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Wave function/Infinite square well confusion

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    "A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a.
    [tex]\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})[/tex],
    E1=[tex]\frac{h^2\pi ^2}{2ma^2}[/tex]*

    What is the initial wave function [tex]\Psi(x,0)?[/tex]

    *[tex]h[/tex] is supposed to be h bar, I just couldn't find it)

    2. Relevant equations


    3. The attempt at a solution
    My attempt: If the general solution is a superposition of all stationary states, [tex]\Psi(x,t)=\sum c_n\psi_ne^\frac{-iE_nt}{h}[/tex], at t=0, [tex]\Psi(x,0)=\sum c_n\psi_n[/tex]. Also, at this time, the particle is in the ground state (n=1), so: [tex]\Psi(x,0)=c_1\psi_1[/tex]. Do I assume c1=1 at this point, because the wave function "collapses" once the energy becomes known? I'm just not sure if I understand exactly what happens when the known data is given.

    The solution itself is supposed to be [tex]\Psi(x,0)=\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})[/tex].
     
  2. jcsd
  3. Dec 8, 2008 #2

    Avodyne

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    Science Advisor

    c1=1 by normalization, assuming that [tex]\Psi(x,t)[/tex] is normalized. For a more general initial state,
    [tex]\sum_{n=1}^\infty |c_n|^2=1.[/tex]
     
  4. Dec 8, 2008 #3

    turin

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    Homework Helper

    You are right to be confused. You cannot know the answer to this question; that is the whole idea behind boundary conditions/initial values: they are INPUTS. For a first-order differential equation (like (d/dt)psi=iHpsi), you need one input BC for each degree of freedom.



    Try \hbar.
     
  5. Dec 9, 2008 #4
    Thanks guys.

    [tex]\hbar[/tex] :redface:
     
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