Vorlesung_28_20171204

$\rightarrow$ (iii)

\frac{1}{2} A\left(1+ \frac{k}{k'} \right)e^{+ik'L}+\frac{1}{2} B\left(1- \frac{k}{k'} \right)e^{+ik'L}+\frac{1}{2} A\left(1- \frac{k}{k'} \right)e^{-ik'L}+\frac{1}{2} B\left(1+ \frac{k}{k'} \right)e^{-ik'L}=Fe^{+ikL}\\ \left(\frac{1}{2} A+\frac{1}{2} B \right)\underbrace{\left( e^{+ik'L}+e^{-ik'L} \right)}_{2\cos(k'L)}+\left(\frac{1}{2}\frac{k}{k'}(A-B) \right)\underbrace{\left( e^{+ik'L}-e^{-ik'L} \right)}_{2i\sin(k'L)}=Fe^{+ikL}\\ \Rightarrow (A+B)\cos(k'L)+i\frac{k}{k'}(A-B) \sin(k'L)=Fe^{+ikL}\text{ (vii)}

$\frac{k'}{k}\left(Ce^{+ik'L}-De^{-ik'L}\right)=Fe^{+ikL}$

$\rightarrow$ (iv)

\frac{k'}{k}\frac{1}{2} A\left(1+ \frac{k}{k'} \right)e^{+ik'L}+ \frac{k'}{k}\frac{1}{2} B\left(1- \frac{k}{k'} \right)e^{+ik'L}- \frac{k'}{k}\frac{1}{2} A\left(1- \frac{k}{k'} \right)e^{-ik'L}- \frac{k'}{k}\frac{1}{2} B\left(1+ \frac{k}{k'} \right)e^{-ik'L}=Fe^{+ikL}\\\left( \frac{k'}{k}\frac{1}{2} A+ \frac{k'}{k}\frac{1}{2} B \right)\underbrace{\left( e^{+ik'L}-e^{-ik'L} \right)}_{2i\sin(k'L)}+\left(\frac{1}{2}A-\frac{1}{2}B \right)\underbrace{\left( e^{+ik'L}+e^{-ik'L} \right)}_{2\cos(k'L)}=Fe^{+ikL}\\ \frac{k'}{k}i(A+B)\sin(k'L)+(A-B){\cos(k'L)}=Fe^{+ikL}\text{ (viii)}

2B\cos(k'L)+ \left[ i\frac{k}{k'}(A-B)-i\frac{k'}{k}(A+B)\right]\sin(k'L)=0

$B$ auflösen

2B\cos(k'L)+ iA\underbrace{\left[ \frac{k}{k'}-\frac{k'}{k}\right]}_{\frac{k^2-k'^2}{k'k}}\sin(k'L)-iB\underbrace{\left[ \frac{k}{k'}+\frac{k'}{k}\right]}_{ \frac{k^2+k'^2}{k'k}}\sin(k'L)=0\\ B\cdot\underbrace{\left(2\cos(k'L) - i\frac{k^2+k'^2}{k'k}\sin(k'L)\right)}_Z=iA\frac{k' ^2-k^2}{k'k}\sin(k'L)\\ \text{mit }\frac{1}{Z}=\frac{Z^*}{ZZ^*}=\frac{Z^*}{\Re(Z)^2+\Im(Z)^2} \\

B=iA\frac{k'^2-k^2}{k'k}\sin(k'L)\cdot \frac{2\cos(k'L) + i\frac{k^2+k'^2}{k'k}\sin(k'L)}{4\cos^2(k'L)+\left(\frac{k^2+k'^2}{k'k}\right)^2\sin^2(k'L) }=A\frac{k'^2-k^2}{k'k}\sin(k'L)\cdot \frac{2i\cos(k'L) -\frac{k^2+k'^2}{k'k}\sin(k'L)}{4\cos^2(k'L)+\left(\frac{k^2+k'^2}{k'k}\right)^2\sin^2(k'L) }

B^*B=(\Re(B))^2 + (\Im(B))^2=A^*A\left( \frac{k^2-k'^2}{k'k}\right)^2 \sin^2(k'L)\frac{4 \cos^2(k'L) +\left( \frac{k^2+k'^2}{k'k}\right)^2\sin^2(k'L)}{[4 \underbrace{\cos^2(k'L)}_{1-\sin^2(k'L)} +\left( \frac{k^2+k'^2}{k'k}\right)^2\sin^2(k'L)]^2}\\ =\frac{A^*A\left( \frac{k^2-k'^2}{k'k}\right)^2 \sin^2(k'L)}{4-4\sin^2(k'L)\frac{k' ^2k^2}{k'^2k^2}+\sin^2(k'L)\frac{k^4+2k^2k'^2+k'^4}{k'^2k^2}}

