Separated smooth-barrier tunnelling: the declared transmission coefficient is exp(-2S/h) up to explicit small-h and barrier-thickness errors

[Hypothesis matching] Under (W2), $W-\lambda$ has one simple zero ${\xi }_a$ with positive derivative and one simple zero ${\xi }_b$ with negative derivative. Since $W$ is compactly supported and $W-\lambda <0$ outside $[{\xi }_a,{\xi }_b]$, $\lambda >0$; choose $L$ with $\mathrm{supp}W\subset (-L,L)$, so the fixed exterior points ${\xi }_{-}=-L$ and ${\xi }_{+}=L$ lie in allowed intervals and $W=0$ near them. Fix the interior point ${\xi }_0=({\xi }_a+{\xi }_b)\mathrm{/}2$. The intervals $[{\xi }_{-},{\xi }_0]$ and $[{\xi }_0,{\xi }_{+}]$ contain respectively only the left and right simple turning point, so [F1] applies on each, with reflection of the coordinate on the right.

[Physical exponent] For $0<\delta <({\xi }_b-{\xi }_a)\mathrm{/}2$, the closed interval $[{\xi }_a+\delta ,{\xi }_b-\delta ]$ lies strictly inside the forbidden region, so $W-\lambda >0$ there and [F3] applies. Both square-root integrands extend continuously to the turning endpoints with value zero. Letting $\delta \downarrow 0$, the omitted endpoint integrals tend to zero because their integrands are continuous and bounded on compact intervals. Thus $\frac{S}{h}=\frac{1}{h}{\int }_{{\xi }_a}^{{\xi }_b}\sqrt{W-\lambda }d\xi =\frac{1}{\mathrm{\hslash }}{\int }_{x_a}^{x_b}\sqrt{2m(E_{\star }W(x\mathrm{/}\mathrm{\ell })-E_{\star }\lambda )}dx=\frac{1}{\mathrm{\hslash }}{\int }_{x_a}^{x_b}\sqrt{2m(V(x)-E)}dx,$ with $x_a=\mathrm{\ell }{\xi }_a$, $x_b=\mathrm{\ell }{\xi }_b$, as in the Statement.

[Exact scattering amplitudes and current ratio] Put $p_{\mathrm{\infty }}=\sqrt{\lambda }$ and define $f_L(\xi )=p_{\mathrm{\infty }}^{-1\mathrm{/}2}{e}^{-i{p}_{\mathrm{\infty }}\xi \mathrm{/}h}$ for $\xi \le -L$ and $f_R(\xi )=p_{\mathrm{\infty }}^{-1\mathrm{/}2}{e}^{i{p}_{\mathrm{\infty }}\xi \mathrm{/}h}$ for $\xi \ge L$, extending each to $[-L,L]$ by its Cauchy data. The first-order system for $(u,u^{\mathrm{\prime }})$ has continuous bounded coefficients on this compact interval for each fixed $h>0$; its Picard iterates converge uniformly, with the $n$th iterate bounded by $M^n(2L{)}^n\mathrm{/}n!$, and the same estimate proves uniqueness. Hence $f_L,f_R$ are exact solutions. Because $W-\lambda$ is real, $J_h^{\mathrm{\prime }}=h\mathrm{Im}(\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2+(W-\lambda )\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{h}^2)=0$. The exterior waves have currents $J_h(f_R)=1$, $J_h(f_L)=-1$, and $W(f_L,\overline{f_L})=2i\mathrm{/}h$. Thus at the left exterior there are unique coefficients $A_{\mathrm{i}\mathrm{n}},A_{\mathrm{o}\mathrm{u}\mathrm{t}}$ with $f_R=A_{\mathrm{i}\mathrm{n}}\overline{f_L}+A_{\mathrm{o}\mathrm{u}\mathrm{t}}{f}_L$; direct expansion of the two plane-wave currents cancels their cross terms and gives $J_h(f_R)=\mathrm{\mid }{A}_{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^2-\mathrm{\mid }{A}_{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\mid }}^2=1$, so $A_{\mathrm{i}\mathrm{n}}\mathrm{\ne }0$. The model's unique unit-incoming solution is $f_R\mathrm{/}{A}_{\mathrm{i}\mathrm{n}}$, with $t=1\mathrm{/}{A}_{\mathrm{i}\mathrm{n}}$ and $r=A_{\mathrm{o}\mathrm{u}\mathrm{t}}\mathrm{/}{A}_{\mathrm{i}\mathrm{n}}$, hence $T=\mathrm{\mid }{A}_{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^{-2}$. Also $W(f_L,f_R)=A_{\mathrm{i}\mathrm{n}}W(f_L,\overline{f_L})$, so $A_{\mathrm{i}\mathrm{n}}=hW(f_L,f_R)\mathrm{/}(2i)$. This derives the exact normalization and the current-ratio formula used in clause (d) of the WKB model.

