# Asymetrical Well (Griffiths)

Solve the asymmetrical well in the figure

General Sch equation is

$-(hbar^2/2m)*(d^2 ψ)/(dx^2 )+V(x)ψ=Eψ$

With

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

The solutions are

$psi(x)=left{begin{matrix} 0 & text{in region 1}\ A*sin(k_1x) & text{in region 2}\ B*exp(-k_2(x-L))) & text{in region 3} end{matrix}right.$

With (deduced from Sch equation written on each region separately)

$k_1=sqrt{2mE/ℏ}$    and $k_2=sqrt{(2m(U_L-E))/ℏ}$
The boundary conditions are the at $x=L$ , the wavefunction $ψ$  and its derivative $dψ/dx$ need to be continuous.

$A*sin⁡ (k_1*L)=B$ and $k_1 A*cos⁡ (k_1 L)=k_2 B$

Dividing the two equations one gets:

$k_1*cotan (k_1*L)=k_2$   or $cotan (k_1 L)=k_2/k_1$

or $cotan^2 (k_1 L)=(k_2^2)/(k_1^2 )$

But

$k_2^2=(2mU_L)/ℏ^2 -2mE/ℏ^2 =k_0^2-k_1^2$

So that

$cotan^2 (k_1*L)=(k_0^2)/(k_1^2 )-1$  or

$cos^2⁡(…)+sin^2⁡(…)/sin^2⁡ (..) =(k_0^2)/(k_1^2)$     or $|sin⁡(k_1*L)|=k_1/k_0$

This is a transcendental equation in $k1$ ($k0$ is fixed and depends only on $UL$) that can be solved by graphing. The graph is at the bottom. With the given data one has $L=1.5*10^{-9} m$  and $k_0=4.432*10^9 (1/m)$ 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

$E=0.125 eV$ & $E=0.237 eV$

The condition of not having bound states (see the graph at the bottom) is that the slope of the line $x/k_0$   at the origin is bigger than the derivative of $sin⁡(x*L)$ at the origin. This way the line will not intersect the sine graph.

$1/k_0 >L*cos⁡(x*L)$    at $x=0$   or $k_0<1/L$    which means  $sqrt{2mU_L}/ℏ<1/L$