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For $U_{0}a\gg 1$, the lowest odd scattering resonance pole lies near the first odd mode trapped between the barriers:in its real part and to leading nonzero order in its imaginary part. Writing $E={\hslash }^2{k}^2/(2m)=E_R-i\Gamma /2$, the survival probability decays as $e^{-\Gamma t/\hslash }$.

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.

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.

Since $\tan {\delta }_0=\gamma /(K_0-k)$,The exact nearby stationary point satisfies $K_0-k={\gamma }^2/K_0+O({\gamma }^4/K_0^3)$, so to leading order in $\gamma /K_0$ the local maximum is at $k=K_0$. This is a scattering resonance withwhere $\Gamma$ is the full energy width of the Breit-Wigner peak.