Rectangular barrier: transmission coefficients and the thick-barrier error

[Shape of bounded solutions] On either exterior interval the equation is ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=-k^2\psi$. At any point $x_0$, the matrix sending $(A,B)$ to the value and derivative of $A\cos (kx)+B\sin (kx)$ has determinant $k\mathrm{\ne }0$, so choose $A,B$ to match any solution's data there. The difference $w$ then solves $w^{\mathrm{\prime }\mathrm{\prime }}=-k^{2}w$ with $w(x_0)=w^{\mathrm{\prime }}(x_0)=0$; differentiating $\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^2+k^2\mathrm{\mid }w{\mathrm{\mid }}^2$ and using [F1] shows it is constant, hence zero, so every solution is $A\cos (kx)+B\sin (kx)$ and is bounded on each half-line. On $(0,a)$ let $v$ solve $v^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^{2}v$ and put $g:=v^{\mathrm{\prime }}-\kappa v$. Then $g^{\mathrm{\prime }}=-\kappa g$, so $(e^{\kappa x}g{)}^{\mathrm{\prime }}=0$ and $g=C{e}^{-\kappa x}$. Solving $(e^{-\kappa x}v{)}^{\mathrm{\prime }}=C{e}^{-2\kappa x}$ gives $v=A{e}^{\kappa x}+B{e}^{-\kappa x}$. Thus each interval has a two-dimensional solution space; a function satisfying the regularity in the statement is described by two constants on each interval, with no boundedness restriction on the exterior coefficients.

[Matching and the transfer matrix] Write ${\psi }_{<},{\psi }_0,{\psi }_{>}$ for the three restrictions. The conditions that $\psi$ be $C^1$ at $0$ and $a$ are the four linear equations ${\psi }_{<}(0)={\psi }_0(0),{\psi }_{<}^{\mathrm{\prime }}(0)={\psi }_0^{\mathrm{\prime }}(0),{\psi }_0(a)={\psi }_{>}(a),{\psi }_0^{\mathrm{\prime }}(a)={\psi }_{>}^{\mathrm{\prime }}(a),$ and the transmission asymptotics of (a) impose the incoming coefficient $1$ and the incoming-from-right coefficient $0$. By [F4], their boundary data are $(1+r,ik(1-r))$ at $0$ and $(t{e}^{ika},ikt{e}^{ika})$ at $a$. The map carrying the interior data $({\psi }_0(0),{\psi }_0^{\mathrm{\prime }}(0))$ to $({\psi }_0(a),{\psi }_0^{\mathrm{\prime }}(a))$ is the matrix $M:=\left(\begin{array}{cc}\cosh (\kappa a)& {\kappa }^{-1}\sinh (\kappa a)\\ \kappa \sinh (\kappa a)& \cosh (\kappa a)\end{array}\right)$: its columns are the values at $a$ of the two solutions $x\mapsto \cosh (\kappa x)$ and $x\mapsto {\kappa }^{-1}\sinh (\kappa x)$ of ${\psi }^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^2\psi$ with data $(1,0)$ and $(0,1)$ at $0$, and by [F1] every solution of the interior equation is a unique linear combination of these two. Hence the matching equations at $0$ and $a$ are equivalent to the single vector equation $\left(\begin{array}{c}t{e}^{ika}\\ ikt{e}^{ika}\end{array}\right)=M\left(\begin{array}{c}1+r\\ ik(1-r)\end{array}\right)\mathrm{.}$ Multiplying by $M^{-1}=\left(\begin{array}{cc}\cosh (\kappa a)& -{\kappa }^{-1}\sinh (\kappa a)\\ -\kappa \sinh (\kappa a)& \cosh (\kappa a)\end{array}\right)$, which is a two-sided inverse because $\det M={\cosh }^2(\kappa a)-{\sinh }^2(\kappa a)=1\mathrm{\ne }0$ by [F2], and reading off the two components gives $r=C_{+}t-1$ and $r=1+Bt$, where $C_{+}:=(\cosh (\kappa a)-i\frac{k}{\kappa }\sinh (\kappa a))e^{ika}$ and $B:=-(\cosh (\kappa a)+i\frac{\kappa }{k}\sinh (\kappa a))e^{ika}$. Since $C_{+}-B=[2\cosh (\kappa a)+i(\frac{\kappa }{k}-\frac{k}{\kappa })\sinh (\kappa a)]e^{ika}=D{e}^{ika}$ and $\mathrm{\mid }D\mathrm{\mid }\ge \mathrm{\mid }\mathrm{Re}D\mathrm{\mid }=2\cosh (\kappa a)\ge 2>0$, these two affine relations determine $(r,t)$ uniquely: $t=2\mathrm{/}(C_{+}-B)=2e^{-ika}\mathrm{/}D$ and $r=C_{+}t-1$. Consequently a transmission solution has at most one pair $(r,t)$ and hence at most one $\psi$, since $(r,t)$ determines the interior data $(1+r,ik(1-r))$ and the two exterior pieces.

