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For positive wavenumber $k$, the odd channel of the symmetric double-delta potential hasFor real $k$ its numerator and denominator are complex conjugates, making the channel unitary. Its poles are the zeros of $k{e}^{-ika}+U_0\sin (ka)$.

Assume $a>0$ and put $\mu =mg/{\hslash }^2$, $q=k/\hslash$, and $\gamma =\mu /q$. At a delta potential located at $x_0$, integration of the Schrödinger equation gives the matching conditionsWriting the adjacent plane wave coefficients as $(A_L,B_L)$ and $(A_R,B_R)$ givesSolving these two linear equations yields the transfer matrix of a delta potential:For the symmetric double-delta potential, use separate plane wave pairs in $x<-a$, $-a<x<a$, and $x>a$. Successive matching gives $M=D(a)D(-a)$ between the two exterior pairs. For real $q$, multiplication givesThe determinant identity $\det M=1$ reads $|d{|}^2-F^2=1$. Left incidence means $(t,0{)}^T=M(1,r{)}^T$, whenceHere the first relation also follows from $t=M_{11}-M_{12}{M}_{21}/M_{22}=\det M/M_{22}$. Therefore the requested transmission probability and quantum reflection probability, as functions of the printed momentum $k$, areThey obey $R+T=1$. As checks, $g=0$ gives $T=1$, and the coincident limit $a\to 0$ gives $T=q^2/(q^2+4{\mu }^2)$, the answer for a single delta potential of strength $2g$.

For the symmetric double-delta potential, take an odd state at $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. On $0<x<a$ odd parity selects $B\sinh (\lambda x)$. For $x>a$, decay and continuity give $B\sinh (\lambda a)e^{-\lambda (x-a)}$; extend to negative $x$ by oddness. At $x=a$ the jump condition isFor a nonzero state this means $\lambda \Delta [1+\coth (\lambda a)]=1$, and thereforeOddness makes the jump at $-a$ follow automatically: both its derivative jump and its required wavefunction value have the corresponding opposite signs.

There is a positive solution precisely when $\Delta <a$. To see this, rewrite the equation asThe right side decreases strictly from $a$ to zero as $\lambda$ increases from zero to infinity: its derivative has the sign of $(1+2a\lambda )e^{-2a\lambda }-1<0$. Hence for $0<\Delta <a$ there is exactly one positive solution, while $\Delta =a$ is a nonnormalizable zero-energy threshold and $\Delta >a$ gives no odd bound state. This verifies the stated sufficiently-small-$\Delta$ condition and identifies its exact boundary, consistent with the odd bound state of a symmetric double-delta potential.

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 Schrödinger 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.