Integrating exp(-i(px+qy)) dxdy over all space

In summary, the conversation discusses an integral that appears in the diffraction notes about Babinet's principle. The speaker is unsure about how the integral is derived and wonders if it has something to do with the completeness relation. The conversation also mentions that the integral can be simplified to ##\delta(p,0)\delta(q,0)##, and a quick analysis shows that this is the case when p and q are both equal to 0. The integral is discussed further in the attached screenshot.
  • #1
I<3NickTesla
12
0
In my diffraction notes, this integral comes up on the page about Babinet's principle:

[itex]\int ^{y=\infty}_{y=-\infty} \int ^{x=\infty}_{x=-\infty} exp(-i(px+qy)) dx dy = \delta (p,q)[/itex]

I'm not sure how this integral is derived as carrying out the integration and putting in the limits seems to give infinity, is it something to do with the completeness relation?

A screenshot of the slide is attached
 

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  • #2
I think that should be ##\delta(p,0)\delta(q,0)##.
Quick analysis:
If you substitute x->x+1, the integral gets multiplied with e(-ip), but it has to keep its value at the same time. Therefore, it is 0 unless p=0. The same applies to q, of course.
If p=0 and q=0, you integrate over 1, this leads to the delta distributions.
 

1. What is the significance of integrating exp(-i(px+qy)) dxdy over all space?

This integral is commonly used in quantum mechanics to solve for the probability amplitude of a particle being found at a particular location in space. It is also used in Fourier analysis to decompose a function into its frequency components.

2. How is this integral solved?

This integral can be solved using techniques such as substitution, integration by parts, or using complex analysis methods. The specific method used will depend on the values of p and q.

3. Is there a specific range for the values of p and q in this integral?

No, there is no specific range for the values of p and q. They can take on any real or complex values, but the integral may not converge for certain values.

4. Can this integral be extended to higher dimensions?

Yes, this integral can be extended to higher dimensions by adding additional variables and integrating over them. The integral becomes more complex as the number of dimensions increases, but the same principles and techniques can be applied.

5. What is the physical interpretation of the result of this integral?

The result of this integral represents the amplitude of the particle wave function at a particular location in space. It can also be interpreted as the contribution of a particular frequency component to the overall function.

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