When you are working with potential barrier of height V0 and width a where E > V0, this means that the particle has enough energy to pass through the potential barrier and end up in the x > a region. This is what the Schrödinger equation looks like in this case:
![image0.png](https://www.dummies.com/wp-content/uploads/396207.image0.png)
The solutions for
![image1.png](https://www.dummies.com/wp-content/uploads/396208.image1.png)
are the following:
![image2.png](https://www.dummies.com/wp-content/uploads/396209.image2.png)
In fact, because there's no leftward traveling wave in the x > a region,
![image3.png](https://www.dummies.com/wp-content/uploads/396210.image3.png)
So how do you determine A, B, C, D, and F? You use the continuity conditions, which work out here to be the following:
![image4.png](https://www.dummies.com/wp-content/uploads/396211.image4.png)
Okay, from these equations, you get the following:
A + B = C + D
ik1(A – B) = ik2(C – D)
Ceik2a + De–ik2a = Feik1a
ik2Ceik2a – ik2De–ik2a = ik1Feik1a
So putting all of these equations together, you get this for the coefficient F in terms of A:
![image5.png](https://www.dummies.com/wp-content/uploads/396212.image5.png)
Wow. So what's the transmission coefficient, T? Well, T is
![image6.png](https://www.dummies.com/wp-content/uploads/396213.image6.png)
And this works out to be
![image7.png](https://www.dummies.com/wp-content/uploads/396214.image7.png)
Whew! Note that as k1 goes to k2, T goes to 1, which is what you'd expect.
So how about R, the reflection coefficient? Without going into the algebra, here's what R equals:
![image8.png](https://www.dummies.com/wp-content/uploads/396215.image8.png)
You can see what the E > V0 probability density,
![image9.png](https://www.dummies.com/wp-content/uploads/396216.image9.png)
looks like for the potential barrier in the figure.
![image10.jpg](https://www.dummies.com/wp-content/uploads/396217.image10.jpg)
That completes the potential barrier when E > V0.