# 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.

Thank you in advance for any replies.

• 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. – Michael Oct 15 '13 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. – Trimok Oct 15 '13 at 10:02

## 1 Answer

Let $[a,a^{\dagger}]~=~1,$ and let $|0\rangle$ be the vacuum state: $a|0\rangle=0$. Define

$$|n\rangle~:=~ \frac{1}{\sqrt{n!}}(a^{\dagger})^n|0\rangle.$$

Then $$a |n\rangle ~=~ \sqrt{n} |n-1\rangle, \qquad a^{\dagger} |n\rangle ~=~ \sqrt{n+1} |n+1\rangle,\qquad\langle n |m\rangle ~=~ \delta_{n,m}.$$

An arbitrary linear operator is of the form

$$T= \sum_{n,m\in\mathbb{N}_0} |n\rangle T_{nm} \langle m|, \qquad T_{nm}~:=~\langle n|T |m\rangle~\in~ \mathbb{C},$$

so it is enough to study operators of the form $|n\rangle \langle m|$. It is straightforward to see that

$$|n\rangle \langle m|~=~\sum_{k\in\mathbb{N}_0} c^{nm}_k (a^{\dagger})^{n+k} a^{m+k},$$

where there exist unique coefficients $c^{nm}_k\in \mathbb{C}$, which can be recursively obtained from the relations

$$\delta_k^0~=~\sum_{r=0}^k c^{nm}_{k-r} \sqrt{ \frac{(n+k)!}{r!} \frac{(m+k)!}{r!} }.$$