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Half infinite well

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data

    4hw7il.jpg

    2. Relevant equations

    -h^2/2m d^2F(x)/dx^2 = EF(x)

    3. The attempt at a solution

    i just need to a part. for E<0 i can find for 0<x<L side F(x) = ACos(Lx) + BSin(Lx)

    at the L<x side, F(x) = e^(Kx) where

    L^2= 2m(E+V)/h^2

    K^2= -2mE/h^2

    but i do not know what will i do. can you help me for a part of question.
     
  2. jcsd
  3. Nov 17, 2012 #2

    Simon Bridge

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    This is an infinite barrier with a finite well in front of it.
    1. Solve the schrodinger equation for the eigenvalues.
    2. Find out what is needed for an eigenvalue to be negative.

    [edit] Me and Mute crossed posts - I've decided to support Mute's approach.

    Have you seen the finite square well? This is similar.
     
    Last edited: Nov 17, 2012
  4. Nov 17, 2012 #3

    Mute

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    You probably want to use a different symbol than L in the Cos(Lx) and Sin(Lx) terms.

    For L<x (using the length L, here), your solution should be a decaying exponential - you have chosen the exponential that grows as x grows.

    Now that you have the solutions in the two regions, you need to apply the boundary conditions. Using the boundary condition that your wave function should go to zero at infinity is what gives you the decaying exponential for L<x. There are two more boundaries: The boundary at x = 0, and the boundary at x = L. Do you know what the boundary conditions are at these boundaries? (Edit: Simon Bridge has given you the conditions, so get to it!)
     
  5. Nov 18, 2012 #4
    aghh i get all A,B,C = 0 :confused::confused:

    i attached the solution;

    http://imgur.com/rtBpCl.jg [Broken]

    where am i doing wrong?
     
    Last edited by a moderator: May 6, 2017
  6. Nov 18, 2012 #5

    Simon Bridge

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    You picked the "trivial" solutions - zero everywhere is, indeed, a solution.
    But you only need the different sections to agree at one point.

    Recap:
    ##\psi_I(0)=\psi_{II}(0)##

    You need ##\psi_{II}(x)=A\cos(kx)+B\sin(kx)## to be 0 at x=0, which must mean that A=0, because ##\cos(0)=1##... so ##\psi_{II}(x)=B\sin(kx)## ... so far so good.

    At the other end it is more complicated:
    ##A\sin(kL)=Ce^{-KL}##
    ##Ak\cos(kL)=-CKe^{-KL}##
    ... which appears to give you four variables in only two equations doesn't it?
    But I think you'll find that k and K have to be related, so that's really only three variables.

    In the end - the entire wavefunction has to be normalized ... so $$\int_0^L \psi_{II}^\star(x)\psi_{II}(x)dx + \int_L^\infty \psi_{III}^\star(x)\psi_{III}(x)dx = 1$$ which should give you the third equation.

    You'll find that only specific values of k (hence K) will satisfy these conditions... (particularly for E<0) so providing discrete energy levels.
    Hint: what are k and K both functions of?

    Like I said before - the method of solving these last two is very similar to that for a finite square well ... which you can look up.
    You should also be able to sketch the basic shape of the (amplitudes) first few bound-states (if they exist) ... eg. the first one starts at 0 when x=0, has a peak between 0 and L, is still >0 at x=L then decays exponentially from there.
     
    Last edited: Nov 18, 2012
  7. Nov 18, 2012 #6
    thank you very much. i think i can solve it after your help.
     
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