty )=2\pi Q$. A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time $\tau$, the energy eigenstate expansion gives $K_Q^E(\tau )\sim C_Q{e}^{-E_Q\tau }$. Thus the exact vacuum-subtracted soliton mass isEquivalently it is $i{T}^{-1}\log (K_1/K_0)$ at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.

With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, ${\varphi }_K^{\prime }=2\mathrm{sech}x$ and $1-\cos {\varphi }_K=2{\mathrm{sech}}^{2}x$, so $M_{\mathrm{cl}}=8m/{\beta }^2$. Write $\eta =\delta \varphi$. Expanding and integrating by parts givesThe first variation is ${\beta }^{-2}\int ({\varphi }_K^{\prime \prime }-\sin {\varphi }_K)\eta$, plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about $\eta$.

The printed expansion omits ${\beta }^{-2}$ despite the stated Lagrangian density. Its displayed form is the expansion of ${\beta }^{2}S$; it is not the physical $S$ at arbitrary coupling. Alternatively, writing $\eta =\beta \chi$ puts the quadratic term in canonical normalization, while leaving the classical term $-8mT/{\beta }^2$ unchanged. This normalization repair does not change ${\Delta }_x$ or the physical fluctuation frequencies.

The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and $A{\psi }_0=0$ gives the normalized translational zero mode of a sine-Gordon kink:A displacement $a$ changes ${\varphi }_K(x-a)$ by $-a{\varphi }_K^{\prime }(x)$. Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.

In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.

For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by $2\pi /L$, replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.

There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is $f_k\propto (\tanh x-ik)e^{ikx}$. Its transmission phase obeyswith the odd phase branch that tends to zero at large $|k|$. For $k>0$, the continuum roots have labels $n=1,2,\dots$; there is no periodic continuum root at $k=0$, since the limiting eigenfunction $\tanh x$ has opposite signs at the two ends. The bound state at ${\omega }_0=0$ replaces the free $k=0$ oscillator with ${\omega }_0^{(0)}=1$. Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite $-m/2$ under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large $L$. Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.

At high momentum, $\delta (k)=2/k+O(k^{-3})$, so both expressions have divergent part $-(m/\pi )\log \Lambda$. The canonical field $\phi =\varphi /\beta$ has a quartic interaction with coupling $-m^2{\beta }^2$. Its vacuum tadpole diagram shifts the squared mass by $-m^2{\beta }^{2}J(\Lambda )/4$, whereThe Sine-Gordon vacuum tadpole counterterm has $\delta m^2=+m^2{\beta }^{2}J/4$ and adds potential energy density $\delta m^2(1-\cos \varphi )/{\beta }^2$. Since $\int (1-\cos {\varphi }_K)dx=4$, its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then $8m/{\beta }^2-m/\pi +O(m{\beta }^2)$ in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.

For a bound state, $0<E<U_0$: the Hamiltonian operator is nonnegative, and an energy at or above the exterior potential does not give a nonzero normalizable wavefunction. The Time-independent Schrodinger equation and the hard wall at zero therefore giveContinuity of the wavefunction and its derivative at the finite step gives $k\cot (ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2m{U}_0}/\hslash$. The matching equation becomesIt requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi ,(j+1)\pi )$ for $j=0,1,\dots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\mathrm{sgn}(\sin z)\cos z)/{\sin }^{2}z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one bound state if and only if $z_0>(j+1/2)\pi$. These are the bound-state thresholds for a square well with one hard wall.

Exactly one normalizable bound state exists if and only ifThe printed strict upper inequality misses the threshold case. At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa =0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one bound state at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.

A bound state has $E<0$ and $\kappa =\sqrt{-2mE}/\hslash >0$, since outside the delta potentials a square-integrable solution must decay exponentially. The wavefunction is continuous at each delta: a jump would produce a delta derivative in ${\psi }^{\prime \prime }$ with no matching term in the equation. Integrating the Time-independent Schrodinger equation across either $x_0=\pm a$ givesThe symmetry of the symmetric double-delta potential permits even and odd solutions. Up to a nonzero normalization factor, the even candidate is $\cosh (\kappa x)$ for $|x|\le a$, and $\cosh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its derivative jump at $a$ givesThe odd candidate is $\sinh (\kappa x)$ inside and $\mathrm{sgn}(x)\sinh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its jump givesParity ensures the jump at $-a$ as well. Put $t=\kappa a$ and $L=2m\lambda a/{\hslash }^2$. The continuous strictly increasing function $t(1+\tanh t)$ ranges from $0$ to infinity, so there is exactly one even bound state for every $L>0$. The continuous strictly increasing function $t(1+\coth t)$ ranges from $1$ to infinity, so the odd bound state of a symmetric double-delta potential exists if and only if $L>1$. Strict increase follows from the allowed monotonicity facts plus the strictly increasing term $t$. ThereforeAt equality the odd limiting solution has $\kappa =0$ and is not square integrable, so it is not a bound state. No nonnegative energy can supply another full-line square-integrable state, since its exterior solutions are oscillatory or affine.

For a normalized trial state $|\psi ⟩={\sum }_n{c}_n|n⟩$ in the form domain of a bounded-below Hamiltonian, the quantum variational principle gives $⟨H⟩={\sum }_n|c_n{|}^2{E}_n\ge E_0$. Thus each trial expectation is an upper bound for the ground-state energy. The same spectral argument covers a continuum: for a decaying potential whose essential spectrum begins at $0$, a negative trial expectation guarantees a negative-energy bound state.

For the stated exponential trial function, normalization follows from $4{\alpha }^3{\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=1$. The kinetic expectation, using $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)$. The potential expectation is $-4A{\alpha }^3{\int }_0^{\infty }r{e}^{-(\mu +2\alpha )r}dr$. ThereforeThe exponential variational bound for the Yukawa potential at $\alpha =\mu /2$ gives $E=(\mu /8)({\hslash }^2\mu /m-A)<0$ if $\mu <Am/{\hslash }^2$. Thus a bound state necessarily exists in the stated range.

For $\mu \ge Am/{\hslash }^2$, this single-exponential family cannot prove binding: the maximum of $4A\alpha /(\mu +2\alpha {)}^2$ is $A/(2\mu )$, so every $E(\alpha )\ge 0$. This does not prove absence of bound states.

The equality case can in fact be strengthened. At $\mu =Am/{\hslash }^2$, use normalized exponentials ${\psi }_{\mu /2},{\psi }_{\mu }$ and write $Q={\hslash }^2{\mu }^2/m$. Their matrix elements areThe Rayleigh numerator for ${\psi }_{\mu /2}+ϵ{\psi }_{\mu }$ is negative for sufficiently small $ϵ>0$, and its norm is nonzero. Hence binding also holds at equality, and by continuity in a small interval beyond it. For arbitrary larger $\mu$, these trials alone do not settle existence; a nonnegative variational upper bound cannot establish a nonnegative ground-state energy.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.

A negative-energy odd bound state has $k=i\kappa$, $\kappa >0$, with $\psi =C\sinh (\kappa x)$ inside and a decaying exponential outside. Its jump condition, equivalently the nontrivial upper-half-plane pole of the preceding S-matrix, givesSet $w=2\kappa a$ and $g=-a{U}_0>0$. The condition is $w=g(1-e^{-w})$. The right side is strictly increasing and strictly concave, has slope $g$ at zero and is bounded by $g$. Hence there is exactly one positive intersection iff $g>1$. ThereforeThe formal zero root is not a positive-decay bound state. At $a{U}_0=-1$ there is a zero-energy threshold solution, not a normalizable bound state.