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

1. Oct 24, 2016

### LSMOG

2. Oct 24, 2016

### Staff: Mentor

3. Oct 24, 2016

### stevendaryl

Staff Emeritus
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)

$\psi_0(x) = F(x)$

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

$F(x) = \sqrt{\frac{1}{2 \pi \lambda}} e^{- \lambda (x-x_0)^2}$

Now, you would like to know what $\psi(x,t)$ will look like at some time $t > 0$. 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 $\psi(x,t) = e^{i (kx - \omega_k t)}$, corresponding to the initial wave function $\psi_0(x) = e^{i k x}$, where for the nonrelativistic Schrodinger equation, $\omega_k = \frac{\hbar k^2}{2m}$ for a particle of mass $m$. So we can use that knowledge to solve the general problem:

Write $\psi_0(x) = \int dk \tilde{F}(k) e^{i kx}$, where $\tilde{F}(k)$ is the Fourier transform of $F(x)$. Then, at a later time $t$, the wave function will be $\psi(x,t) = \int dk \tilde{F}(k) e^{i (kx - \omega_k t)}$. Voila! The Fourier transform allows you to write down the general solution.

4. Oct 25, 2016

### LSMOG

Thanks, why don'y we consider the solution to the Schrodinger equation as a general solution

5. Oct 25, 2016

### stevendaryl

Staff Emeritus
The issue is how to come up with a solution to the Schrodinger equation. The Fourier transform gives you a way to do that.

6. Oct 25, 2016

### stevendaryl

Staff Emeritus
If you are trying to solve the equation

$-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x,t) = i \hbar \frac{\partial}{\partial t} \psi(x,t)$

subject to the initial condition

$\psi(x,0) = F(x)$

then the unique solution is

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

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

This solution can also be written in the form:

$\psi(x,t) = e^{-i \frac{H t}{\hbar}} F(x)$

where $H = \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial x^2}$