Why the closure relation is equivalent to the completeness of a set of eigenfunctions: $\{\psi _k\}$ which is complete:
$\{\psi _k\}$ is a complete basis $\iff \sum_k \psi^*(x) \psi(x') = \delta(x-x')$
What is the relation between completeness and the closure relation?
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eigenfunctions
complete-spaces
1 Answers
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Let $\{\psi_k\}$ be a set of eigenfunctions that satisfies the closure relation. For a given function $f$ you want to be able to express it as a sum of the $\{\psi_k\}$.
Starting with the closure relation multiply by $f(x)$ and integrate with respect to $x$:
$$\int \sum_k f(x)\psi_k^*(x) \psi_k(x') dx= \int f(x) \delta(x-x')dx$$
Interchanging the integral and finite sum over $k$ we get:
$$ \sum_k \psi_k(x')\Bigg[\int f(x) \psi_k^*(x)dx\Bigg] = f(x')$$
The integral $\displaystyle\int f(x) \psi_k^*(x)dx$ is a constant $c_k$. $f$ is now expressed in terms of the basis $\{\psi_k\}$.