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* Partial Differential Equations and Fourier Transform

The Schrödinger equation for a free particle,
$\displaystyle i\hbar\frac{\partial}{\partial t} \Psi(x,t)
=-\frac{\hbar^2\partial_x^2}{2m} \Psi(x,t), \quad \Psi(x,t=0)=\Psi_0(x)$     (53)

can be solved by Fourier transformation:
$\displaystyle i\hbar\frac{\partial}{\partial t} \int_{-\infty}^{\infty}\frac{dk}{2\pi} e^{ikx} \tilde{\Psi}(k,t)$ $\displaystyle =$ $\displaystyle - \frac{\hbar^2\partial_x^2}{2m} \int_{-\infty}^{\infty}\frac{dk}...
...hbar^2}{2m} \int_{-\infty}^{\infty}\frac{dk}{2\pi}
k^2e^{ikx} \tilde{\Psi}(k,t)$  
$\displaystyle \leadsto$ $\displaystyle 0=$ $\displaystyle \int_{-\infty}^{\infty}\frac{dk}{2\pi} e^{ikx} \left[
i\hbar\frac...
...\partial t} \tilde{\Psi}(k,t) - \frac{\hbar^2 k^2}{2m}\tilde{\Psi}(k,t) \right]$  
$\displaystyle \leadsto$ $\displaystyle 0=$ $\displaystyle i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(k,t) - \frac{\hbar^2 k^2}{2m}\tilde{\Psi}(k,t).$ (54)

The last equation is an ordinary differential equation in $ t$ that can be solved easily:
  $\displaystyle 0=$ $\displaystyle i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(k,t) - \frac{\hbar^2 k^2}{2m}\tilde{\Psi}(k,t)$  
  $\displaystyle \leadsto$ $\displaystyle \tilde{\Psi}(k,t)=\tilde{\Psi}(k,t=0)e^{-i\omega(k)t},\quad
\omega(k):=\frac{\hbar k^2}{2m}.$ (55)

Here, the initial value $ \tilde{\Psi}(k,t=0)$ appears; it is given by our definition for the Fourier transform
$\displaystyle \tilde{\Psi}(k,t=0)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}dx \Psi(x,t=0) e^{-ikx}=
\int_{-\infty}^{\infty}dx \Psi_0(x) e^{-ikx}=\tilde{\Psi}_0(k)$ (56)

Therefore, if we know the initial wave function $ \Psi_0(x)$, we know its Fourier transform $ \tilde{\Psi}(k,t=0)$ and can calculate the solution $ \Psi(x,t)$ at a later time $ t>0$ by Fourier back transformation,
$\displaystyle {\Psi(x,t)}$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\int_{-\infty}^{\infty}dk \tilde{\Psi}(k,t)e^{ikx}
...
...int_{-\infty}^{\infty}dk e^{ikx} e^{-i\frac{\hbar k^2t}{2m}} \tilde{\Psi}_0(k)}$  
  $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\int_{-\infty}^{\infty}dk e^{ikx} e^{-i\omega(k) t} \int_{-\infty}^{\infty}dx'
\Psi(x',t=0) e^{-ikx'}$  
  $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}dx' \underline{ \frac{1}{2\pi}\int_{-\infty}^{\infty}dk e^{ik(x-x')-i\omega(k) t} }
\Psi(x',t=0)$  
  $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}dx' \underline{G(x,x';t,t=0)} \Psi(x',t=0),$ (57)

where we defined the propagator of the particle which propagates the wave function from its initial form at $ t=0$ to its form at a later time $ t>0$.


next up previous contents
Next: Position and Momentum in Up: Fourier Transforms and the Previous: The Delta Functional (`Delta   Contents
Tobias Brandes 2004-02-04