# Eigenvalue of an Euler product type operator?

## 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?

• What's the little $s$ mean? Transpose/adjoint? – user223391 Jun 17 '17 at 2:46
• It means raised to the power ... Sorry for the confusion ... – drewdles Jun 17 '17 at 2:50
• Aside from $\hat 1$, none of the operators $\hat n$ appear normal (i.e. $\hat n^\dagger\cdot\hat n\neq\hat n\cdot\hat n^\dagger$, where $\dagger$ is the adjoint). So I would be hesitant to call $\hat H$ similar to a Hamiltonian operator, or a physical observable. But I think this construction is interesting, and you might be able to do something else with it. I wish you success! – user254433 Jun 17 '17 at 4:11
• @user254433 Thank you very much ... I think I managed to construct a hermitian Hamiltonian from this ... I would be interested in your thoughts – drewdles Jun 17 '17 at 5:05
• The 2x2 matrix Hamiltonian reminds me of a coupled two-state system. Another similarity to this is that $H$ is bounded (continuous in $L^2$), which means, in particular, the eigenvalues (energies) are bounded. This is unlike the usual unbounded operator $-\nabla^2+V$, which has arbitrarily large energies. Also, another Hermitian operator you could try is $H+H^\dagger$. – user254433 Jun 17 '17 at 6:22