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Barrier Potential

  1. Dec 13, 2006 #1
    Question: What happens when the energy of the state is exactly equal to the barrier it hits?

    Say the particle is traveling to the right and hits the barrier. Before hitting, you have your typical solutions, say Aexp(ikx) + Bexp(-ikx). After hitting, the V(psi) and E(psi) terms cancel. Integrating then yields some Cx + D, which diverges at infinity, so C=0 and you have constant wave function, psi(x)=D

    Here's where my question is. You can either ignore normalization and find A,B in terms of D, the compute things like transmission/reflection coefficients. When I did this I got A=B=D/2.

    I computed R=(|B|/|A|)^2 = 1, and also T = (|D|/|A|)^2 = 4.

    So I normalize these by hand and say R=1/5, T=4/5

    On the other hand, a constant wave function is non-normalizable. So what does this mean? My friend thinks it is a non physical solution, and thus all the wave must be reflected. But the free particle is also initially non-normalizable until we express it as an integral over a continuous wave number. But that's not exactly the same thing here, and the non-normalizable argument seems strong.

    Our problem was to find the SE for E=V, without normalizing, then find the reflection coefficient.

    So basically, what happens when E=V for a barrier potential
  2. jcsd
  3. Dec 13, 2006 #2


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    Last edited: Dec 14, 2006
  4. Dec 14, 2006 #3
    hey acme37 hope u can help me if u did get the right answer bcoz i have to give it as homework so i can higher my grades on midterm ... i have the same problem when E=V with finite potential ...and plz olderdan the url u present is not active ...do u have any other ... thanks anwyay
  5. Dec 14, 2006 #4


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    Add a bracket on the end.. for some reason this doesn't come up in the link
  6. Dec 14, 2006 #5
    thank you :)
  7. Dec 14, 2006 #6
    Well I was wrong then, bummer. A linear wave function in x was something I hadn't seen before so that was tricky. I think the non-normalizable argument works here, or you can solve it as a barrier that you then extend to infinity (like the wikipedia article does).

    It's weird to me that the transmission coefficient can depend on the width of the barrier, (and go to zero for the step). The particle would then have to know the total width of the barrier right as it is incident upon it.
  8. Dec 14, 2006 #7


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    Seems the parser didn't like the ) at the end of the link. I don't know why that happened. I edited the link. It works now.
  9. Dec 14, 2006 #8


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    Think about what you are saying. If the particle knew exactly when it was incident at the barrier, what must it NOT know?
  10. Jan 9, 2007 #9
    the wikipedia article seems ligic for me ...but i didn't know how to get the transmission coefficient ...while i was trying i didn't get to their answer ...do u have the steps for that ?
  11. Jan 9, 2007 #10
    sorry i mean logic
  12. Jan 11, 2007 #11
    hey guyz plz can anyone help???? i have to present my report for extrea point
  13. Jan 11, 2007 #12


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    Show us what you've tried, and we'll help you finish it correctly.
  14. Jan 11, 2007 #13
    ok it looked logic to me that psi(x) is linear on the region where the electron hit the barrier ... and i applied the boundary condition ... and i got 4 equations:

    here are the 4 equations from apllying boundary conditions on x=0 and x =a and now i have to get the transmission coeff. i know it should be FF*/AA*
    but i am not getting to a realtion like the answer on the wikipedia article.. sorry i don't have a scanner to show u my work but i did some work ...
  15. Jan 11, 2007 #14
    i would solve it by myself but i am facing some problems because i missed 5 successive lectures so i am a little bit lost where i am ... and i am trying to recover what i last and no one can help in my university because they all didn't understand the basics they only apply rules ...
  16. Jan 12, 2007 #15
    so.... should i try any other method?
  17. Jan 14, 2007 #16
    any help guyz... not the answer where am i goin' wrong?
  18. Jan 14, 2007 #17
    i just saw the FAQ ... guys try to ask me question to be sure that i am qorking on that ... but i have no scanner and i tried to capture the work on the cam ...it is not clear... try to ask me some question and i am ready to answer .
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