# Number theory to estimate lower bound of S-Matrix?

I recently worked on the following idea: https://math.stackexchange.com/questions/2325724/eigenvalue-of-an-euler-product-type-operator

Edit: And realised it was more probable to be used in Quantum field theory.

## Summary of the idea

We represent numbers by infinite dimensional matrices such as $3$ will have all $0$s except the $1$st row will have a $1$ in the third column, the $2$nd row will have a $1$ in the $6$th row and the $r$th row and $3r$th column also have $1$s:

$$\hat 3 = | 1 \rangle \langle 3 | + | 2 \rangle \langle 6 | + | 3 \rangle \langle 9 | + \dots = \begin{bmatrix} 0 &0 & 1 &0 & 0& \dots & 0 \\ 0 &0 & 0 &0 &0 & 1 & \dots & \\ \vdots \\ 0 & 0 & 0 & \dots \end{bmatrix}$$

Similarly we can define $\hat 2$:

$$\hat 2 = | 1 \rangle \langle 2 | + | 2 \rangle \langle 4 | + | 3 \rangle \langle 6 | + \dots =\begin{bmatrix} 0 &1 & 0 &0 & 0& \dots & 0 \\ 0 &0 & 0 &1 &0 & 0 & \dots & \\ 0 &0 & 0 &0 &0 & 1 & \dots & \\ \vdots \end{bmatrix}$$

One notices that these numbers obey multiplication

$$\hat 6 = \hat 3 . \hat 2 = \hat 2 . \hat 3$$

where the $r$th row of the $6r$th column has a $1$

One can also use this to define an Euler like product formula:

$$\hat \zeta (s) = \hat 1 + \hat 2^s + \hat 3^s + \dots = (1- \hat 2^s)^{-1}(1- \hat 3^s)^{-1}(1- \hat 5^s)^{-1} \dots$$

Note: $\zeta(1) |\lambda \rangle = \sum|\text{factors of } \lambda \rangle$

## My Observation

Let us define the following ladder operators $a$ and $a^\dagger$ from quantum mechanics:

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

Now we make the following observation:

$$\langle m | A^{\dagger s} (1 - A^\dagger)^{-1}|n \rangle \geq \langle m |(\hat \zeta(1) - \sum_{r=1}^s \hat r)|n \rangle$$

## Question

Given the field operators can be expressed as annihilation and creation operators. Can we get a lower bound on the $S$ matrix?

$$S(a^\dagger, a) \geq S'(\hat \zeta,\hat \zeta^\dagger)$$