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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.

For a one-dimensional wavefunction, the probability current isA plane wave $D{e}^{ikx}$ carries current $(\hslash k/m)|D{|}^2$ to the right, while $D{e}^{-ikx}$ carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude $A$, a reflected wave of amplitude $B$, and a transmitted wave of amplitude $C$. Their currents areCurrent conservation for a real potential gives $j_{\mathrm{in}}+j_{\mathrm{ref}}=j_{\mathrm{tr}}$, and thereforewith $P_{\mathrm{ref}}+P_{\mathrm{tr}}=1$.

A normalizable bound state has $E<0$. Writingits decaying asymptotic behaviour is

LetUsing $\sin (2ka)=2ka+O(k^3)$ and $\cos (2ka)=1+O(k^2)$ in the phase-shift formula givesBy the definition of the scattering length from a partial-wave S-matrix,Thus $a_s$ diverges exactly whenAt each such value, an S-wave bound state lies at the continuum threshold: the exterior zero-energy radial solution becomes constant, its logarithmic derivative vanishes, and the system has a zero-energy scattering resonance. As the well deepens through the critical value, a new bound state appears and the scattering length passes through infinity and changes sign.

As $a\downarrow 0$,so $E(a)<0$ for all sufficiently small positive $a$. On the other hand,The derivative satisfiesThe extremum equationhas exactly one positive solution because its left side is strictly increasing. Thus $E$ has one global minimum, and that minimum is negative.

The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable bound state. Hence every $V_0>0$ produces at least one bound state.

As $|x|\to \infty$, the equation approachesIf $E>U_0$, the asymptotic solutions are oscillatory and do not decay, so there are no bound states. If $0<E<U_0$, the asymptotic solutions are exponential; one decaying branch can be selected at each end. Bound states are therefore possible, although only energies satisfying the global matching conditions actually occur.

The time-independent Schrodinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.

At small $k$,Since $S_0=e^{2i{\delta }_0}$,The scattering length and low-energy total cross-section are therefore

The poles at $k=2i\lambda$ and $k=3i\lambda$ lie on the positive imaginary axis. They represent two bound states, with energiesThere are no resonance poles away from the imaginary axis in the lower half-plane. Positivity of $\lambda$ puts the poles on the physical bound-state axis, where $e^{ikr}$ decays; changing its sign would move them to the unphysical half-plane. This illustrates the bound-state poles of a partial-wave S-matrix.

Write $E=-{\hslash }^2{\kappa }^2/(2m)$. An even bound state is proportional to $\cos (kx)$ inside the well and to $e^{-\kappa |x|}$ outside, whereContinuity of the logarithmic derivative at $a$ givesFor sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow 0$, $\kappa \sim k^{2}a\to m/{\hslash }^2$. HenceThe wells converge distributionally to $V_0(x)=-\delta (x)$, whose normalized even bound state isIts derivative jump is precisely the one imposed by the delta potential.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.

For a reflectionless scattering wavefunction whose two asymptotic phases differ by $\delta (k)$, a large periodic box gives $k_{n}L+\delta (k_n)=2\pi n$. Compared with the free value $k_n^{(0)}=2\pi n/L$, $k_n-k_n^{(0)}=-\delta (k_n^{(0)})/L+O(L^{-2})$ away from thresholds. Discrete bound states and phase-branch changes alter mode indexing and must be counted when comparing complete spectra.

At relative rapidity $iu$, the analytically continued sum of the constituent four-momentum vectors is on the bound-state mass shell with the displayed mass. For equal constituents, rapidities $\chi \pm iu/2$ sum to $2m_a\cos (u/2)(\cosh \chi ,\sinh \chi )$. A pole at $u=0$ is at threshold rather than a strictly bound state.

The direct channel combines the two incoming four-momentum vectors. A one-particle pole in this channel has the intermediate particle's squared mass equal to $s$. Imaginary rapidity differences can satisfy this mass relation for a bound state even though there is no pole at real physical scattering rapidity.

For the unit-modulus hyperbolic scattering block with elementary mass $m$, a pair bound at rapidity difference $iu$ has mass $m_2=2m\cos (u/2)$. Bootstrap fusion with another elementary particle gives a further direct pole at $3iu/2$. It is inside the physical rapidity strip with positive imaginary residue only for $0<u<2\pi /3$. In that range the new bound state has rest-frame constituent rapidities $iu,0,-iu$, so its energy is $m(1+2\cos u)=m\sin (3u/2)/\sin (u/2)$. Charges add at the bound-state vertex. At $u=2\pi /3$ the apparent pole cancels; beyond that point it lies outside the physical strip. The formula is conditional on this particular scattering block and its direct-pole interpretation, not a universal three-body binding law.