QM: Need Help Applying Fourier Transform

AI Thread Summary
The discussion revolves around the application of the Fourier transform in quantum mechanics, with a focus on understanding its mathematical representation. The original poster expresses difficulty in reproducing the Fourier transform despite grasping its conceptual basis of decomposing waves into sinusoidal components. A detailed mathematical explanation is provided, illustrating the relationship between discrete sums and integrals in the context of Fourier transforms. The conversation emphasizes the importance of clarifying specific points of confusion to facilitate better assistance. Overall, the exchange aims to bridge the gap between conceptual understanding and mathematical application of the Fourier transform.
Radarithm
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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
 
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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 said:
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
 
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stevendaryl said:
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! :biggrin:
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...

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