Summary

I recently defined some numbers which obey multiplication but not addition. To my surprise after some heuristic manipulations (ignoring convergence), it seems I can express the creation and annihilation operators from quantum mechanics in terms of these numbers! I am now interested in these numbers in their own right.

Question

Using the below arguments how does one calculate the $( \sum_{0 < R \leq 1} \hat R)^\dagger | \phi \rangle$ where one can define:

$$ I - (I + \underbrace{(\sum_{0 < R < 1}\hat R)^\dagger}_{\hat O} )^{-1} = A^\dagger$$

where $I$ is the identity.

Hat Numbers

We define the following numbers $$\hat 1 = |1 \rangle \langle 1 | + |2 \rangle \langle 2 | + |3 \rangle \langle 3 | + \dots $$

$$\hat 2 = |1 \rangle \langle 2 | + |2 \rangle \langle 4 | + |3 \rangle \langle 6 | + \dots $$

$$\vdots$$

In general,

$$ \hat n = |1 \rangle \langle n | + |2 \rangle \langle 2n | + |1 \rangle \langle 3n | + \dots $$

We notice the following:

$$ \hat x \hat y = \hat y \hat x = \hat (xy)$$

For example:

$$ \hat 2 \cdot \hat 2 = \hat 4$$

Now we also their can create fractions by taking hermitian conjugate of the hat numbers: For example, $$ \hat 2^\dagger = |2 \rangle \langle 1 | + |4 \rangle \langle 2 | + |6 \rangle \langle 3 | + \dots$$

Then we can define rational numbers now as (again for example):

$$ \hat {\frac{3}{2}} = \hat 3 \hat 2^\dagger $$

Creation and Annihilation operators

One can define a creation operator:

$$ A^\dagger | n \rangle = | n+1 \rangle$$

In fact, $$ A^\dagger = |1 \rangle \langle 2 | + |2 \rangle \langle 3 | + |3 \rangle \langle 4 | + \dots$$

Now, $\hat 1$ is the identity element and if one allows for the geometric series:

$$(\hat 1 - A^\dagger)^{-1} = \hat 1 + \hat A^{\dagger} + \hat A^{\dagger 2} + \hat A^{\dagger 3} + \dots $$

One also can express the above series using the number operators:

$$ (\hat 1 - A^\dagger)^{-1} = ( \sum_{0 < R \leq 1} \hat R)^\dagger $$

where $( \sum_{0 < R \leq 1} \hat R)$ represents the sum of all rational hat numbers less than $1$. I don't have a rigorous proof of the above equation but am very confident it is correct (in the sense the LHS and RHS have the bra-ket notation).