Well potential, barrier potential and Schrodinger equation The following potential is given:
$ V\left(x\right)=\begin{cases}
V_{0} & x<-a\\
0 & -a<x<0\\
\infty & 0<x
\end{cases} $
I have to find the general solution for Schrodinger equation for this potential with energy $ E>V_{0} $, and then I have to use the boundary conditions in order to find the probability of particles being reflected (for particles that comes from the direction $ - \infty $
Now I have tried some things but it dosent seem to add up. Here's what I've done:
First of all the solution to Schrodinger equations are:
$ \psi=0 $ $ \thinspace\thinspace\ $
for$ \thinspace\thinspace\ $ $ x>0 $
For $ -a<x<0 $
The solution is $ \psi_{1}\left(x\right)=Ae^{-ik_{1}x}+Be^{-ik_{1}x}\thinspace\thinspace\thinspace\thinspace $ where $ \thinspace\thinspace\thinspace k_{1}=\frac{\sqrt{2mE}}{\overline{h}} $
And for $ x<-a $
The solution is $ \psi_{2}\left(x\right)=Ce^{ik_{2}x}+De^{-ik_{2}x}\thinspace\thinspace\thinspace\thinspace $ where $ \thinspace\thinspace\thinspace k_{2}=\frac{\sqrt{2m\left(E-V_{0}\right)}}{\overline{h}} $
(The solution for each part composed of a wave moving to the right and a wave moving to the left).
In order to find the probability $ R $ of particles being reflected I have to find $ R=\frac{|\psi_{reflected}|^{2}}{|\psi_{incident}|^{2}} $
Which will be given by the correspond amplitude ratio.
Im not sure if my solutions are correct or maybe Im missing more details, because it gets really complicated - I tried to find an expression for the amplitudes $ A,B,D $ using $ C $ as parameter.
Also for particles that comes from  $ -\infty $ I have 2 reflected waves, First at the $ V_0 $ barrier we'll get one transmitted wave (Thats the wave with the amplitude D in the solutions I wrote  and a reflected wave (Thats the wave with the amplitude C). and then after the $ V_0 $ barrier in the interval $ -a<x<a $ we have the wave that was transmitted and another wave that will be reflected because of the infinite well wall.
I'll be glad for any hints or guidance.
Thanks in advance
 A: Let us label the wavefunction solutions for the two regions from left to right:
$$\psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x}\quad \text{for} \quad x<-a$$
and
$$\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x}\quad \text{for} \quad -a<x<0,$$
where $k_1=\sqrt{2m(E-V_0)}/\hbar\,$ and $k_2=\sqrt{2mE}/\hbar$.
Simply imposing that at the infinite barrier $\psi_2$ must vanish leads to
$$0=\psi_2(0) = C + D$$
so that $D=-C$.
You can picture a particle being shot from the far left and the question is how much (amplitude-wise) do you get back. This means we want to compute the ratio $\left|\frac{B}{A}\right|^2$. We have only two boundary conditions left, continuity at $x=-a$ and continuity of the first derivative at $x=-a$ which is enough to determine the ratio above but not the individual coefficients (for that we would need to impose normalization, but this is not possible for the free particle in the region $x<-a$).
From continuity, $\psi_1(-a)=\psi_2(-a)$ we have
$$Ae^{-ik_1a}+Be^{ik_1a}=C(e^{-ik_2a}-e^{ik_2a}).\tag{1}$$
From the continuity of the derivative $\psi'_1(-a)=\psi'_2(-a)$:
$$k_1Ae^{-ik_1a}-k_1Be^{ik_1a}=k_2C(e^{-ik_2a}+e^{ik_2a}).\tag{2}$$
You can take the ratio of equations (1) and (2) to arrive
$$\left|\frac{B}{A}\right|^2 = \frac{1}{\sin^2(k_2 a)+\frac{k_2^2}{k_1^2}\cos^2(k_2 a)}$$
A: First, let's write the wave function
$$ \psi(x) = \begin{cases} 
      A \ e^{ik_1x} +B \ e^{-ik_1x} & x\leq -a \\
      C\ e^{ik_2 x}+De^{-ik_2x}& -a\leq x\leq 0 \\
      0 & 0< x 
   \end{cases}
$$
where $$k_1=\frac{\sqrt{2m(E-V_0)}}{\hbar} \ \ \ \mathrm{and }\ \ \ \  \ k_2=\frac{\sqrt{2mE}}{\hbar}$$
There are $4$ constant that can be determine from the boudrary codition :
$$\begin{align} \psi(x)\biggr\rvert_{x=-a-0} &=  \psi(x)\biggr\rvert_{x=-a+0} \\
 \psi(x)\biggr\rvert_{x=-0}=0 \\
\psi'(x)\biggr\rvert_{x=-a-0} &=  \psi'(x)\biggr\rvert_{x=-a+0} \\
\end{align}$$
Use the following condition to determine the ratio's of amplitude.
