Does Summation Over n from -∞ to +∞ in Quantum Mechanics Equal Ψ(x)?

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SUMMARY

The summation notation in Quantum Mechanics, specifically ∑cnexp(iknx) = Ψ(x), requires that the index n ranges from -∞ to +∞ to encompass all possible combinations of n. This is essential for consistency with the Fourier series representation using complex exponentials. When n is limited to 1 to +∞, it corresponds to the transformation exp(ikNx) into sine and cosine functions. Clarity regarding the range of n is crucial for accurate interpretation.

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Homework Statement
Clarification of Summation index
Relevant Equations
see below
I have a (trivial) question regarding summation notation in Quantum mechanics. Does

∑cnexp(iknx) = Ψ(x) imply that n ranges from -∞ to +∞ (i.e. all possible combinations of n)? i.e.
n

∑exp(iknx)
-∞

I believe it does to be consistent with the Fourier series in terms of complex exponentials.
n = 1 to +∞ would then be used when exp(ikNx) -> sinx/cosx.

Just want to be absolutely sure. Thanks.
 
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knowwhatyoudontknow said:
Homework Statement:: Clarification of Summation index
Relevant Equations:: see below

I have a (trivial) question regarding summation notation in Quantum mechanics. Does

∑cnexp(iknx) = Ψ(x) imply that n ranges from -∞ to +∞ (i.e. all possible combinations of n)? i.e.
n

∑exp(iknx)
-∞

I believe it does to be consistent with the Fourier series in terms of complex exponentials.
n = 1 to +∞ would then be used when exp(ikNx) -> sinx/cosx.

Just want to be absolutely sure. Thanks.
The range of ##n## should be stated or clear from the context.
 

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