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With outgoing-channel index first, the scattering matrix convention is ${\psi }_b\sim I_b+{\sum }_{a=\pm }{S}_{ab}{O}_a$ as $|x|\to \infty$. An even potential commutes with parity, so its incoming even and odd channels remain separate: $\boxed{S_{+-}=S_{-+}=0}$.

For the odd channel, on $0<x<a$ write ${\psi }_{-}(x)=C\sin (kx)$, while on $x>a$ its prescribed normalization is $e^{-ikx}-S_{--}{e}^{ikx}$. A delta potential keeps the wavefunction continuous and imposes ${\psi }^{\prime }(a+)-{\psi }^{\prime }(a-)=U_0\psi (a)$, with $U_0=2m{V}_0/{\hslash }^2$. Thus the exterior logarithmic derivative is $k\cot (ka)+U_0$. Matching givesMultiplying numerator and denominator by $2\sin (ka)$ and using exponential forms yieldsFor real $k,U_0$, the numerator of the fraction is the complex conjugate of its denominator, and the prefactor has modulus one. Hence $\boxed{|S_{--}{|}^2=1}$, conservation of probability flux in this single elastic parity channel. In particular the minus sign built into $O_{-}$ makes the free odd-channel matrix element equal to one.

For an odd bound state, continue to $k=i\kappa$, $\kappa >0$. A pole means the outgoing solution is exponentially decaying on both sides. Its denominator vanishes whenThe function $2\kappa /(1-e^{-2\kappa a})$ rises strictly from $1/a$ to infinity: differentiation reduces positivity to $e^{2\kappa a}>1+2\kappa a$. Thus there is exactly one positive solution if $-U_0>1/a$, and none otherwise. The numerator does not vanish at that solution, so the pole is genuine. ConsequentlyAt equality the zero-energy threshold solution is not square integrable.

For repulsive strong barriers put $u=U_{0}a\gg 1$ and $z=ka=\pi +\delta$. The pole equation is $z\cot z+u-iz=0$. Since $\cot \delta ={\delta }^{-1}-\delta /3+O({\delta }^3)$, multiplication by $\delta$ gives $\pi +(u+1-i\pi )\delta +O({\delta }^2)=0$. Iterating in $u^{-1}$ yieldsThis lower-half-plane resonance pole is an odd standing wave trapped between the barriers and slowly leaking by tunnelling. Its complex energy is $E_R-i\Gamma /2$, with $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)>0$. The factor $e^{-iEt/\hslash }$ then decays rather than grows; $\gamma >0$ encodes the positive escape rate.