How does $\sum_k \psi_k^*(\vec{r})\psi_k(\vec{r}')=\delta(\vec{r}-\vec{r}')$ express the completeness of a basis?

Question

The annihilation field operator is defined as $$\hat{\psi}(\vec{r})=\sum_k \hat{b}_k \psi_k(\vec{r})$$ Two of these operators satisfy the commutation relations $$[\hat{\psi}(\vec{r}),\hat{\psi}^{\dagger}(\vec{r}')]=\delta(\vec{r}-\vec{r}')$$ $$[\hat{\psi}(\vec{r}),\hat{\psi}(\vec{r}')]=[\hat{\psi}^{\dagger}(\vec{r}),\hat{\psi}^{\dagger}(\vec{r}')]=0$$

My textbook (Nazarov and Danon) states that these can be derived from the definition of the field operator if one notes the completeness of the basis which is expressed by $$\sum_k \psi_k^*(\vec{r})\psi_k(\vec{r}')=\delta(\vec{r}-\vec{r}')\tag{1}$$ How is the above equation an expression of completeness though? Similarly, given a complete basis $\{\psi_k\}$, how would one go about proving that condition (1) is satisfied for the given complete basis? The way that I would usually express the completeness of a basis is with the relation $$\sum_k |\psi_k\rangle \langle \psi_k |=\hat{1}$$ But this expression is seemingly nothing alike the expression in eq 1. So how does the condition shown in eq 1 express the completeness of a basis?

Answer 1

If $$ \sum_k |\psi_k\rangle \langle \psi_k|= {\mathbb I} $$ then, sandwiching this expression between $\langle x|$ and $|x'\rangle$, we have $$ \sum_k \langle x|\psi_k\rangle \langle \psi_k|x'\rangle= \langle x|x'\rangle $$ or, since $\langle x|\psi_k\rangle\equiv \psi_k(x)$ and $\langle \psi_k|x'\rangle = (\langle x|\psi_k\rangle)^*$, and also $\langle x|x'\rangle= \delta(x-x')$, your eq 1 becomes
$$ \sum_k \psi_k(x) \psi^*_k(x')= \delta(x-x') $$