Betrachte Nenner:

4+\sin^2(k'L)\frac{k^4-2k^2k'^2+k'^4}{k'^2k^2}=4+\frac{\left(k'^2-k^2\right)^2}{k'^2k^2}sin^2(k'L)\\ \Rightarrow R=\frac{B^*B}{A^*A}=\frac{\sin^2(k'L)}{4\frac{k'^2k^2}{(k'^2-k^2)^2}+\sin^2(k'L)}\text{, mit }k=\frac{\sqrt{2mE}}{\hbar}\text{ und }k'=\frac{\sqrt{2m(E-U_0)}}{\hbar}\\ \Rightarrow R=\frac{\sin^2\left(\frac{\sqrt{2m(E-U_0)}}{\hbar}L\right)}{\sin^2\left(\frac{\sqrt{2m(E-U_0)}}{\hbar}L\right)+4\frac{E}{U_0}\left(\frac{E}{U_0}-1 \right)}

$T=1-R$

Resonante Transmission

$\sin^2\left(\sqrt{\frac{2m(E-U_0)}{\hbar^2}}L\right)=0$

\Rightarrow\sqrt{\frac{2m(E-U_0)}{\hbar^2}}L=n\pi\Rightarrow E=U_0+\frac{n^2\pi^2\hbar^2}{2mL^2}

Analog zu Antireflexbeschichtung bei Licht

$E<U_0$ Tunneleffekt

$x=0$ reflektiert

$0$ $L$

Lösungen: $e^{+\alpha x}$ $e^{-\alpha x}\ \surd$ , weil Barriere endlich breit

\psi_{x<0}(x)=Ae^{+ikx}+Be^{-ikx}\\ \psi_{0<x<L}(x)= Ce^{+\alpha x}+De^{-\alpha x}\\ \psi_{x>L}(x)=Fe^{+ikx}

Stetigkeitsbedingung: $x= 0$

\psi_{x<0}(0)=\psi_{o<x<L}(0)\Rightarrow Ae^{+ik0}+Be^{-ik0}=Ce^{\alpha 0}+De^{-\alpha 0}\Rightarrow A+B=C+D

Stetigkeit der Ableitungen:

ikAe^{+ik0}-ikBe^{-ik0}=\alpha Ce^{\alpha 0}-\alpha De^{\alpha 0}\Rightarrow k(A-B)=\alpha (C-D)

$k'=-i\alpha$

mit vorheriger Lösung:

\Rightarrow R=\frac{\sin^2\left(-i\alpha L\right)}{\sin^2\left(-i\alpha L\right)+4\frac{-\alpha^2k^2}{\left(k^2+\alpha ^2\right)^2}}

i\sin(x)=\frac{e^{ix}-e^{-ix}}{2}\text{ mit }x=iz\Rightarrow i\sin(iz)=\frac{e^{-z}-e^{z}}{2} \\\Leftrightarrow \sin(iz)=i\frac{e^z-e^{-z}}{2}=i\sinh(z)\\ \Rightarrow R=\frac{-\sinh^2(\alpha L)}{-\sinh^2(\alpha L)+4\frac{-\alpha^2k^2}{\left(k^2+\alpha ^2\right)^2}}\\ \boxed{R=\frac{\sinh^2(\alpha L)}{\sinh^2(\alpha L)+4\frac{\alpha^2k^2}{\left(k^2+\alpha ^2\right)^2}} }

mit

\alpha=\frac{\sqrt{2m(U_0-E)}}{\hbar}\text{ und }k=\frac{\sqrt{2mE}}{\hbar}\\ \Rightarrow \boxed{R=\frac{\sinh^2\left(\frac{\sqrt{2m(U_0-E)}}{\hbar}L\right)}{\sinh^2\left(\frac{\sqrt{2m(U_0-E)}}{\hbar}L\right)+4\frac{E}{U_0}\left(1-\frac{E}{U_0}\right)}}\\ T=1-R

Effekt: analog zur verhinderten ('frustrated') Totalreflexion