[Exterior-to-turning-point basis conversion] Let $p(\xi )=\sqrt{\lambda -W(\xi )}$ on the allowed sides, ${\sigma }_{-}=-1$, ${\sigma }_{+}=1$, and define the outward phases [F1, algebra] ${\mathrm{\Theta }}_{-}(\xi )=\frac{1}{h}{\int }_{\xi }^{{\xi }_a}p(s)ds+\frac{\pi }{4},{\mathrm{\Theta }}_{+}(\xi )=\frac{1}{h}{\int }_{{\xi }_b}^{\xi }p(s)ds+\frac{\pi }{4},$ so ${\mathrm{\Theta }}_j^{\mathrm{\prime }}={\sigma }_{j}p\mathrm{/}h$. Divide the exact local decaying and growing solutions from [F1] by their nonzero forbidden-side constants and call them $d_j,g_j$. At ${\xi }_j\in \{{\xi }_{-},{\xi }_{+}\}$, put $Y_j(u)=(p_{\mathrm{\infty }}^{1\mathrm{/}2}u({\xi }_j),h{p}_{\mathrm{\infty }}^{-1\mathrm{/}2}{u}^{\mathrm{\prime }}({\xi }_j){)}^{\mathsf{T}}$. The fixed-section value and derivative errors in [F1] are $O(h)$: there ${\zeta }_j({\xi }_j)\mathrm{\ne }0$ is fixed, ${\epsilon }_h\le Ch$, and $h\mathrm{/}\mathrm{\mid }{\zeta }_j({\xi }_j){\mathrm{\mid }}^{3\mathrm{/}2}=O(h)$. The Airy derivative asymptotics and chain rule $u^{\mathrm{\prime }}=a^{\mathrm{\prime }}w+a{\zeta }^{\mathrm{\prime }}{w}^{\mathrm{\prime }}$ give for the normalized allowed-side forms $h{p}^{-1\mathrm{/}2}{g}_j^{\mathrm{\prime }}=-{\sigma }_j\sin {\mathrm{\Theta }}_j+O(h),h{p}^{-1\mathrm{/}2}{d}_j^{\mathrm{\prime }}=2{\sigma }_j\cos {\mathrm{\Theta }}_j+O(h),$ because the amplitude derivative contributes only $h{p}^{-1\mathrm{/}2}(p^{-1\mathrm{/}2}{)}^{\mathrm{\prime }}=O(h)$ at a fixed section. Hence, after the displayed scaling of $u^{\mathrm{\prime }}$, $Y_j(g_j)=\left(\genfrac{}{}{0px}{}{\cos {\mathrm{\Theta }}_j}{-{\sigma }_j\sin {\mathrm{\Theta }}_j}\right)+O(h),Y_j(d_j)=2\left(\genfrac{}{}{0px}{}{\sin {\mathrm{\Theta }}_j}{{\sigma }_j\cos {\mathrm{\Theta }}_j}\right)+O(h)\mathrm{.}$ The exact exterior outgoing plane wave satisfies $Y_j(f_j)={\gamma }_j(e^{i{\mathrm{\Theta }}_j},i{\sigma }_j{e}^{i{\mathrm{\Theta }}_j}{)}^{\mathsf{T}}$ for some $\mathrm{\mid }{\gamma }_j\mathrm{\mid }=1$, since it has the same constant amplitude and phase derivative there. The leading basis matrix and its determinant are $M_j^0=\left(\begin{array}{cc}\cos {\mathrm{\Theta }}_j& 2\sin {\mathrm{\Theta }}_j\\ -{\sigma }_j\sin {\mathrm{\Theta }}_j& 2{\sigma }_j\cos {\mathrm{\Theta }}_j\end{array}\right),\det M_j^0=2{\sigma }_j,$ and $M_j^0(1,i\mathrm{/}2{)}^{\mathsf{T}}=(e^{i{\mathrm{\Theta }}_j},i{\sigma }_j{e}^{i{\mathrm{\Theta }}_j}{)}^{\mathsf{T}}$. Since the entries of $(M_j^0{)}^{-1}$ are uniformly bounded and the exact matrix is $M_j=M_j^0+O(h)$, for all sufficiently small $h$ it is invertible with uniformly bounded inverse; solving its exact two-component system yields coefficients $({\alpha }_j,{\beta }_j)={\gamma }_j(1,i\mathrm{/}2{)}^{\mathsf{T}}+O(h)$. Equality of Cauchy data and ODE uniqueness then give $f_L={\gamma }_L[(1+O(h))g_a+(i\mathrm{/}2+O(h))d_a],f_R={\gamma }_R[(1+O(h))g_b+(i\mathrm{/}2+O(h))d_b]\mathrm{.}$ This derives the incoming/outgoing coefficient formulas from the explicit matrices.

