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Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrodinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

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.

In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

The required dipole matrix element iswhere differentiating the elementary Laplace integral for $\sin (kx)$ evaluates the last integral. Hence

Put $E_f=E_0+\hslash \omega >0$ and $k_f=\sqrt{2m{E}_f}/\hslash$. The continuum transition probability at long times usesSincethe total escape probability becomesFor the attractive delta potential,Substitution gives the requested leading-order result:

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.