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Help with Potential Barrier question

  1. Feb 21, 2014 #1
    I have always understood (As well as I can) potential barrier questions, but this one has stumped me, and I was hoping someone could point out where my thinking has gone wrong.

    The wave function of an electron of mass m incident to the step from x = -∞ with energy E < V is = eikx + ρe-ikx for x≤0, and ψ=τe-κx for x > .

    Now consider an electron of the same energy incident from x=-∞ to a barrier of width L consisting of two potential steps described by U(x)=0 for x≤0 and U(x)=V for 0 < x ≤ L, U(x) = 0, for x > L. The electron can be considered to undergo multiple reflections within the barrier before being transmitted. Show that the amplitude for transmission through the barrier after a single pass is

    t1= (1 + ρ)e-κL(1 - ρ)

    and after a double pass with two reflections

    t2=(1 + ρ)e-κL(-ρ)e-κL(-ρ)e-κL(1 - ρ)

    I assumed I worked out the Transmission coefficient T = 1 - R at x=0 and x=L and square-root to get the transmission amplitudes, but this does not seem to be working. If anyone can shed any light, I would be much grateful. I could write out my scribbles and attempts, but it would be pretty fruitless as they're not taking me anywhere
     
  2. jcsd
  3. Feb 21, 2014 #2

    Simon Bridge

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    How are you using the transmission and reflection coefficients?
    Are you accounting for the amplitude loss traversing the barrier?
    Are you accounting for the difference between going outside→inside vs inside→outside?

    When attempting QM problems, it is best practice to learn LaTeX :)

    You need to show:$$t_1=(1+\rho)e^{-\kappa L}(1-\rho)$$... for one pass.

    I'm guessing you got something like: $$t_1=(1-\rho)^2$$
     
    Last edited: Feb 21, 2014
  4. Feb 21, 2014 #3
    Hi Simon, thanks for the reply and yes, LaTeX is something I really should learn. Yeah I wasn't too sure about how to accommodate the difference between going inside and going outside other than that there shouldn't be a $$e^{-ikx}$$ coefficient. I had the transmission coefficient at x=0 to be 1 - R or $$1 - \rho^2$$ and then I assumed the transmission amplitude would be the root of this. Otherwise if that's wrong, my other line of thinking was that the transmission amplitude at x=0 is just $$1-\rho$$. I'm still not to sure where to start with the boundary conditions for the transmission coefficient at x=L
     
  5. Feb 21, 2014 #4

    Simon Bridge

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    At the first interface, if the incoming amplitude is 1, then the amplitude that gets through is ##1-\rho##.
    You've already written this down. It's a matter of going back and looking at the wavefunctions as travelling plane waves.

    so at ##x=0##, just inside the barrier, the amplitude is ##1-\rho## (do you see why this has to be?)
    then at ##x=L##, still inside the barrier, but at the other side, what is the amplitude?
    don't worry about the reflection just yet - the amplitude changes as the wave penetrates the barrier: what does it do?

    In general, the reflection coefficient at x=L will be different, but the symmetry should suggest something.
    If not, then the form of the two correct equations you are given should give you a hint.

    Considering you have to prove them, then, if it's not already been done earlier in the coursework, you'll probably have to do the math.

    Note: it is a good idea to treat transmissions, reflections, and translations, as operators ... so the first one to happen to the wavefunction appears to the right, the rest are premultiplied in order, just like matrix optics.
     
  6. Feb 22, 2014 #5
    Hi thanks, again. I have to admit I have no idea what you mean by treating the reflections as operators, but I'd happily read up on it if you could link me to any articles or forum posts.

    I think I see why just inside the barrier the amplitude is ##(1-\rho)## presumably this is because 1 is the incident amplitude and ##\rho## the reflected. At x=L presumably the amplitude has decreased exponentially meaning you can just multiply the initial amplitude by ##e^{-\kappa L}##. However I can't work out what the amplitude of the reflected wave inside the potential barrier should be at x=L, but would it take some coefficient multiplied by ##e^{+\kappa L}##?
     
  7. Feb 22, 2014 #6

    Simon Bridge

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    Doing well.
    Quite right - if the incoming wave has amplitude 1, and the reflected wave has amplitude r, then the transmitted wave has amplitude 1-r for the boundary condition to add up: ##\psi_{in}(0)=\psi_r(0)+\psi_t(0)##

    Remember "waves on a string"?

    When the wave went from a fast medium to a slow medium, the reflected wave inverted?
    So if the wave approaching the boundary had amplitude A, then the reflected wave would be -rA ... the minus sign indicating the inversion and r<1 : i.e the reflected wave is upside down wrt the incoming wave.

    It's kinda like that in QM. the reflected wave from outside a barrier has amplitude ##\rho## (times the incoming amplitude), then, by symmetry, the reflected wave inside the barrier would have amplitude ##-\rho##.
    The transmitted wave has the same relationship to the reflected wave as before, so ##t=1-(-\rho)=1+\rho## when going from inside to outside.

    You can see this fits in with the model answer.

    I don't know if your marker will accept "by symmetry" as an argument - which means you'll have to prove it. It will be difficult for me to help you there, it means looking at how the wavefunction changes at the interfaces very carefully.


    Next: here's what I mean by "treat like operators".
    I mean - you write down the effect of the reflections and transmissions from right to left in the order they happen (not left to write like you normally write)

    If you have two operations A and B and A happens before B then they can be represented by a single operator of form T=BA.

    transmission and reflection etc could be represented by operators.
    loo up "matrix optics" to see how this can be done classically

    anyway, the first one to happen, goes on the right.

    i.e. you have a transmission that has effect ##(1-\rho)##, then you have a translation inside a barrier that has effect ##e^{-kL}##.

    Then you write those down as ##e^{-kL}(1-\rho)## for the overall effect.

    In this case it doesn't really matter for the final equation because these are all scalars.
    It will help you understand the model answers, and it will hep when you have more complicated operations.

    Aside: looking at ##t_1=(1+\rho)e^{-kL}(1-\rho)##, what happens to ##t_1## when ##L=\frac{1}{k}\ln|(1-\rho)##?(1+\rho)|## - what does it mean?

    All this ties in with your earlier work in optics, and also with thin films.
     
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