[Interior Wronskian] Put $\rho (\xi )=W(\xi )-\lambda >0$ on the barrier and [step 1.4, F1, F2, algebra] $A_a=\frac{1}{h}{\int }_{{\xi }_a}^{{\xi }_0}\sqrt{\rho }d\xi ,A_b=\frac{1}{h}{\int }_{{\xi }_0}^{{\xi }_b}\sqrt{\rho }d\xi ,A=A_a+A_b=S\mathrm{/}h\mathrm{.}$ Both $A_a,A_b$ are positive. At ${\xi }_0$ use the common state $Y_0(u)=(\rho ({\xi }_0{)}^{1\mathrm{/}4}u({\xi }_0),h\rho ({\xi }_0{)}^{-1\mathrm{/}4}{u}^{\mathrm{\prime }}({\xi }_0){)}^{\mathsf{T}}$. The normalized forbidden-side value and slope forms of [F1] give $Y_0(g_a)=e^{A_a}[(1,1{)}^{\mathsf{T}}+O(h)],Y_0(d_a)=e^{-A_a}[(1,-1{)}^{\mathsf{T}}+O(h)],$ $Y_0(g_b)=e^{A_b}[(1,-1{)}^{\mathsf{T}}+O(h)],Y_0(d_b)=e^{-A_b}[(1,1{)}^{\mathsf{T}}+O(h)]\mathrm{.}$ Indeed $A_a^{\mathrm{\prime }}=\sqrt{\rho }\mathrm{/}h$ and $A_b^{\mathrm{\prime }}=-\sqrt{\rho }\mathrm{/}h$ at the fixed point. Differentiating only the explicit leading forbidden envelopes gives $h{\rho }^{-1\mathrm{/}4}({\rho }^{-1\mathrm{/}4}{e}^{\pm A_a}{)}^{\mathrm{\prime }}=e^{\pm A_a}(\pm 1-\frac{h{\rho }^{\mathrm{\prime }}}{4{\rho }^{3\mathrm{/}2}}),h{\rho }^{-1\mathrm{/}4}({\rho }^{-1\mathrm{/}4}{e}^{\pm A_b}{)}^{\mathrm{\prime }}=e^{\pm A_b}(\mp 1-\frac{h{\rho }^{\mathrm{\prime }}}{4{\rho }^{3\mathrm{/}2}})\mathrm{.}$ The last terms are $O(h)$ because $\rho ({\xi }_0)>0$ and ${\rho }^{\mathrm{\prime }}$ is fixed there; the derivative error from [F1] has the same scaled order. Thus the value and slope vector errors above are indeed $O(h)$ without differentiating an unspecified error term. Since $\det (Y_0(u),Y_0(v))=hW(u,v)$, expand it using the exact identities of step 1.4. The $g_a{g}_b$ determinant is $-2e^A(1+O(h))$. The leading vectors in each mixed determinant coincide, so those terms are $O(h)e^{A_a-A_b}$ and $O(h)e^{-A_a+A_b}$; relative to $e^A$ these are at most $O(h)e^{-2A_b}$ and $O(h)e^{-2A_a}$. The $d_a{d}_b$ term is $O(e^{-A})$. After including the coefficients from step 1.4 and the phases $\mathrm{\mid }{\gamma }_L{\gamma }_R\mathrm{\mid }=1$, this yields $hW(f_L,f_R)=-2{\gamma }_L{\gamma }_R{e}^A[1+O(h)+O(e^{-2A})]\mathrm{.}$ All constants depend only on the fixed barrier data and sections; no exponentially ill-conditioned matrix is inverted.

[Quantitative transmission error] Write the relative error in step 2.1 as ${\delta }_h=O(h+e^{-2A})$; then $\mathrm{\mid }{A}_{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^2{e}^{-2A}=\mathrm{\mid }1+{\delta }_h{\mathrm{\mid }}^2=1+{\eta }_h$ with ${\eta }_h\in \mathbb{R}$ and $\mathrm{\mid }{\eta }_h\mathrm{\mid }\le C_0[h+e^{-2A}]$ after enlarging a fixed constant. Because $T=\mathrm{\mid }{A}_{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^{-2}$, $T{e}^{2A}=(1+{\eta }_h{)}^{-1}$, so whenever $1-C_0[h+e^{-2A}]>0$, [step 1.3, step 2.1, algebra] $\mathrm{\mid }T{e}^{2A}-1\mathrm{\mid }\le \frac{C_0[h+e^{-2A}]}{1-C_0[h+e^{-2A}]}\mathrm{.}$ Choose $\epsilon (h)=C_{1}h$ with $C_1$ large enough to dominate the fixed matching errors in steps 1.4-2.1; after increasing $C_0$ as needed, this is the quantitative bound of the Statement with $A=S\mathrm{/}h$. It implies $T=e^{-2S\mathrm{/}h}(1+O(\epsilon (h)+e^{-2S\mathrm{/}h}))$ and ${\lim }_{h\downarrow 0}T{e}^{2S\mathrm{/}h}=1$ for the fixed barrier.

[Scope] The proof uses only the declared smooth compactly supported profile, its two separated simple turning points, the local Airy connections and the exact boundary-defined scattering solution. It uses no rectangular wall, coalescing or multiple turn, barrier-top uniformity, resonant multiwell, wave packet, detector, apparatus or sampling assumption. Thus the claimed exclusions and scope hold.