Applying the Fourier transform in quantum mechanics

[Radarithm]

I understand the Fourier transform conceptually, but I am unable to reproduce it mathematically; I am very familiar with calculus and integration, but I am taking a QM course and I need to know how to apply it. No websites or videos are able to give me a good explanation as to how I can use it, so I decided to ask for help here.
Thanks in advance.

EDIT: Sorry if this is in the wrong section

 

[jtbell]

It would greatly help people here if you could show an example that you don't understand, and say what you don't understand about it. Otherwise we're "shooting blindly in the dark."

 

[Radarithm]

I am unable to understand how this: http://gyazo.com/5c78dc5774850d609ce200efa446cfbf
and this: http://gyazo.com/f182937cb662f5e2241bea977f2929ea
are equal, and how the Fourier transform is done. I do however understand what the Fourier transform is: A way to break down a certain wave into its sinusoidal wave components (ie. WAVE = sin wave1 + sin wave2 + ...)

 

[stevendaryl]

I am unable to understand how this: http://gyazo.com/5c78dc5774850d609ce200efa446cfbf
and this: http://gyazo.com/f182937cb662f5e2241bea977f2929ea
are equal, and how the Fourier transform is done. I do however understand what the Fourier transform is: A way to break down a certain wave into its sinusoidal wave components (ie. WAVE = sin wave1 + sin wave2 + ...)

The proof of that is a little complicated, but I can give you an intuitive argument.

Start with the discrete version. Suppose we have a function $f(x)$ that is created by adding up a bunch of sines and cosines:

• $f(x) = \sum_n C_n e^{\frac{n \pi i x}{L}}$
• $C_n = \frac{1}{2 L} \int_{-L}^{L} f(x) e^{\frac{-n \pi i x}{L}} dx$

Now, to make the forward and reverse transforms look more alike, let me introduce a new variable, $k_n$ defined by $k_n = \frac{n \pi}{L}$, then $\delta k = k_{n+1} - k_n = \frac{\pi}{L}$

In terms of $k_n$, you can write the forward and reverse transforms as:

• $f(x) = \frac{1}{2 \pi} \sum_n (2 L C_n) e^{i k_n x} \delta k$
• $2L C_n = \int_{-L}^{L} f(x) e^{-i k_n x} dx$

Now, let's define $F(k_n) = 2 L C_n$, so we have:

• $f(x) = \frac{1}{2 \pi} \sum_n F(k_n) e^{i k_n x} \delta k$
  
• $F(k_n) = \int_{-L}^{L} f(x) e^{-i k_n x} dx$

Now, as $L \rightarrow \infty$, $\delta k \rightarrow 0$. Then the discrete sum for $f(x)$ approaches an integral:

$f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk$

So in this limit, the forstward and reverse transformations look like:

• $f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk$
  
• $F(k) = \int f(x) e^{-i k x} dx$

 

[Radarithm]

The proof of that is a little complicated, but I can give you an intuitive argument.

Start with the discrete version. Suppose we have a function $f(x)$ that is created by adding up a bunch of sines and cosines:

• $f(x) = \sum_n C_n e^{\frac{n \pi i x}{L}}$
• $C_n = \frac{1}{2 L} \int_{-L}^{L} f(x) e^{\frac{-n \pi i x}{L}} dx$

Now, to make the forward and reverse transforms look more alike, let me introduce a new variable, $k_n$ defined by $k_n = \frac{n \pi}{L}$, then $\delta k = k_{n+1} - k_n = \frac{\pi}{L}$

In terms of $k_n$, you can write the forward and reverse transforms as:

• $f(x) = \frac{1}{2 \pi} \sum_n (2 L C_n) e^{i k_n x} \delta k$
• $2L C_n = \int_{-L}^{L} f(x) e^{-i k_n x} dx$

Now, let's define $F(k_n) = 2 L C_n$, so we have:

• $f(x) = \frac{1}{2 \pi} \sum_n F(k_n) e^{i k_n x} \delta k$
  
• $F(k_n) = \int_{-L}^{L} f(x) e^{-i k_n x} dx$

Now, as $L \rightarrow \infty$, $\delta k \rightarrow 0$. Then the discrete sum for $f(x)$ approaches an integral:

$f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk$

So in this limit, the forstward and reverse transformations look like:

• $f(x) = \frac{1}{2 \pi} \int F(k) e^{i k x} dk$
  
• $F(k) = \int f(x) e^{-i k x} dx$

Made things a lot clearer, thanks!