Can one calculate the following operator? [closed]

Question

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).

Answer 1

I disagree with the previous comments. As Christian says, the sums all converge strongly if we interpret $|x\rangle \langle y|$ as a rank 1 operator. In particular, $\hat{n}$ is a coisometry which takes the orthonormal set $\{|n\rangle, |2n\rangle, |3n\rangle, \ldots\}$ to the standard basis $\{|1\rangle, |2\rangle, |3\rangle, \ldots\}$. The "rational operator" $\frac{\hat{m}}{n}$ takes the basis vector $|km\rangle$ to $|kn\rangle$ when $kn$ is an integer. This is math.

I am not sure exactly what the question is --- maybe whether one can make rigorous the series involving the sum over $0 < R \leq 1$? I don't have time to check it now, but at first blush it looks to me like the series converges strongly to the given answer. Maybe it isn't deep, but I think it's interesting and it looks like research math to me.

Edit: I had a chance to take another look, and the sums do not all converge strongly. The operator $(\hat{1} - A^\dagger)^{-1}$ is unbounded, but the series for it clearly makes sense and is correct on finite linear combinations of basis vectors. So this gets even more interesting.

Answer 2

comment
So far it doesn't look like math to me. Let me guess some math meaning. $$ |1 \rangle,\qquad |2 \rangle, \qquad |3 \rangle, \qquad \cdots $$ are an orthonormal basis for a Hilbert space $H$. And $$ \langle1 |,\qquad \langle2 |,\qquad \langle3 |,\qquad $$ are the biorthogonal basis in the dual space $H^\dagger$. Then products like $$ \langle 2 |\;\ |5 \rangle := \langle 2 |5 \rangle $$ are inner products, so thay are scalars, but things like $$ |5 \rangle\langle 2 | $$ are tensors; belonging to a tensor product $H \otimes H^\dagger$. I guess the set of all $|m \rangle\langle n |$ is an orthonormal basis for $H \otimes H^\dagger$.

If this is right, then of course $$ \hat 1 = |1 \rangle \langle 1 | + |2 \rangle \langle 2 | + |3 \rangle \langle 3 | + \dots $$ does not converge in the Hilbert space sense.

But as an alternative, maybe we can think of things like $$ |5 \rangle\langle 2 | $$ as operators. Or infinite matrices (with respect to the orthonormal bases chosen). That is, $|m \rangle\langle n |$ is the matrix where the entry in row $m$ column $n$ is $1$, and all other entries are $0$. Or maybe the adjoint of that? Then the other things mentioned may be computed as infinite matrices also.