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 $a^0=I$)


2 Answers 2


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. $$\tag*{$\square$}$$


  1. Weinberg - Quantum theory of fields, Vol.1, §4.2.
  • $\begingroup$ Why is the only numbered equation in this post numbered 4.2.8? o_O $\endgroup$
    – DanielSank
    Jul 13, 2018 at 16:33
  • 3
    $\begingroup$ @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) $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$
    – lurscher
    Jul 13, 2018 at 13:40
  • $\begingroup$ 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? $\endgroup$ Jul 13, 2018 at 13:52

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