# Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis?

**Theorem**: any operator $\mathcal O$ may be expressed as a sum of products of creation and annihilation operators:
$$
\mathcal O=\sum_{n,m\in\mathbb N} (a^\dagger)^n(a)^m c_{nm}\tag{4.2.8}
$$
for some coefficients $c_{nm}\in\mathbb C$.

This theorem can be generalised to field theory, where $a,a^\dagger$ are indexed by continuous parameters. The proof of the generalised theorem can be found on ref.1.

For completeness, we sketch the proof here. We proceed by induction. Given $\mathcal O$, we set $$ c_{00}:=\langle 0|\mathcal O|0\rangle $$

We now claim that if we are able to fix $c_{nm}$ for all $(n,m)\leq(\ell,k)$ with $(n,m)\neq (\ell,k)$ so that $(4.2.8)$ holds for all matrix elements with $n$- and $m$-particle states, then we can fix $c_{\ell k}$ so that the same holds true for the matrix elements with $\ell$- and $k$-particle states. This is easy to see, because sandwiching $(4.2.8)$ between $\langle \ell|$ and $|k\rangle$, we get $$ \langle\ell|\mathcal O|k\rangle=\ell! k!c_{\ell k}+\text{terms involving $c_{nm}$ with $(n,m)\leq(\ell,k)$ and $(n,m)\neq(\ell,k)$} $$ whence the claim follows. By induction, the theorem is proven. $$\tag*{$\square$}$$

**References.**

- Weinberg -
*Quantum theory of fields*, Vol.1, §4.2.

@Accidental reminds you this is a theorem. To actually see it in your terms, use the infinite matrix representation of $a, \quad a^\dagger$ of Messiah's classic QM, v 1, ChXII, § 5. Specifically, your vacuum projection operator has a 1 in the 1,1 entry and zeros everywhere else.

The operator you chose is freaky to represent, but, purely formally, the diagonal operator for $N\equiv a^\dagger a$, $$ |0\rangle\langle 0|=(1+N) (1-N) \frac{2-N}{2} \frac{3-N}{3} \frac{4-N}{4} ... $$ would do the trick, once anti-normal ordered.