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#1
05-21-2003, 07:41 PM
 Mines Mystique Guest Join Date: Mar 2003
Calling all physics gods

I recently had this problem as one on my final exam for a modern physics class. I managed to get all the points available, but I don't think I did it right, so I thought I would ask the guys here.

The question:
A beam of particles with energy E is incident on a potential barrier of height 2E. The wave function for these particles is psi(x,t)=1*exp[i(kx-wt)]. Using the Schrodinger equation and the continuity rules derive the wave equation for the reflected wave and the transmitted wave.

I would give you the answer I got, but I don't have the test with me.

So, I would like to see how you guys who seem to be physics gods would solve this. Thanks in advance.

Andrew
#2
05-21-2003, 08:01 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
I have no idea, as this sort of thing is beyond me, but be advised that most members of this board are adverse to doing homework for people. You'll probably get some tips to lead you in the right direction, but unless you show a serious effort at solving it yourself, no one's going to work out the problem for you.
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SnUgGLypuPpY -- TakE BaCk tHe PiT!
#3
05-21-2003, 08:05 PM
 Mines Mystique Guest Join Date: Mar 2003
That's the thing. This was a test question on my final and I already solved and got the points. I just want to know if I did it right and didn't just fool the grader by putting down a bunch of symbols and making it look like I know what I'm doing with this stuff.

Andrew
#4
05-21-2003, 08:29 PM
 Rabid_Squirrel Guest Join Date: Feb 2003
[Sidles up]

*Psst*!

So how did you go about solving it? If you could give a rough outline, it would let others see where you when wrong.
#5
05-21-2003, 08:43 PM
 Achernar Guest Join Date: Aug 1999
This problem, or at least something close to it, is a classic problem. I've never solved this using time-dependent Schr&ouml;dinger, but I don't see what it adds to the problem compared with the time-independent form.

This page - Penetration of a Barrier - seems to cover what I've seen before. Inside the barrier you have a decaying exponential, while outside you have a sinusoidal wave function. The amplitudes (called R and T on that page) are a bit complicated, but you get them from the boundary conditions: make psi(x) and psi'(x) continuous. I'd be surprised if you were expected to derive them on a timed exam.

Your quantum text may have this in it somewhere; I think most do. (Sakurai covers it in &sect;A.3.) It's right up there with the harmonic oscillator as an archetypal quantum problem.
#6
05-21-2003, 08:54 PM
 Mines Mystique Guest Join Date: Mar 2003
My first assumption was to use the time independent Schrodinger equation. If I remember correctly this is:

(-Hbar^2/2*m)*(second derivitive of Psi with respect to x)+U*Psi=(this is the one part I'm not sure about)E*Psi

where U is the potential energy. This also involves breaking the intial wave equation up using exponential laws ie Psi(x)=1*exp[ikx]. I then did the derivatives and solved the equation for k. I don't remember what I got. Then I assumed that the transmited wave equation would take the form Psi=exp(k'x). I used this equation in the Schrodinger equation and solved for k'. This is another part I'm not sure about. Anyway, I knew that the function had to be continuous at the barrier, so I could say that

Psi(incident)-Psi(reflected)=Psi(transmitted), with Psi(reflected)

being A*exp(ikx) and Psi(transmitted) being B*exp(k'x). I put the barrier at x=0. Then I knew that the derivatives have to be continuous at the barrier so

dPsi(incident)/dx-dPsi(reflected)/dx=dPsi(transmitted)/dx

I then plugged in what I got for k and k', and solved for A and B. I am thinking I may have gotten some signs wrong somewhere, but the grader didn't think so. Anyway, there is my method. Any ideas if I did anything wrong?

Andrew
#7
05-21-2003, 08:59 PM
 Mines Mystique Guest Join Date: Mar 2003
Looking at that page, (thanks by the way), I can see a mistake I made in the signs. And yes, we were expected to do this on a two hour exam for the final. Of course, most of my classmates had cheat sheets and probably had all the equations on it. Anyway, thanks for the help.

Andrew
#8
05-21-2003, 09:15 PM
 Achernar Guest Join Date: Aug 1999
Okay, I see this is somewhat different than the page I linked to, because your barrier is infinite.

What you did sounds more or less right to me. Where did you get k and k' from, though? If you did what I'm used to, you would have defined k to make the S-equation look like this:

psi''(x) + k2 psi(x) = 0

Then they should be the same, shouldn't they? Sqrt(2mE)/h-bar, no?

I'm still not sure about this solution. According to that page, the limit of R and T (as x0 goes to infinity) is 1 and 0. Which would mean psi(x) = 0 inside the barrier. Maybe someone can clear this up for me.
#9
05-21-2003, 09:44 PM
 Don Roberto Guest Join Date: Dec 2002
Andrew,

you certainly approached the problem correctly (i.e. solving the S.E. in each region and matching logarithmic derivatives at the boundary), so it shouldn't be surprising you got the right answer.

Achernar,

psi(x) is not zero inside the barrier. If the barrier is of finite height V, then k in the barrier region is Sqrt(2m(E-V))/hbar. When V > E, this becomes imaginary, and psi(x) inside the barrier is an exponential decay.
#10
05-21-2003, 11:08 PM
 Achernar Guest Join Date: Aug 1999
What's the amplitude, though? According to that page, if T = 0, then C = 0.
#11
05-22-2003, 01:25 AM
 g8rguy Guest Join Date: Jun 2001
Achernar, C goes to zero because it's the amplitude of the wavefunction on the other side of the barrier. When x0 goes to infinity, there is no other side of the barrier (that is, there's no wavefunction for x>infinity), and hence, T = C = 0. Inside the barrier, of course, the wavefunction has some standard exponential decay.
#12
05-22-2003, 01:46 AM
 Achernar Guest Join Date: Aug 1999
Ah oops! I was misreading B as C. I hope that makes what I said more clear.

Okay, so the boundary conditions are as follows, right?

A + B = 1
k A + k' B = k
#13
05-22-2003, 01:52 AM
 Achernar Guest Join Date: Aug 1999
Er, with an i in the... thingy... ah, screw it. The OP is satisfied, and I can work this out some other time. Nevermind me....

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