Integration question re Fourier Transform

[celal777]

Hello Listmembers,

I am trying to make some progress in my understanding of
Schrodinger's equation.I have been trying to teach myself about
Fourier transforms in the hope that this will help me understand the
derivation of Schrodinger's equation.

My question has to do with slide number 16 from a presentation which can be found at :

http://www.physics.gatech.edu/gcuo/UltrafastOptics/3803/OpticsI14FourierTransformI.ppt

My question is : I know about integrating the product of two
functions [see Section 6.1 at
http://mathews.ecs.fullerton.edu/c2002/ca0601.html] but i don't see
how you do end up with an expression containing "e" [i.e. the "exp"
factor] by performing integrations on the Fourier coefficients above
it which contain sine and cosine functions ?

Your help gratefully appreciated.
Sincerely
Celal Berker
London, England

 

[Justin Lazear]

Maybe I'm reading too quickly (or probably not at all), but that just looks like an application of the complex exponential.

$$e^{-i \theta} = \cos{\theta} - i\sin{\theta}$$

--J

 

[M98Ranger]

Dear Celal Berker,

Thank you for your question about Fourier transforms and their application in understanding Schrodinger's equation. It is great that you are taking the initiative to teach yourself about this topic and I am happy to help clarify your understanding.

Firstly, let's review what a Fourier transform is. It is a mathematical operation that decomposes a function into its constituent frequencies. This is useful in analyzing signals and waveforms, as it allows us to see the different frequencies present in a given signal. In the context of Schrodinger's equation, Fourier transforms are used to represent the wavefunction, which describes the probability of finding a particle at a given position and time.

Now, let's take a look at slide number 16 from the presentation you mentioned. This slide shows the Fourier transform of a function f(x). The expression on the left side of the equal sign is the Fourier transform, while the expression on the right side is the inverse Fourier transform. The integral in the Fourier transform is the product of the function f(x) and the complex exponential function, which contains the sine and cosine functions you mentioned.

To understand how this integral results in an expression containing "e", we need to look at the properties of the complex exponential function. The complex exponential function can be written as e^(ix) = cos(x) + i sin(x), where i is the imaginary unit. This means that the integral in the Fourier transform can be rewritten as the sum of two integrals, one containing the cosine function and the other containing the sine function. This is where the "e" comes from in the final expression.

I hope this explanation helps to clarify your understanding of Fourier transforms and their application in Schrodinger's equation. Keep up the good work in learning and exploring this topic. Best of luck in your studies.