(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(*I have solved similar problems before and know the method, but my results don't make sense, so I want to make sure I understand this particular potential. Also, I am new to Latex so ignore the equal portion of the less/greather than or equal two signs.*)

I made a diagram of the potential I am working with.

http://img2.freeimagehosting.net/image.php?8baf7057bb.jpg [Broken]

I need to find the transmission (T) and reflection (R) coefficients.

Just to make sure, what have you seen. The potential V(x) is as follows

[tex]0[/tex] for x[tex]\le[/tex] -a

[tex]V_{1}[/tex] for -a[tex]\leq[/tex]x[tex]\leq[/tex]0

[tex]V_{0}[/tex] for 0[tex]\leq[/tex]x[tex]\leq[/tex]+b

[tex]0[/tex] for x[tex]\ge[/tex] +b

2. Relevant equations

3. The attempt at a solution

I don't need to worry about bound states E < 0. So, there are two remaining cases

Case 1: [tex] V_{1}[/tex][tex]\le[/tex][tex]E[/tex][tex]\le[/tex][tex]V_{0}[/tex]

Case 2: [tex] V_{0}[/tex][tex]\le[/tex][tex]E[/tex]

Now I found R and T for case one and currently working on case two. From what other problems I have done R + T always equals 1, but for case one I actually got an interesting expression. If I assume that [tex] V_{1}[/tex] is small (shallow well) and if [tex] V_{0}[/tex] is small or b is small (the sums depends on the exponent of V0 and b) , then the sum approaches one. I have gone over my algebra a few times. Is it possible they are times that R+T, I have never encountered a problem that has. It's not like I have done that many either. I have done a simple barrier, step potential, drop potential. I have seen the finite well and delta function well examples.

Without the assumptions, it is implying that there is some absorption, but that doesn't make sense to me... where would it be absorbing... the well? I figure it kinda just bounces back and forth... so R + T still equals one.

Since I know the method of solving these problems and I don't want to bother you with equations. I am just giving you those little constants (think they are called eigenvalues) that determine the solutions to the SchrÃ¶dinger's eq. for each separate religion.

[tex]\frac{\sqrt{-2mE}}{\hbar}[/tex] for x[tex]\le[/tex] -a

[tex]\frac{\sqrt{2m(E+V_{1})}}{\hbar}}[/tex] for -a[tex]\leq[/tex]x[tex]\leq[/tex]0

[tex]\frac{\sqrt{2m(V_{0}-E)}}{\hbar}[/tex] for 0[tex]\leq[/tex]x[tex]\leq[/tex]+b

[tex]\frac{\sqrt{-2mE}}{\hbar}[/tex] for x[tex]\ge[/tex] +b

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# Homework Help: Quantuam Mechanics: Discussion of unbound states of a combination well and barrier

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