3
$\begingroup$

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 $\iff \sum_k \psi^*(x) \psi(x') = \delta(x-x')$

$\endgroup$
5
$\begingroup$

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 $\{\psi_k\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.