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Shifted infinite well

  1. Sep 17, 2009 #1
    I can solve the infinite will with length a. But what happens when the coordination is shifted, namely:
    [tex]V(x)=0,\frac{ -a}{4}<x<\frac{3a}{4}[/tex]

    I use the usual solution:
    [tex]\psi=Asin(kx)+Bcos(kx)[/tex]
    Now when I apply the first boundary condition:
    [tex]\psi(\frac{ -a}{4})=\psi(\frac{3a}{4})=0[/tex]
    I can't get rid of one of the terms (namely the B=0 like when the well was defined from 0 to a).
    I know that A will be the same as if the well was defined from 0 to a and I can get k by applying the other boundary condition, but I can't do anything if I can't get rid of the cos term.
     
  2. jcsd
  3. Sep 17, 2009 #2

    Avodyne

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    Science Advisor

    If psi(x) is a solution to the Schrodinger equation with potential V(x), then psi(x-b) is a solution to the Schrodinger equation with potential V(x-b). So all you need to do is shift the solution you have by the correct amount.

    You can also do it your way. Use psi(-a/4)=0 to solve for B in terms of A. Plug this into psi(3a/4)=0; every term will have an A in it, so cancel out A, since you know it isn't zero. Now you have an equation whose only unknown is k. You should be able to show that the solutions are k = pi n/a, where n=1,2,3,...
     
  4. Sep 18, 2009 #3
    This is not working too well either.
    Solving for B:
    [tex]\psi(\frac{-a}{4})=Asin(k\frac{-a}{4})+Bcos(k\frac{-a}{4})=0[/tex]
    [tex]B=\frac{-Asin(k\frac{-a}{4})}{cos(k\frac{-a}{4})}=\frac{Asin(k\frac{a}{4})}{cos(k\frac{a}{4})}[/tex]
    Then:
    [tex]\psi(\frac{3a}{4})=Asin(k\frac{3a}{4})+\frac{Asin(k\frac{a}{4})}{cos(k\frac{a}{4})}cos(k\frac{3a}{4})=0[/tex]
    [tex]\frac{sin(\frac{3ak}{4})}{cos(\frac{3ak}{4})}+\frac{sin(\frac{ak}{4})}{cos(\frac{ak}{4})}=0[/tex]
    [tex]\frac{sin(\frac{3ak}{4})cos(\frac{ak}{4})+cos(\frac{3ak}{4})sin(\frac{ak}{4})}{cos(\frac{3ak}{4})cos(\frac{ak}{4})}=0[/tex]
    [tex]sin(\frac{3ak}{4})cos(\frac{ak}{4})+cos(\frac{3ak}{4})sin(\frac{ak}{4})=0[/tex]
    [tex]sin(\frac{3ak}{4}+\frac{ak}{4})=sin(ak)=0[/tex]
    Then I indeed recover:
    [tex]k=\frac{n\pi}{a}[/tex]
    Plugging it back in:
    [tex]\psi(x)=Asin(\frac{n\pi}{a}x)+Bcos(\frac{n\pi}{a}x)[/tex]
    This is not what I expected and I am quite sure that it is wrong since if I plug in the limits, psi is non-zero at the end points.


    Further more, this should be just the same as the infinite well with length a shifted over. So shouldn't the solution be the same but with a phase constant added? Namely:
    [tex]\psi=Asin(\frac{n\pi}{a}x)\rightarrow\psi=Asin(\frac{n\pi}{a}(x+\frac{a}{4}))[/tex]
    I think that this is the answer, but that's kind of cheating since it is reverse engineered from another infinite well. I can't figure out why I can't do it top down.
     
    Last edited: Sep 18, 2009
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