# Is it always possible to express an operator in terms of creation/annihilation operators?

I'm referring to "Path integral approach to birth-death processes on a lattice", L. Peliti, J. Physique 46, 1469-1483 (1985), available at: http://people.na.infn.it/~peliti/path.pdf

The article is about a reformulation of the master equation for a Markov process in terms of the path-integral formalism. However my question is mainly about Quantum Mechanics.

The author defines a Hilbert space $$\mathcal{H}$$, an orthogonal basis of which is given by $$\mid n \rangle$$, $$n \in \mathbb{N}$$, with:

$$\langle n \mid m \rangle = n! \delta_{n,m}.$$

The creation/annihilation operators are defined on $$\mathcal{H}$$ as follows:

$$a \mid n \rangle = n \mid n - 1 \rangle,$$ $$\pi \mid n \rangle = \mid n + 1 \rangle.$$

and they are easily to be seen each other's hermitean conjugates, according to the scalar product just defined.

The conventions are a little bit different from Quantum Mechanics, but this is not really relevant for my question. The author implies that it is possible to rewrite every operator $$O: \mathcal{H} \rightarrow \mathcal{H}$$ only in terms (sums of products) of creation/annihilation operators.

I cannot demonstrate this assertion. I have tried taking the matrix elements of a generic operator $$O$$, and demonstrating that everything can be rewritten in terms of $$a$$ and $$\pi$$ but actually this is not working.

• If you write the generic operator $O=\sum_{n,m} o_{n,m} a^{\dagger n} a^m$ and compute its matrix elements you'll get an infinite linear system. You need to prove that this system is solvable. Note that it has structure because any term with $n+m>i+j$ doesn't contribute to the $ij$ matrix element. So I would recommend an inductive strategy grouping the equations by $n+m=0,1,2,\ldots$. All you need to do is prove that there are enough new coefficients $o_{n,m}$ at each order to solve the new equations that arise at that order. Commented Oct 15, 2013 at 1:18
• Starting from an expression : $O = \sum_{m,n=0}^{+\infty} A_{m,n} \Pi^m a^n$, we get : $O_{m' n'} = \langle m'|O|n' \rangle = \sum_{m-n=m'-n', n \leq n'} \frac{n'! m'!}{n!}A_{m,n}$. Finally, one has to express the $A_{m,n}$ in function of the $O_{m' n'}$. It is worth beginning with $n′=0$, increase $m′$ at fixed $n′$ ,then increase $n′$, and so on. Commented Oct 15, 2013 at 10:02

1. Let us first review the setting to fix the notation. Let $$[a,a^{\dagger}]~=~1,\tag{1}$$ and let $$|0\rangle$$ be the vacuum/ground state: $$a|0\rangle=0.\tag{2}$$ Define $$|n\rangle~:=~ \frac{1}{\sqrt{n!}}(a^{\dagger})^n|0\rangle.\tag{3}$$ Then \begin{align} a |n\rangle ~=~& \sqrt{n} |n-1\rangle, \cr a^{\dagger} |n\rangle ~=~& \sqrt{n+1} |n+1\rangle,\cr \langle n |m\rangle ~=~& \delta_{n,m}. \end{align}\tag{4} Consider the corresponding Fock space $${\cal H}$$ with the completeness relation $${\bf 1}~=~\sum_{n\in\mathbb{N}_0}|n\rangle\langle n|.\tag{5}$$
2. Now let us address OP's question. An arbitrary linear operator is of the form \begin{align} T~=~& \sum_{n,m\in\mathbb{N}_0} |n\rangle T_{nm} \langle m|, \cr T_{nm}~:=~&\langle n|T |m\rangle~\in~ \mathbb{C},\end{align}\tag{6} so it is enough to study operators of the form $$|n\rangle \langle m|$$.
3. Example: The projection operator on the vacuum/ground state is $$|0\rangle\langle 0|~=~:e^{-a^{\dagger}a}:~\equiv~\sum_{n\in\mathbb{N}_0}\frac{(-1)^n}{n!}(a^{\dagger})^n a^n,\tag{7}$$ cf. e.g. this Phys.SE post.
4. We can now derive a bijective correspondence with creation/annihilation operators: \begin{align} |n\rangle\langle m|~\stackrel{(3)}{=}~& \frac{1}{\sqrt{n!}} (a^{\dagger})^n |0\rangle\langle 0| a^m\frac{1}{\sqrt{m!}}\cr ~\stackrel{(7)}{=}~&\frac{1}{\sqrt{n! m!}} (a^{\dagger})^n :e^{-a^{\dagger}a}: a^m\cr ~=~&\sum_{k\in\mathbb{N}_0} \frac{(-1)^k}{k!\sqrt{n! m!}} (a^{\dagger})^{n+k} a^{m+k}, \end{align}\tag{8} and conversely, \begin{align} (a^{\dagger})^n a^m~\stackrel{(5)}{=}~& (a^{\dagger})^n \sum_{k\in\mathbb{N}_0}|k\rangle\langle k|a^m\cr ~\stackrel{(3)}{=}~& \sum_{k\in\mathbb{N}_0} \frac{\sqrt{(n+k)!(m+k)!}}{k!} |n+k\rangle\langle m+k|. \end{align}\tag{9}