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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:

Note: there are other choices of basis that satisfy these multiplicative properties such as the above

$$ \hat a \cdot \hat b = \hat b \cdot \hat a = (\hat{ab}) $$

For example:

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

We notice that it has nice multiplicative properties and hence, define the following operator:

$$ \hat O_{euler} = \hat 1^s + \hat 2^s + \hat 3^s + \hat 4^s + \dots $$

Using the Euler product formula (which seems to hold in this case as well):

$$ \hat O_{euler} = (\hat 1- \hat 2^s)^{-1} \cdot (\hat 1- \hat 3^s)^{-1} \cdot (\hat 1- \hat5^s)^{-1} \cdot \dots $$

We also notice that $O_{euler} | \lambda \rangle = \sum_{i} a_{\lambda, i}(s)| \text{factors of } \lambda \rangle $

Hence, we can now construct an observable:

$$ O_{euler} + O^{\dagger}_{euler} = O $$

Hence, the probabilities of beginning with a state $n$ and ending with a state $m$ is given by:

$$ \kappa_m \kappa_n \langle n | O | m \rangle = $$ $$ \kappa_m \kappa_n (\langle n | \sum_{i} a_{n,i}(s)| \text{factors of } m \rangle) + (\sum_{i} a_{m,i}(s) \langle \text{factors of } n |) |m \rangle)$$

where $\kappa_m$ is related to the normailzation constant of $m$ such that $|\kappa_m|^2=1$

Questions

Is it possible realize such an observable in nature? What would be the corresponding physical system if so?

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  • $\begingroup$ What are $\lambda$ and $a$ in $O_{euler} | \lambda \rangle = \sum_{i} a_{\lambda, i}(s)| \text{factors of } \lambda \rangle$, what what do you mean by "$\text{factors of } \lambda$"? is your Hilbert space finite-dimensional? what is your precise definition of $A^s$ for an operator $A$ and a complex number $s$? $\endgroup$ – AccidentalFourierTransform Dec 1 '17 at 19:14
  • $\begingroup$ @AccidentalFourierTransform $\lambda$ is an arbitrary number and $a_{\lambda,i} (s)$ are the coefficients as a function of integer $s$. So let $\lambda =4$ and $s=1$ then the output is $|1 \rangle + |2 \rangle + |4 \rangle $ $\endgroup$ – drewdles Dec 1 '17 at 19:17
  • $\begingroup$ Also, I don't think the Hilbert space can be finite dimensional (I maybe wrong) as there are infinite-dimensional matrices $\endgroup$ – drewdles Dec 1 '17 at 19:22
  • $\begingroup$ Your question title asks about a Hamiltonian, but I don't see any notion of time evolution or energy in your question; just a particular operator on an infinite-dimensional Hilbert space. $\endgroup$ – Michael Seifert Dec 1 '17 at 19:40
  • $\begingroup$ I see ... The reason I wrote Hamiltonian was because I "intially" thought as the operator is infinite dimensional it may be related to the momentum and position of the system. But I guess your right perhaps I should rewrite the tittle. Any suggestions? $\endgroup$ – drewdles Dec 1 '17 at 19:44

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