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A I've tried to read the reason for using Fourier transform

  1. Oct 24, 2016 #1
    I've tried to read the reason for using Fourier transform in wave packets, I don't understand why. Please help me with this.
     
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  3. Oct 24, 2016 #2

    DrClaude

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  4. Oct 24, 2016 #3

    stevendaryl

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    Is this specifically a physics question, or a question about Fourier transforms?

    The issue is this: Suppose we have a particle that is initially distributed in a packet described by (in one-dimension)

    [itex]\psi_0(x) = F(x)[/itex]

    where [itex]F(x)[/itex] is some function that is strongly peaked at [itex]x=x_0[/itex]. For example, it might be Gaussian:

    [itex]F(x) = \sqrt{\frac{1}{2 \pi \lambda}} e^{- \lambda (x-x_0)^2}[/itex]

    Now, you would like to know what [itex]\psi(x,t)[/itex] will look like at some time [itex]t > 0[/itex]. You can try to solve the Schrodinger equation for that particular initial condition, but here's an easier way:

    You know that a solution to Schrodinger's equation is [itex]\psi(x,t) = e^{i (kx - \omega_k t)}[/itex], corresponding to the initial wave function [itex]\psi_0(x) = e^{i k x}[/itex], where for the nonrelativistic Schrodinger equation, [itex]\omega_k = \frac{\hbar k^2}{2m}[/itex] for a particle of mass [itex]m[/itex]. So we can use that knowledge to solve the general problem:

    Write [itex]\psi_0(x) = \int dk \tilde{F}(k) e^{i kx}[/itex], where [itex]\tilde{F}(k)[/itex] is the Fourier transform of [itex]F(x)[/itex]. Then, at a later time [itex]t[/itex], the wave function will be [itex]\psi(x,t) = \int dk \tilde{F}(k) e^{i (kx - \omega_k t)}[/itex]. Voila! The Fourier transform allows you to write down the general solution.
     
  5. Oct 25, 2016 #4
    Thanks, why don'y we consider the solution to the Schrodinger equation as a general solution
     
  6. Oct 25, 2016 #5

    stevendaryl

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    The issue is how to come up with a solution to the Schrodinger equation. The Fourier transform gives you a way to do that.
     
  7. Oct 25, 2016 #6

    stevendaryl

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    If you are trying to solve the equation

    [itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x,t) = i \hbar \frac{\partial}{\partial t} \psi(x,t)[/itex]

    subject to the initial condition

    [itex]\psi(x,0) = F(x)[/itex]

    then the unique solution is

    [itex]\psi(x,t) = \int dk e^{i (kx - \omega_k t)} \tilde{F}(k)[/itex]

    where [itex]\tilde{F}(k) = \frac{1}{2\pi} \int dx F(x) e^{-ikx}[/itex], and where [itex]\omega_k = \frac{\hbar k^2}{2m}[/itex].

    This solution can also be written in the form:

    [itex]\psi(x,t) = e^{-i \frac{H t}{\hbar}} F(x)[/itex]

    where [itex]H = \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial x^2}[/itex]
     
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