Two-turn barrier tunnelling error

[Exact exterior solutions and flux] Define $f_L(x)=p_0^{-1\mathrm{/}2}{e}^{-i{p}_{0}x\mathrm{/}h}$ on $x\le -L$ and $f_R(x)=p_0^{-1\mathrm{/}2}{e}^{i{p}_{0}x\mathrm{/}h}$ on $x\ge L$, and extend each by its Cauchy data through the compact interaction interval. Existence and uniqueness follow directly from the first-order integral equation for $(u,u^{\mathrm{\prime }})$: its continuous coefficient matrix is bounded on each compact interval, and the ordered Picard iterates are bounded by $M^n{\mathrm{\ell }}^n\mathrm{/}n!$; their uniformly convergent sum solves the equation, and the same iterates applied to a difference give uniqueness. Thus no interval Dirichlet spectral/domain supplier is required. Since $q$ is real, $J_h^{\mathrm{\prime }}=h\mathrm{Im}(\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2+q\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{h}^2)=0$. On the left write uniquely $f_R=A_h^{\mathrm{i}\mathrm{n}}\overline{f_L}+A_h^{\mathrm{o}\mathrm{u}\mathrm{t}}{f}_L$. The currents are $J_h(f_R)=1$ and $J_h(f_L)=-1$, so $\mathrm{\mid }{A}_h^{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^2-\mathrm{\mid }{A}_h^{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\mid }}^2=1$, in particular $A_h^{\mathrm{i}\mathrm{n}}\mathrm{\ne }0$. Consequently $u_h=f_R\mathrm{/}{A}_h^{\mathrm{i}\mathrm{n}}$ has precisely the prescribed boundary form, with $t=1\mathrm{/}{A}_h^{\mathrm{i}\mathrm{n}}$ and $r=A_h^{\mathrm{o}\mathrm{u}\mathrm{t}}\mathrm{/}{A}_h^{\mathrm{i}\mathrm{n}}$, uniquely. With $W(u,v):=u{v}^{\mathrm{\prime }}-u^{\mathrm{\prime }}v$, $W(f_L,\overline{f_L})=2i\mathrm{/}h$ and therefore $A_h^{\mathrm{i}\mathrm{n}}=hW(f_L,f_R)\mathrm{/}(2i)$.

[A single error budget and outgoing coefficients] Choose fixed allowed matching sections on either side of each turn and a fixed interior point $x_0\in (a,b)$. Let $\epsilon (h)$ be the sum of the two Volterra errors $e^{{\epsilon }_{h,l}}-1$, the allowed-region coefficient errors $e^{h\int \mathrm{\mid }{q}_{\mathrm{W}\mathrm{K}\mathrm{B}}\mathrm{\mid }\mathrm{/}p}-1$ on the fixed intervals connecting these sections to $\pm L$, and the Airy matching remainders $h\mathrm{/}\mathrm{\mid }{\zeta }_l{\mathrm{\mid }}^{3\mathrm{/}2}$ at the sections and $x_0$. All integrals and matching locations are fixed, hence $\epsilon (h)\le Ch$ for small $h$. In the allowed region use the scaled state $(p^{1\mathrm{/}2}u,h{p}^{-1\mathrm{/}2}{u}^{\mathrm{\prime }})$. The leading sine/cosine states of [F2] form an invertible two-by-two matrix with inverse bounded independently of $h$, since their phases enter only through sine and cosine; the amplitude-derivative terms are $O(h)$. [F1] transports each exterior outgoing wave to this section with coefficient error $O(\epsilon )$. Since $e^{i{\mathrm{\Theta }}_l}=\cos {\mathrm{\Theta }}_l+i\sin {\mathrm{\Theta }}_l$, solving the perturbed bounded matrix gives, for constants ${\gamma }_L,{\gamma }_R$ of modulus one, $f_L={\gamma }_L[(1+O(\epsilon ))g_a+(i\mathrm{/}2+O(\epsilon ))d_a],f_R={\gamma }_R[(1+O(\epsilon ))g_b+(i\mathrm{/}2+O(\epsilon ))d_b]\mathrm{.}$ Both outward phase conventions have positive ${\mathrm{\Theta }}_l$ derivative toward the corresponding exterior, so $e^{i{\mathrm{\Theta }}_l}$ denotes the outgoing wave on each side. Inversion is justified for small $h$ by the finite-matrix Neumann series; no exponentially weighted matrix is inverted here.

[The interior Wronskian] Write $S_a=S_a(x_0)$, $S_b=S_b(x_0)$, so $S_a+S_b=S=B\mathrm{/}h$. In the common state coordinates $Y(u):=(q(x_0{)}^{1\mathrm{/}4}u(x_0),hq(x_0{)}^{-1\mathrm{/}4}{u}^{\mathrm{\prime }}(x_0){)}^{\mathsf{T}}$, [F2]-[F3] give $Y(g_a)=e^{S_a}[(1,1{)}^{\mathsf{T}}+O(\epsilon )],Y(d_a)=e^{-S_a}[(1,-1{)}^{\mathsf{T}}+O(\epsilon )],$ $Y(g_b)=e^{S_b}[(1,-1{)}^{\mathsf{T}}+O(\epsilon )],Y(d_b)=e^{-S_b}[(1,1{)}^{\mathsf{T}}+O(\epsilon )]\mathrm{.}$ The $O(\epsilon )$ terms include the fixed-section amplitude-derivative errors $O(h)$. Since $\det (Y(u),Y(v))=hW(u,v)$, the growing/growing term in step 1.2 is $-2{\gamma }_L{\gamma }_R{e}^S(1+O(\epsilon ))$. Each mixed determinant has zero leading determinant (its two leading vectors coincide), hence is $O(\epsilon )e^{S_a-S_b}$ or $O(\epsilon )e^{-S_a+S_b}$. Relative to $e^S$ these are bounded by $O(\epsilon )e^{-2S_b}$ and $O(\epsilon )e^{-2S_a}$, respectively. The decaying/decaying term is $O(e^{-S})$. Thus $hW(f_L,f_R)=-2{\gamma }_L{\gamma }_R{e}^S[1+O(\epsilon )+O(e^{-2S})]\mathrm{.}$ This calculation controls the dominant determinant directly and avoids unstable inversion of exponentially separated columns.

[Quantitative transmission error] By steps 1.1 and 2.1 and $\mathrm{\mid }{\gamma }_L{\gamma }_R\mathrm{\mid }=1$, $\mathrm{\mid }{A}_h^{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^2{e}^{-2S}=1+{\eta }_h$, where ${\eta }_h$ is real and $\mathrm{\mid }{\eta }_h\mathrm{\mid }\le C_0[\epsilon (h)+e^{-2S}]$ after enlarging a fixed $C_0$ to absorb the square of the amplitude error for sufficiently small $h$. Since $T=\mathrm{\mid }{A}_h^{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^{-2}$, one has $T{e}^{2S}=(1+{\eta }_h{)}^{-1}$ and consequently $\mathrm{\mid }T{e}^{2S}-1\mathrm{\mid }\le \frac{C_0[\epsilon (h)+e^{-2S}]}{1-C_0[\epsilon (h)+e^{-2S}]}$ whenever its denominator is positive. This is the claimed quantitative bound and implies $T=e^{-2B\mathrm{/}h}[1+O(\epsilon (h)+e^{-2B\mathrm{/}h})]$. Constants depend only on $m,V,E$ and the fixed sections, and the excluded cases do not satisfy the hypotheses used.