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Why is $\sum_a u_a(y)u_a^*(x)=\delta(x-y)$ (the completeness relation) equivalent to saying that for any arbitrary $f(x)$ s.t. $\int_{-\infty}^{\infty}|f(x)| = 1$, $f(x)$ can be expanded as $\sum_ac_au_a(x)$, where $c_a$'s are constants (the expansion theorem)?

Thanks.

Symbols:

{$u_a(x)$} are orthonormal eigenfunctions

$\delta(x-y)$ is the Dirac delta function

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    Sure, specification added.2011-08-27

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For any such $f(x)$, multiply both sides by $f(x)$ and integrate with respect to $x$, doing the left hand side term by term. You obtain $\sum_a c_a u_a(y) = f(y)$ Here $c_a = \int f(x)u_a^*(x)\,dx$.

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    Thanks! I am having a brain-dead day... sigh...2011-08-27