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

It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

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.

1. Weinberg - Quantum theory of fields, Vol.1, §4.2.
• Why is the only numbered equation in this post numbered 4.2.8? o_O Jul 13, 2018 at 16:33
• @DanielSank Oh, I have a personal code: the four means I was happy when I wrote this; the two is for 2Pac; and the eight is a secret. (The equation was taken from the reference, and it has that number in the book: i.imgur.com/bMEWTfl.png) Jul 13, 2018 at 16:49

@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.

• but can this 'purely formal' expression get arbitrarily close to the operator under some norm or topology sense as we take in more terms in the expansion? Jul 13, 2018 at 13:40
• You are asking for a hidebound proof out of a formal wisecrack? The factors are all diagonal matrices. If you truncate at m, of course it works--- if you trusted your normal ordering. But then you are reduced to @Pisanty & Nahmad-Achar 2012. Surely a different question, no? Jul 13, 2018 at 13:52