1
$\begingroup$

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

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $$\left|\frac{B}{A}\right|^2 = \frac{1}{\sin^2(k_2 a)+\frac{k_2^2}{k_1^2}\cos^2(k_2 a)}$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ What is the ratio that Im looking for in order to find the reflected probability? Is it $D/A$ or $B/A$ $\endgroup$
    – FreeZe
    Nov 14, 2020 at 12:18
  • $\begingroup$ As there is no transmission through the infinite wall, so $C=D$ should be true. You are looking for $A/C$ and $A/B$. $\endgroup$ Nov 14, 2020 at 12:30
  • $\begingroup$ How come Im looking for 2 different ratios in order to find specific reflection probability? $\endgroup$
    – FreeZe
    Nov 14, 2020 at 12:37
  • $\begingroup$ Also if C=D we have 3 equations with 3 variables and we could easily find them, which is very unlikely dfor this sort of problems where we only able to find the ratios $\endgroup$
    – FreeZe
    Nov 14, 2020 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.