[Existence of the transmission solution] Conversely, let $r:=C_{+}t-1$ and $t:=2\mathrm{/}(C_{+}-B)=2e^{-ika}\mathrm{/}D$ be the unique pair of step 2.1, let ${\psi }_0$ be the solution of ${\psi }^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^2\psi$ on $(0,a)$ with $({\psi }_0(0),{\psi }_0^{\mathrm{\prime }}(0))=(1+r,ik(1-r))$ given explicitly by [F1], and define $\psi (x):=e^{ikx}+r{e}^{-ikx}$ for $x\le 0$, $\psi :={\psi }_0$ on $(0,a)$ and $\psi (x):=t{e}^{ikx}$ for $x\ge a$. The relations $r=C_{+}t-1$ and $r=1+Bt$ are exactly the two components of the vector equation of step 2.1, so $({\psi }_0(a),{\psi }_0^{\mathrm{\prime }}(a))=(t{e}^{ika},ikt{e}^{ika})$; hence the three pieces match in value and derivative at $0$ and at $a$, the glued function is $C^1$ on $\mathbb{R}$ and $C^{\mathrm{\infty }}$ on each of the three open intervals with $-\frac{{\mathrm{\hslash }}^2}{2m}{\psi }^{\mathrm{\prime }\mathrm{\prime }}+V\psi =E\psi$ by [F1, F4], and it is bounded, with exactly the prescribed transmission asymptotics. Together with the uniqueness of step 2.1 this proves (a) and gives the first display of (b), $t=2e^{-ika}\mathrm{/}D$.

[Transmission coefficient] Since [F5] gives $\mathrm{\mid }{e}^{-ika}\mathrm{\mid }=1$, $T=\mathrm{\mid }t{\mathrm{\mid }}^2=4\mathrm{/}\mathrm{\mid }D{\mathrm{\mid }}^2$, and by [F2] $\mathrm{\mid }D{\mathrm{\mid }}^2=4{\cosh }^2(\kappa a)+(\frac{\kappa }{k}-\frac{k}{\kappa }{)}^2{\sinh }^2(\kappa a)=4+[4+(\frac{\kappa }{k}-\frac{k}{\kappa }{)}^2]{\sinh }^2(\kappa a)\mathrm{.}$ Now $(\frac{\kappa }{k}-\frac{k}{\kappa }{)}^2=\frac{({\kappa }^2-k^2{)}^2}{k^2{\kappa }^2}=\frac{(2m(U-2E)\mathrm{/}{\mathrm{\hslash }}^2{)}^2}{k^2{\kappa }^2}$ because ${\kappa }^2-k^2=\frac{2m}{{\mathrm{\hslash }}^2}((U-E)-E)$, and $k^2{\kappa }^2=(2mE)(2m(U-E))\mathrm{/}{\mathrm{\hslash }}^4$, so $4+(\frac{\kappa }{k}-\frac{k}{\kappa }{)}^2=\frac{4E(U-E)+(U-2E{)}^2}{E(U-E)}=\frac{U^2}{E(U-E)}\mathrm{.}$ Therefore $\mathrm{\mid }D{\mathrm{\mid }}^2=4+\frac{U^2}{E(U-E)}{\sinh }^2(\kappa a)=4[1+\frac{U^2{\sinh }^2(\kappa a)}{4E(U-E)}]$ and $T=[1+\frac{U^2{\sinh }^2(\kappa a)}{4E(U-E)}{]}^{-1}$.

