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?


1 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
$$ \sum_k \psi_k(x) \psi^*_k(x')= \delta(x-x') $$

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    $\begingroup$ +1: Lol, I initially thought that this is simply a restatement of the original relation without explaining as to why it implies completeness but apparently that's what the OP is asking for! ;) $\endgroup$
    – user87745
    Aug 28, 2021 at 12:39
  • $\begingroup$ Thanks for the great answer! All cleared up now. In line with what Dvij D.C. said though, is there a way to intuitively understand why eq (1) implies completeness? If I imagine the basis set to be the basis of the infinite square well $\{\psi_n\}={\sqrt(2/a)\sin(n\pi x/a)}$, Its not immediately obvious to me that applying eq (1) to this set will yield zero for $x\neq x'$. I can see how the summation over all n will yield infinity when $x=x'$ though. $\endgroup$ Aug 28, 2021 at 13:07
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    $\begingroup$ Completness is subtle. See pages 62 onwards in goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf and also later where we consider completness rlations. $\endgroup$
    – mike stone
    Aug 28, 2021 at 13:30

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