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Getting Psi(x,t) from Psi(x,0)

  1. Mar 1, 2007 #1
    Hey guys, first time poster.

    I am doing some quantum physics homework, and I came across the following problem:

    A particle in an infinite square well has the initial wave function
    Psi(x,0) = Ax 0?x?a/2
    A(a-x) a/2 ?x?a

    Find Psi (x,t)

    Now after normalizing it, I tried plugging it in Schrödinger's equation, however I'm still having problems.

    Thanks in advance, Rob.
     
  2. jcsd
  3. Mar 1, 2007 #2

    Meir Achuz

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    Psi(x,t)=Sum exp{-iE_n hbar t}phi_n(x)<phi_n|Psi(x,0)>,
    where phi_n are the estates and E_n the eignevalues of the square well.
     
  4. Mar 1, 2007 #3
    Why did you do the summation of the function instead of integrating it??
     
  5. Mar 1, 2007 #4

    nrqed

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    The energies are discrete, hence the summation.

    You must write
    [tex] \Psi(x,t=0) = \sum_n c_n \psi_n(x) [/tex]
    where the [itex] \psi_n(x) [/itex] are the energy eigenstates, [itex] \sqrt{2/a} \, sin(n \pi x/a) [/itex] (for a well located between x=0 and x=a). What you have to do is to find the coefficients c_n using the orthonormality of the sine wavefunctions. Once you have that, the wavefunction at any time is

    [tex]\Psi(x,t) = \sum_n c_n e^{-i E_n t / \hbar} \psi_n(x) [/tex]
     
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