Originally Posted by

**oldguynewstudent** Let the states $\displaystyle {|n>}$ form a discrete ONB for the space of single particle, and let $\displaystyle \phi_n (\vec{r})$ and $\displaystyle \phi_n (\vec{r}^{'})$ be the wavefunctions for the state $\displaystyle {|n>}$ in the position and wavevector representations, respectively. Prove the so-called closure relations:

$\displaystyle \sum_n {\phi_n^{*}(\vec{r})\phi_n(\vec{r}^{'})=\delta(\ve r-\vec{r}^{'})$ and $\displaystyle \sum_n {\phi_n^{*}(\vec{k})\phi_n(\vec{k}^{'})=\delta(\ve {k}-\vec{k}^{'})$

The second part for k should be the same as the first.

$\displaystyle \sum_n {\phi_n^{*}(\vec{r})\phi_n(\vec{r}^{'})=\sum_n <{\phi_n^{*}(\vec{r})|n><n|\phi_n(\vec{r}^{'})>$

I am pretty sure that $\displaystyle \phi_n^{*}(\vec{r})$ can be rewritten as $\displaystyle |\vec{r}>$ which would give

$\displaystyle \sum_n <{\vec{r}|n><n|\vec{r}^{'}>$

Since $\displaystyle {|n>}$ form a discrete ONB we should get $\displaystyle \sum_n <{\vec{r}|(\textbf{1}\vec{r}^{'}>)$

I don't know how to justify $\displaystyle \sum_n <{\vec{r}|\vec{r}^{'}>=\delta(\ve r-\vec{r}^{'})$

Or am I totally off in left field again?

Thank you for any help, clarification, or sanity check.