Potential Barrier in Infinite Well: Equal Amplitudes Both Sides?

[Oliman]

Homework Statement

The well is set up exactly like this https://www.physicsforums.com/showt...ential+barrier+inside+an+infinite+square+well. It is basically a infinite potential well with a barrier in the center.

Basically I've got solutions for regions 1, 2 and 3, found some allowed energy levels. In region one we have Asin(k1x) and in region 3 I have Dsin(k1x), when I use the continuity conditions to solve for the constants I find that A=-D. This means that when I plot the wave functions I end up with equal amplitudes on either side of the barrier. I would have thought that the amplitude in region three is smaller than the amplitude in region one. Can anyone tell me which is correct?

My apologies for not having it in the normal form but I thought because I'm not asking for help with a question but merly asking whether my result is feasible this form makes more sense. Also my apologies for my lack of latex but for some reason it wasn't working :-(. Sorry this is my first post (if you hadn't already guessed! ;-)

 

[nickjer]

Why is region 3 described by only D\sin(k_1 x)?

You would want \Psi(2a+2b) = 0, which doesn't necessarily throw out the cosine term.

In fact it might be easier to just change the coordinate system so that the origin is in the center of the well. Then due to the symmetry of the potential, you just look for symmetric and antisymmetric wavefunctions.

 

[Oliman]

Hi Nickjer, Thanks for the reply. In actual fact I did translate the cordinate system so the center of the well was at the origin and then looked at odd and even functions within the well but left region three as sin. Shall I solve for my constants in the usual way except using a linear comination of sin and cos in region three? Should I take into account the symmetric and antisymetric properties just in region two or region three as well?

 

[vela]

This means that when I plot the wave functions I end up with equal amplitudes on either side of the barrier. I would have thought that the amplitude in region three is smaller than the amplitude in region one.

Why would you expect that? Is there something different about region 3 that would cause the amplitude to be smaller there than in region 1?

 

[Oliman]

Why would you expect that? Is there something different about region 3 that would cause the amplitude to be smaller there than in region 1?

I thought that when something transmits through a barrier the amplitude gets smaller. I guess I thought that was a kind of rule. Am I wrong in thinking this?

 

[nickjer]

There is no transmitting here. A particle isn't incident from the left and transmitting through the right. You have a standing wave with a barrier in the center.

And yes, you should start off with a linear combination of sin and cos on both sides of the well. If it is a symmetric function then you will want a mirror image of the function on both sides (replacing x->(-x)). If it is antisymmetric, then you will want symmetry about the origin.

In both cases the amplitude on either side will be related. Just don't forget to solve for the BC's at the boundary.