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

[LSMOG]

I've tried to read the reason for using Fourier transform in wave packets, I don't understand why. Please help me with this.

 

[DrClaude]

Have you looked at the relation between position space and momentum space?

 

[stevendaryl]

I've tried to read the reason for using Fourier transform in wave packets, I don't understand why. Please help me with this.

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.

 

[LSMOG]

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

 

[stevendaryl]

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

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

 

[stevendaryl]

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

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}