- #1

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i've found the following expression:

How do they get that? They somehow used the kronecker delta Sum_k exp(i k (m-n))=delta_mn. But in the expression above, they're summing over i and not over r_i??

Best

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- Thread starter Faust90
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- #1

- 20

- 0

i've found the following expression:

How do they get that? They somehow used the kronecker delta Sum_k exp(i k (m-n))=delta_mn. But in the expression above, they're summing over i and not over r_i??

Best

- #2

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Do you need more background or is the question not precise enough? :-)

- #3

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The index ##i## under the sum refers to the subscript ##i## under the ##r##. The other i is the imaginary number. The Kronecker delta gets rid of the exponential and thus the sum on ##i## anyway. Try this website and see if it clears any confusion you are having. http://www.physicspages.com/2014/11/09/discrete-fourier-transforms/

- #4

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thanks for your answer. Yes, but my problem is that the sum is not running over r_i but over i.

Let's assume the r_i are an set of positions, for example always the same position, i.e. r_i={1,1,1,1,1,1,....}. Then in the end, that's just a product

prod_n=0^\infity e^{i(k-q)}

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