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Infinite Square Well, finding Psi(x,t)

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle in the infinite square well has the initial wave function ## \Psi (x, 0) = Ax(a-x), (0 \le x \le a) ##, for some constant A. Outside the well, of course, ## \Psi = 0 ##.
    Find ## \Psi (x,t)

    2. Relevant equations:
    Equation [1.0]:


    ## \displaystyle c_n = \sqrt{\dfrac{2}{a}} \int_0^a \sin \left( \dfrac{n \pi x}{a}\right) \Psi (x,0)dx ##


    3. The answer (that is given):
    First, determine A by normalising ## \Psi (x,0) ##

    ## \displaystyle 1 = \int_0^a | \Psi(x,t)|^2 dx##

    Therefore ## A = \sqrt{\dfrac{30}{a^5}} ##

    The nth coefficient is (Equation [1.0])

    ## \displaystyle c_n = \sqrt{\dfrac{2}{a}} \int_0^a \sin \left( \dfrac{n \pi x}{a}\right) \sqrt{\dfrac{30}{a^5}}x(a-x)dx ##

    ... skipped working...

    ## \displaystyle c_n = \dfrac{4 \sqrt{15}}{(n \pi)^3}[\cos(0) - \cos(n \pi)] ##

    Therefore:

    ## c_n = 0 ## if n is even

    ## c_n = \dfrac{8 \sqrt{15}}{(n \pi)^3} ## if n is odd


    Thus:

    ## \displaystyle \Psi (x,t) = \sqrt{\dfrac{30}{a}} \left( \dfrac{2}{a} \right) \sum \limits_{n=1, 3, 5...} \dfrac{1}{n^3} \sin \left( \dfrac{n \pi x}{a} \right) e^{(-in^2 \pi^2 \hbar t) / (2ma^2)} ##



    4. The attempt at understanding it:

    I get the normalisation part, and I get how they do the nth coefficient (hence I omitted the working) but I don't get why.

    I thought all I had to do was normalise the ## \Psi (x,0)## and then add the time dependent factor to it, so shouldn't I get:

    ## \Psi (x,t) = \sqrt{\dfrac{30}{a^5}} x (a-x) \cdot e^{(-in^2 \pi^2 \hbar t) / (2ma^2)}##

    Can someone explain this to me (in simple terms)?
     
  2. jcsd
  3. May 6, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Cogswell.

    If you look at your proposed solution, you can see a difficulty. What value of ##n## are you going to use in the time dependent factor?

    Only the energy eigenstates ## \sin \left( \dfrac{n \pi x}{a}\right) ## evolve in time according to an exponential factor Exp##(-iE_n t/\hbar)##. Each energy eigenstate has its own time factor. So, the idea is to write the initial wavefunction as a superposition of energy eigenstates and then let each of those eigenstates evolve in time with its own time factor.
     
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