[Flux conservation] Put $J(x)=\mathrm{Im}(\overline{\psi (x)}{\psi }^{\mathrm{\prime }}(x))$. On each of the three intervals $J$ is constant by [F3], and by continuity of $\psi ,{\psi }^{\mathrm{\prime }}$ it is the same constant on all of $\mathbb{R}$. By [F5], $\overline{e^{ikx}}=e^{-ikx}$ and $\mathrm{\mid }{e}^{ikx}\mathrm{\mid }=1$. On $(-\mathrm{\infty },0)$, $\psi =e^{ikx}+r{e}^{-ikx}$ gives $J=\mathrm{Im}[ik-ikr{e}^{-2ikx}+ik\bar{r}{e}^{2ikx}-ik\mathrm{\mid }r{\mathrm{\mid }}^2]=k(1-\mathrm{\mid }r{\mathrm{\mid }}^2)$ (the two oscillating terms sum to a real number and contribute no imaginary part). On $(a,\mathrm{\infty })$, $\psi =t{e}^{ikx}$ gives $J=k\mathrm{\mid }t{\mathrm{\mid }}^2$. Hence $k(1-\mathrm{\mid }r{\mathrm{\mid }}^2)=k\mathrm{\mid }t{\mathrm{\mid }}^2$, i.e. $R+T=1$, which is the second display of (b).

[Thick-barrier form and controlled error] With $q=e^{-2\kappa a}$ one has ${\sinh }^2(\kappa a)=(e^{\kappa a}-e^{-\kappa a}{)}^2\mathrm{/}4=(q^{-1}-2+q)\mathrm{/}4$, so with $C=U^2\mathrm{/}(4E(U-E))$, $T=[1+\frac{C}{4}(q^{-1}-2+q){]}^{-1}=\frac{(4\mathrm{/}C)q}{1+(\frac{4}{C}-2)q+q^2},$ multiplying numerator and denominator by $4q\mathrm{/}C$. Since $16E(U-E)\mathrm{/}{U}^2=4\mathrm{/}C$, the exact difference is $T-\frac{4}{C}q=-\frac{4}{C}q\frac{\delta }{1+\delta },\delta =(\frac{4}{C}-2)q+q^2\mathrm{.}$ Put $\eta =\mathrm{\mid }\frac{4}{C}-2\mathrm{\mid }q+q^2$, so $\mathrm{\mid }\delta \mathrm{\mid }\le \eta$. If $\eta <1$, then $\mathrm{\mid }1+\delta \mathrm{\mid }\ge 1-\mathrm{\mid }\delta \mathrm{\mid }\ge 1-\eta >0$ and $\frac{\mathrm{\mid }\delta \mathrm{\mid }}{\mathrm{\mid }1+\delta \mathrm{\mid }}\le \frac{\eta }{1-\eta }$, which gives the first error bound in (c). If $\mathrm{\mid }\delta \mathrm{\mid }\le \frac{1}{2}$, then $\mathrm{\mid }1+\delta \mathrm{\mid }\ge \frac{1}{2}$ and $\mathrm{\mid }T-\frac{4}{C}q\mathrm{\mid }\le \frac{4}{C}q\cdot 2\mathrm{\mid }\delta \mathrm{\mid }$, which is the second error bound in (c) after substituting $4\mathrm{/}C=16E(U-E)\mathrm{/}{U}^2$.