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FINITE potential step - SOS - got lost in system of equation

  1. Aug 8, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    I know that energy mentioned in the statement is kinetic energy so keep in mind when reading that ##E\equiv E_k##.

    In our case the kinetic energy is larger than the potential energy ##\boxed{E>E_p}## and this is why stationary states for the regions 1, 2, 3 (picture) are as folows:

    \begin{align}
    \psi(x)_1 &=Ae^{iLx}+B^{-iLx} && L=\sqrt{\tfrac{2mE}{\hbar^2}}=1.619\times 10^{10}\tfrac{1}{m}\\
    \psi(x)_2 &=Ce^{iKx} +De^{-iKx} && K=\sqrt{\tfrac{2m(E-E_p)}{\hbar^2}}=0.724\times 10^{10}\tfrac{1}{m}\\
    \psi(x)_3 &=Ee^{iLx}
    \end{align}

    [Broken]


    3. The attempt at a solution

    I first used the boundary conditions for the border 1-2 and got a system of equations:

    \begin{align}
    A+B &= C+D\\
    iLA-iLB&=iKC-iKD\\
    &\downarrow\\
    A+B &= C+D\\
    A-B &= \tfrac{K}{L}(C-D)\\
    &\left\downarrow \substack{\text{I used the assumption that}\\\text{$E \gg E_p$ and got the simplified}\\ \text{relation between L and K:}\\ \tfrac{L}{K}=\sqrt{E/E_p}\gg 1 \Longrightarrow L\gg 1\\ \text{and hence}\\ \tfrac{K}{L}=\sqrt{E_p/E}\ll 1 \Longrightarrow K\ll 1} \right.\\
    A + B &= C+ D\\
    A-B &= 0\\
    &\downarrow~\Sigma\\
    2A &=C+D\\
    &\downarrow\\
    C&=2A-D
    \end{align}

    Now i used the boundary conditions for the border 2-3 and got a system of equations in which i inserted the above result:

    \begin{align}
    Ce^{iKd} + De^{-iKd} &= Ee^{iLd}\\
    iKCe^{iKd} -iK De^{-iKd} &= iLEe^{iLd}\\
    &\downarrow\\
    Ce^{iKd} + De^{-iKd} &= Ee^{iLd}\\
    Ce^{iKd} -De^{-iKd} &= \tfrac{L}{K}Ee^{iLd}\longleftarrow\substack{\text{Things get weird}\\\text{when i insert }C=2A-D}\\
    &\downarrow\\
    2Ae^{iKd} -De^{iKd} + De^{-iKd} &= Ee^{iLd}\\
    2Ae^{iKd}-De^{iKd} -De^{-iKd} &= \tfrac{L}{K}Ee^{iLd}\\
    &\downarrow \substack{ \text{$L$ is very big while $K$ is very small}}\\
    2A &= Ee^{iLd}\\
    2A -2D &= \tfrac{L}{K}Ee^{iLd}
    \end{align}

    Now this result is what i don't understand. First equation gives me a ratio ##E/A=2\exp[-iLd]## which i think i need to calculate the transmissivity, but it is a complex exponential. Is this possible? How do i continue? Is this the way to solve this kind of case or did i go completely wrong?

    This is my first finite potential well problem and i have been searching the web for a loong time to even find 2 videos (video 1 and video 2) for a similar problem but for the case when ##E<E_p##. The professor in the video allso uses similar approximation - but in his case ##E \ll E_p##.

    In the appendix there is allso a scan of the problem solved on the paper just in case anyone likes it this way. It is in Slovenian language but i added some headings to help you reading.
     

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 8, 2013 #2
    You are on the right track. As you said, you need to calculate the transitivity, that will give you the desired probability.

    ##T=\left| \frac{E}{A} \right|^2##

    You have some mistake on the result you've found anyway. I don't know where the mistake is, you should check it by your self, the transmission can never be bigger than one. You can find this problem solved in Cohen Tannoudji, if that helps. Page 72.
     
  4. Aug 8, 2013 #3
    Good to know that i am at least on the right track. But do you think my approximation ##E \gg E_p## is a bit too idealistic - because i have ##E=10eV## and ##E_p=8eV##. A bit idealistic I would say...

    I will chech this for sure. I hope i get some more detail. Is there any book which shows all the algebra for the case when ##E>E_p##(scattering problem) and for the case when ##E<E_p## (tunelling). I know that there is a lot of algebra to do but all of the books just avoid it - thats like letting reader down at the hardest part of the 1-D QM! Long storry short i need this and if anzone has ny papers please scan them and post here.
     
