# Help Needed With Sketching Eigenfunctions

1. Oct 26, 2004

### Claire84

Hi there, I was hoping someone could maybe give me a hand with sketching eigenfunctions (although I reliase it can be difficult over the net!). I we have a particle at n=2 inside a finite potential well but we have a potential barrier in the middle of it, how do we draw this (the particle has energy lower than the sides of the potential well and lower than the potential barrier)? I have exponential decay into the sides of the well and I have exponential decay going through the potential barrier, but I'm unsure of how the wave itself should look. I know it will only have one node which is fine, but my problem is is there one node at each side of the step or just one for the entire way across the finite well?

I'm sorry I can't post up a picture or anything but I've just discovered that my scanner has mysteriously broken and heads are about to roll here. Not a happy bunny.

Anyway, any help would be much appreciated. Thank you.

2. Oct 26, 2004

### Pieter Kuiper

Last edited by a moderator: May 1, 2017
3. Oct 27, 2004

### Claire84

The thing that is confusing me now is about the amplitude of the eigenfunctions. Our lecturer says that the amplitude is proportional to the wavelength, but when we're looking at the potential step the amplitude seems to decrease as the wavelength increases which has me super-confused.

4. Oct 27, 2004

### Pieter Kuiper

As you can see from the particle-in-a-box, the amplitude of the wave function does not depend on the quantum number or wavelength.

5. Oct 27, 2004

### Claire84

Weird, he gave us that on Monday (mind you, it was 9 on Monday morning and most of us were still waking up). If I'd just thought that though one of my drawings I'd have def thought I had made a mistake but he said that amplitude was proportional to wavelength. Like if we have a potential barrier the wavelntgh before the barrier is shorter and the amplitude is smaller, then the amplitude and wavelength increase as we go over it. I think his reasoing was that we'd have a smaller velocity over the barrier so the probability would therefore increase of locating the particle.

6. Oct 27, 2004

### Claire84

Plus it was odd cos I asked him about the question I posed here and he drew something kinda different (he didn't seem to treat it like a double well). All he did was draw what we would have had for our normal n=2 and then rammed a barrier in the middle of it and made it decay exponentially. So it still kinda looks like somthing from the potential well.

7. Oct 27, 2004

### Claire84

Oh, and sorry, one other thing. In our notes we have for a particle in a box that the amplitude doesn't depend on wavelength (well, the drawings don't indicate that anyway), but for everything else the amplitude seems to depend on wavelength.

8. Oct 27, 2004

### Claire84

http://www.physics.qub.ac.uk/pastexams/0203phy211exam.pdf

If you look at 1 e) here this is effectively what he started bumbling on about (not the same as homework qs, but just to explain where I'm coming from about potentials etc). He had a larger amplitude and larger wavelength in the diagram on the right until it came to the point where the potential of the well sort of 'dropped' , and then the amplitude and wavlength of the eigenfunction in n=8 both decreased.

9. Oct 27, 2004

### Pieter Kuiper

Ah, this is a bit counterintuitive. At first sight one might expect that the particle is most likely to be in the most attractive regions of the potential. But, if you look at it classically, it has the highest kinetic energy there, and highest velocity (and the de Broglie wavelength is inversely proportional to the velocity). So the probability amplitude is higher at the shallow end of the well, and the wavelength is longer there.

It is like the pendulum (harmonic oscillator): the probability amplitude is highest at the turning points.

(I hope I am not "bumbling" now...)

10. Oct 27, 2004

### Claire84

No, no, not bumbling at all, just checking that what I had copied down wasn't rubbisn! Thank you!

See, I get where you're coming from with that, but what I don't then get that why when we have a potential step does the amplitude seem to decrease even though the wavelength increases/

One other wee thing. For the diagram on the right of 1 e on that link, as you move from left to right, do the amplitude and wavelength both increase? I'm pretty sure that the wavelength does, but I'm not too sure about the amplitude. I'm presuming it also increases but I've doubts in my mind because of my confusion over the potential step.

11. Oct 27, 2004

### Pieter Kuiper

The step is more difficult, because the wave functions cannot be normalized. Only the relative amplitudes have a meaning. Also the waves are travelling waves, not standing waves as in the well.

So at the step, there is some reflection, whether you are coming from the left or from the right (compare with optics).

And yes, in the sloping potential also the amplitude increases, with a maximum at the classical turning point. Because there you have a standing wave again, a particle rolling from a slope and bouncing at a wall, back and forth.

12. Oct 27, 2004

### Claire84

Ahhhhh I think I understand better now. It just got confusing between the standing and travelling waves. The one thing is, as I mentioned with the barrier earlier, does the amplitude of the wave above the barrier change on whether it's standing or travelling? For instance, with what we're given in the h/work where it's standing, would the amplitude increase? Would this still be the same for when we didn't have the well and just had the barrier? Thanks for clarifying this stuff for me btw, I really appreciate it.

13. Oct 27, 2004

### Pieter Kuiper

I do not think you can have a standing wave above the barrier (or above a well). Anyway, such wave functions extend to infinity, and cannot be normalized. You can only interpret the squared amplitude as a flux, positive if the wave travels to the right, negative if it goes to the left. In general you have sums of incident and reflected waves.

But I only remember the solution of the tunneling barrier, which has its optical analogy in the not quite complete internal reflection of light at a gap.

It was a pleasure discussing with you - goodnight now.