Here is another solution. This is a wave with twice as much momentum, as you can see by the shorter wavelength. That means it has four times as much kinetic energy. The higher energy is reflected in the fact that the wave oscillates more rapidly in time.
If the potential energy in the x>0 region is more than the particle's energy, then the particle is forbidden to be in that region. Classically, we would expect a particle approaching that region from the left to bounce off the region and be reflected back.
When we solve the Schroedinger equation for this particle, we find that the solution comes in three parts:
Here are the three pieces, seen separately:
And here they are all added together:
The part of the wavefunction that extends into the forbidden zone may not come as too much of a surprise: it's really just like the finite potential well we examined earlier.
In the previous example, we assumed that the high-potential region extended forever (or at least for a very long way). Now suppose that there's just a short region that's "forbidden", like this:
Since the wavefunction persists some distance into the forbidden region, it has some chance of making it all the way to the other side. Once it does, it's not "forbidden" anymore. At that point, it just looks like a traveling wave again. The solution to the Schroedinger equation for this scenario looks like this:
Note: the part on the left is a superposition of two traveling waves, one going left and one going right, just as before, only this time the amplitude of the reflected part is less than the amplitude of the incoming part, because some of the wave gets through.
These wavefunctions are not normalizable -- that is, they extend forever in the positive and negative x directions, instead of dying away to zero. A wavefunction for a single particle can't behave this way. One way to understand this wavefunction is to imagine that it is associated with a beam of particles rather than a single particle. Strictly speaking, in that case, you have to imagine that the beam has been going on forever, with particles streaming in from infinitely great distance. But as long as you're willing to imagine that the beam has been around for a long time, with particles coming in from a great distance, then this wavefunction will work well as an approximation. In this case the absolute square of the wavefunction is proportional to the number of particles found at a given place and time, whereas for a single particle it'd be the probability of finding the particle there.
If you want to model a single particle, your wave function really should look like a wave packet. Here's an animation of a wave packet striking a potential barrier and partially tunneling through:
(The animation here is on a continuous loop, but it's really just one wave packet coming through.)
It's worthwhile to think a bit about how this wave packet solution is related to the infinitely-extending, non-normalizable particle-beam solutions we looked at earlier. The key is the notion of the Fourier transform, which says that any nice function can be expressed as a linear combination of a bunch of sine and cosine functions. This means that the wave packet is just the sum of a bunch of the mathematically simpler non-normalizable guys.
One way to make this seem plausible is to take the wave packet animation above and zoom in on the region right near the boundary:
During the time that the peak of the wave packet is passing through (for a few seconds right near the middle of this animation), the solution looks a lot like one of the non-normalizable solutions we saw above.
(By the way, the last two animations aren't synchronized with each other. The second one runs slower than the first.)