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1-D infinite potent. well

  1. Oct 29, 2009 #1
    Ok, I have a 1-D box confined at at x = 0 and x = L. So, points between 0 and L distances are the continuum state and otherwise distances be discontinous.
    a) I need to find the egien functs: Un(x) and related egien values: En .... n are the excited levels represented as postive whole numbers.

    The wave funct is: φ(x, t = 0) = (1/(3^1/2))U_2(x) + ((2/3)^1/2)U_3(x)


    b) As time progresses, what will the function look like?
    c) What is the prob. density (φ squared) and P(x,t) = total probability.


    What I have so far...

    (-(h/2pi)^2)/2m * (d^2/dx^2)Psi(x) = E*Psi(x)
    Psi(x)|x=0 = Asin(0) + Bcos(0) = B = 0 ?
    Psi(x)|x=L = Asin(kL) + Bcos(kL) = 0 ?

    [0 0; sin(kL) cos(kL)] *[A;B] = [0 0]

    set KnL/2 = n*pi
    En = (h/2pi)^2 *k^2]/2m
    = [(h/2pi)^2] /2m * (2n*pi/L)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 2, 2009 #2
  4. Nov 2, 2009 #3
    Hi Panic_Attack!

    In the second boundary condition you made a mistake, because you
    already know that B=0, You really have:

    [tex]\Psi(x=L) = ASin(kL) = 0 [/tex]

    And this condition say to you:

    [tex] kL = n\pi [/tex]

    where

    [tex] k^{2}\equiv\frac{2mE}{\hbar^{2}} [/tex]

    Then you got [tex]E_{n}[/tex]

    To find the wave function you don't know A yet, but try to normalize the wave function.

    On (b) part.. Have you heard about the Evolution Operator? Maybe this simplify your problem.

    (Sorry my english sucks)
     
    Last edited: Nov 2, 2009
  5. Nov 5, 2009 #4
    Thanks so much for replying to my question. Fortunately I was able to find an answer without using the evolution operator. I basically went through solving with the schrodinger equation with setting up the solutions of the differential equations based on the regions. And had the same K value you got too. Then I normalised the wave function with it squared over the integral and found A too... I really apreciate your help, sorry I couldnt reply sooner.

    Your english sounds better than mine!! lol
     
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