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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).

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closed as unclear what you're asking by YCor, Andy Putman, Jan-Christoph Schlage-Puchta, Ben McKay, András Bátkai Jul 22 '18 at 8:18

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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.

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  • $\begingroup$ We can define the rational operators and creation operators as operators on the vector space $\mathbb R^{\mathbb N}$. Then we can ask whether there exists a Hilbert norm (or Banach norm) such that all these operators are bounded and the sums converge. In this case they extend nicely to the completion. Maybe it's better if the basis vectors are not orthonormal. $\endgroup$ – Will Sawin Jul 1 '18 at 17:28
  • $\begingroup$ That's not how I would have thought about it ... all these operators are closable and we have a lot of technology for dealing with closed unbounded operators on Hilbert space. That's the approach that seems natural from my background. $\endgroup$ – Nik Weaver Jul 1 '18 at 21:40
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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.

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    $\begingroup$ We're of course dealing with Dirac's bra/ket notation here, and $|x\rangle\langle y|$ is the rank one operator $\langle y, \cdot\rangle x$. The series does converge to the identity strongly. The question is hardly suitable for MO though. $\endgroup$ – Christian Remling Jun 1 '18 at 1:44

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