  5. Aug 8, 2013 #4

    TSny

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    I don't think it's valid to consider 10 as being much greater than 8.
     
  6. Aug 8, 2013 #5
    I know, but i haven't found any derivation (in any of the books i have read) for the case when ##E>E_p## and not ##E\gg E_p##. Noone does the hard part - the algebra.
     
  7. Aug 8, 2013 #6
    I think Griffiths works those details, I should check it, but I'm almost sure. Perhaps he doesn't treat this specific problem, but he put everything on the examples he gives. Anyway, with the help that Cohen gives it's enough for me. There are not many ways of getting this wrong. I didn't see that approximation that you did there. The only thing that's left on Cohen to the reader is the matching of the boundary conditions, if you work it carefully you can't get it wrong.
     
  8. Aug 8, 2013 #7
    I would prefer to get my hands on the cases when ##E>E_p##, ##E<E_p## and not the extremes like ##E\gg E_p##, ##E\ll E_p##. I was only trying to derive the extreme epproximation because i didn't know how to do it properly i guess...
     
  9. Aug 8, 2013 #8
    He just describes the result and doesn't do the algebra. He only posts the final equation for the tunneling effect (##E<E_p##) and scattering problem when (##E>E_p##). Same level of detail as Wikipedia.
     
  10. Aug 8, 2013 #9

    TSny

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    Right. The algebra is tedious, but fairly straight forward. I don't recall any texts where the algebra steps are shown.
     
  11. Aug 8, 2013 #10
    I found a nice detailed algebra explaination for the case ##E < E_p## here (he is solving the system of eq. using matrix form - this is what i am not so familiar with). I am almost satisfied now. But i still have to find the algebra for ##E>E_p##... If anyone knows any similar sites please post a link...

    After some studying i think that on the webpage i posted that guy used the procedure described in QM book from Merzbacher - page 97
     
    Last edited: Aug 8, 2013
  12. Aug 8, 2013 #11

    TSny

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    That link still looks kind of formidable to me. It's really not that hard to just push it through by elementary algebra. It helps to simplify the notation.

    Let ##a = e^{iKd}##, ##b = e^{iLd}##, and ## r = \frac{K}{L}##. Then you can write your four boundary equations as

    ##A+B = C+D \ \ \ \ \ ## (1)

    ##A-B = rC - rD \ \ \ ## (2)

    ##aC+a^*D = bE \ \ \ \ \ ## (3)

    ##raC - ra^*D = bE \ \ \ ## (4)

    Use (3) and (4) to solve for ##C## and ##D## in terms of ##E##.

    Eliminate ##B## from (1) and (2) to get an equation that relates ##A## to ##C## and ##D##. Then substitute your expressions you found for ##C## and ##D## to get the relation between ##A## and ##E##.
     
  13. Aug 9, 2013 #12
    I will try this. Thank you. In case i get into trouble i ll post here.
     
  14. Aug 9, 2013 #13
    After some algebra i did get the equation which gives the ratio between ##A## and ##E##:

    $$\frac{A}{E}=\frac{e^{iLd}}{2LK}\left[ K^2 \cosh(iKd)-L^2\sinh(iKd)+KLe^{-iKd} \right]$$

    I am a bit worried because i have a complex number in the hyperbolic cosine and sine. Is it still possible i am on the right track?

    I know that for my case the transmission coeficient is calculated using this ratio i have obtained ##T=|E/A|^2## which means i will have to multiply the result by its complex conjugate, but will i get rid of those complex hyperbolic functions discussed earlier?

    In the result i should get there is no complex numbers:
    [Broken]

    Ignore ##t## while ##k_1\equiv L## and ##k_2\equiv K## so that zou will see the conection with my result.
     
    Last edited by a moderator: May 6, 2017
  15. Aug 9, 2013 #14

    TSny

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    You can use the identities ##cosh(ix) = cos(x)## and ##sinh(ix) = isin(x)## to replace the hyperbolic functions with trig functions.

    You appear to be on the right track. Most of your terms look like what you should get. However, your result is not yet correct. I believe that E/A should equal 1 in the limit d → 0.

    Yes, you will get rid of the complex numbers when multiplying by the complex conjugate.

    If you want to check some intermediate steps, I got the following expressions for C and D in terms of E

    $$C = \frac{b(1+r)E}{2ar}$$ and $$D = -\frac{b(1-r)E}{2a^*r}$$
    Also,
    $$2A = (1+r)C+(1-r)D$$
     
  16. Aug 9, 2013 #15
    Ok i did it. Thank you all :)
     
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