# States of Asymmetrical Well

For the asymmetric quantum well in the figure please find the first two bound states and energies. prove that it is possible for an asymmetrical well to have no bound states.

The general Schrodinger equation is $-frac{hbar}{2m}*frac{d^2psi}{dx^2} +V(x)psi=Epsi$

where $V(x)=left{begin{matrix}infty&text{ in region 1}\0&text{ in region 2}\U_L&text{ in region 3}end{matrix}right.$

The solutions are $psi(x)=left{begin{matrix}0&text{ in region 1}\Asin(k_1x)&text{ in region 2}\Bexp[-k_2(x-L)],&text{ in region 3}end{matrix}right.$

having the wave vectors $k_1=frac{sqrt{2mE}}{hbar}$ and $k_2=frac{sqrt{2m(U_L-E)}}{hbar}$

The boundary conditions are at $x=L$, the wave function $psi$ and its derivative $dpsi/dx$ need to be continuous:

$Asin(k_1L)=B$ and

$k_1Acos(k_1L)=k_2B$.

We divide the two equations and obtain:

$k_1cot(k_1L)=k_2$ or

$cot(k_1L)=k_2/k_1$ or

$cot^2(k_1L)=k_2^2/k_1^2$

But $k_2^2=frac{2mU_L}{hbar^2}-frac{2mE}{hbar^2}=k_0^2-k_1^2$ so that

$cot^2(k_1L)=(k_0^2/k_1^2)-1$ or $frac{cos^2(…)+sin^2(…)}{sin^2(…)}=k_0^2/k_1^2$ or $|sin(k_1L)|=k_1/k_0$

This is a transcendental equation in $k_1$ ($k_0$ is fixed and depends only on $U_L$) that can be solved by graphing. From the graph one has $L=1.5 nm$ and $k_0=4.43*10^9 (1/m)$ so that the first two solutions are

$k_1=left{begin{matrix}1.81*10^9 (1/m)\2.49*10^9(1/m)end{matrix}right.$ and the bounded energies $E=left{begin{matrix}E=0.125 eV\E=0.237 eVend{matrix}right.$.

The sketches of the wave functions are also in the figure.The condition of not having bound states (see the graph) is that the slope of the line $x/k_0$ at the origin is bigger than the derivative of $sin(xL)$ at the origin. This way the line will not intersect the $sin(…)$ graph.

$frac{1}{k_0} >Lcos(xL)$ at $x=0$ or $k_0<frac{1}{L}$ which means $frac{sqrt{2mU_L}}{hbar} <frac{1}{L}$