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How does one find the time dependent wavefunction?

  1. Dec 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Random given wavefunction,say $$\Psi (x) = N e^{- \mu x}$$ in a V(x) e.g. infinite well .Find ## \Psi (x,t) ##.

    2. Relevant equations
    -

    3. The attempt at a solution
    If the wavefunction is given as the sum of eigenfunctions,you just multiply them by ## e^{-i \frac{tE}{\hbar}}##.But in the case above there isn't one defined energy so you can't do this.
     
  2. jcsd
  3. Dec 31, 2016 #2
    Well, first of all, the wavefunction that you proposed is incompatible with the infinite square well as it does not live within the Hilbert space of the infinite square well potential - the boundary conditions clearly don't agree.

    That aside, you already recognised that if the wavefunction were to be written in the form of a sum of energy eigenfunctions, then you would simply affix the relevant exponential factor to each term i.e.
    [tex]\Psi(x,t) = \sum_{n} c_{n} \psi_{n}(x) e^{- iE_{n}t / \hbar} [/tex]
    So then, if you were given an arbitrary wavefunction, the first thing that needs to be done is none other than to decompose it into the sum of the relevant eigenfunctions of the system:
    [tex]\Psi(x) = \sum_{n} c_{n} \psi_{n}(x) [/tex] i.e. you need to solve for the coefficients ##c_{n}##.
     
  4. Dec 31, 2016 #3

    hilbert2

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    It would have to be ##\psi(x) = Ne^{-\mu |x|}## to be normalizabe, and even then the wavefunction wouldn't be exactly zero outside any interval ##x\in [a,b]## like it has to be in an infinite well.

    The options for solving ##\psi (x,t)## are either the decomposition to eigenfunctions of ##H## and multiplication with the ##e^{-iEt}## factors, or discretization of the considered time-space domain ##\psi (m\Delta x, n\Delta t)=\psi_{m,n}## and solution of the numbers ##\psi_{m,n}## with Crank-Nicolson method (this works even if the spectrum of ##H## isn't known).
     
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