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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – Michael
    Commented Oct 15, 2013 at 1:18
  • 1
    $\begingroup$ 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. $\endgroup$
    – Trimok
    Commented Oct 15, 2013 at 10:02
  • $\begingroup$ January 13th, 2024: 404 Link Not Found. $\endgroup$
    – Qmechanic
    Commented Jan 13 at 8:12

1 Answer 1

  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, $$\begin{align} (a^{\dagger})^n a^m~\stackrel{(5)}{=}~& (a^{\dagger})^n \sum_{k\in\mathbb{N}_0}|k\rangle\langle k|a^m\cr ~\stackrel{(3)}{=}~& \sum_{k\in\mathbb{N}_0} \frac{\sqrt{(n+k)!(m+k)!}}{k!} |n+k\rangle\langle m+k|. \end{align}\tag{9}$$

  • $\begingroup$ I am concerned about the completeness of the basis. How can we be sure that all possible states of the theory can be expressed as a linear combination of the |n> states ? btw thank you for replying. $\endgroup$
    – M111
    Commented Oct 5, 2022 at 12:47
  • 1
    $\begingroup$ We can't. Completeness is an (implicit) assumption. $\endgroup$
    – Qmechanic
    Commented Oct 5, 2022 at 12:54

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