Background
Let us have the following orthonormal basis such that:
$$ \langle m | n \rangle = \delta_{mn}$$
Consider the following operators defined as:
$$ \hat 1 = | 1 \rangle \langle 1 | + | 2 \rangle \langle 2 | + | 3 \rangle \langle 3 | + \dots $$ $$ \hat 2 = | 1 \rangle \langle 2 | + | 2 \rangle \langle 4 | + | 3 \rangle \langle 6 | + \dots $$ $$ \hat 3 = | 1 \rangle \langle 3 | + | 2 \rangle \langle 6 | + | 3 \rangle \langle 9 | + \dots $$ $$ \vdots $$ $$ \hat n = | 1 \rangle \langle n | + | 2 \rangle \langle 2n | + | 3 \rangle \langle 3n | + \dots $$
Hence, we notice it these operators have the following properties:
$$ \hat a \cdot \hat b = \hat b \cdot \hat a = (\hat{ab}) $$
For example:
$$ \hat 2 \cdot \hat 2 = \hat 4 $$
We notice that it has nice multiplicative properties and hence, define the following operator:
$$ \hat H = \hat 1^s + \hat 2^s + \hat 3^s + \hat 4^s + \dots $$
Using the Euler product formula (which seems to hold in this case as well):
$$ \hat H = (\hat 1- \hat 2^s)^{-1} \cdot (\hat 1- \hat 3^s)^{-1} \cdot (\hat 1- \hat5^s)^{-1} \cdot \dots $$
Questions
What is the eigenvalues and eigenvectors of this operator $\hat H$? Can one define $H$ as some sort of Hamiltonian or density operator?
Edit
I just thought of the following idea. What if we define the Hamiltonian as:
$$ H' = \left[ {\begin{array}{cc} 0 & H \\ H^\dagger & 0 \\ \end{array} } \right]$$
The above is a Hermitian operator. What would it physically correspond to?