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\begin{document}

\title{Revisited: A Representation for Numbers with Quantum Bra-Kets}



\author[1]{Benedict Irwin}%
\affil[1]{University of Cambridge}%


\vspace{-1em}



  
  \date{May 26, 2024}


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\section{Introduction}
If a number is expressed as the sum of all permutations of it's divisors, then when we divide the number we can use a Kronecker delta to keep to the representation. We keep to the following representation of base states, where primes are unit vectors and composites are higher dimensional objects made with the outer product of the unit vectors corresponding to the prime factors of the composite.

\begin{align}
\ket{2}=(1,0,\cdots)\\
\ket{3}=(0,1,\cdots)\\
\ket{4}=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\\
\ket{5}=(0,0,1,\cdots)\\
\end{align}
However, this notation is cumbersome for highly composite numbers, as drawing cubes and hypercubes is tricky. We can express the these composites $N$ in a different form, $\net{N}$, which is the sum of the permutations of the divisors, for example for $6$ which is $2\times3$ and $3\times 2$ only we have
\begin{align}
\langle 6\rangle=\ket{2}\bra{3} + \ket{3}\bra{2}\\
6\ket{2} = [\ket{2}\bra{3} + \ket{3}\bra{2}]\ket{2}\\
6\ket{2} = \ket{2}\braket{3}{2}+\ket{3}\braket{2}{2}\\
\braket{3}{2}=0\\
\braket{2}{2}=1\\
6\ket{2} = \ket{3}\\
6\ket{2} \equiv 6/2
\end{align}

There is again a problem with notation on dimensionality, if a number has three divisors it is not clear how to draw the bra-ket vectors, so we introduce the following rules, where the outer numbers (leftmost and rightmost) in the bracket are the only ones accessible by left or right division \begin{align}
[a,b,c,...,y,z]\ket{q}=[a,b,c,...,y]\braket{z}{q}\\
\bra{q}[a,b,c,...,y,z]=\braket{q}{a}[b,c,...,y,z]\\
\braket{a}{b}=\delta_{ab}
\end{align}
Then for example a highly composite such as $30$ can be written in a finite, but still fairly long expression as
\begin{align*}
\net{30}=[2,3,5]+[2,5,3]+[3,2,5]+[3,5,2]+[5,2,3]+[5,3,2]\\
30\ket{5} = [2,3,5]\ket{5}+[2,5,3]\ket{5}+[3,2,5]\ket{5}+[3,5,2]\ket{5}+[5,2,3]\ket{5}+[5,3,2]\ket{5}\\
30\ket{5} = [2,3]+[3,2] = \net{6}\\
30\ket{5}\equiv 30/5
\end{align*}
here we saw right division by $5$.

We can add these composite states together, and due to the linearity of operations the representation holds \begin{align*}
\net{4}+\net{6}=[2,2]+[2,2]+[2,3]+[3,2]\\
(\net{4}+\net{6})\ket{2}=\ket{2}+\ket{2}+\ket{3}=(2,1,0,\cdots)\\
(2,1,0,\cdots):=(0,2,0,\cdots):=(0,0,1,\cdots)=\ket{5}
\end{align*}
however carrying was required. [Is this a fluke?]

What if we wish to divide composites by other composites? Then we have \begin{align}
\net{30}\net{6}=([2,3,5]+[2,5,3]+[3,2,5]+[3,5,2]+[5,2,3]+[5,3,2])([2,3]+[3,2])\\
\net{30}\net{6}=[3,5,2][2,3]+[5,3,2][2,3]+[2,5,3][3,2]+[5,2,3][3,2]\\
\net{30}\net{6}=[3,5][3]+[5,3][3]+[2,5][2]+[5,2][2]\\
\net{30}\net{6}=\ket{5}+\ket{5}\\
\end{align}
seems there are too many $5$'s


\section{Fractions and Differences}
Some things \begin{align}
([2,3]+1)\ket{2}\equiv \frac{7}{2} \to \\\ket*{7/2}=\ket{2}+\ket{3}\\
([2,5]+1)\ket{2}\equiv \frac{11}{2} \to \\\ket*{11/2}=\ket{2}+\ket{5}\\
\ket*{15/2}=\ket{2}+\ket{7}
\end{align}
Then \begin{align}
\ket{2}+\ket{5}-\ket{2}-\ket{3}=\frac{11}{2}-\frac{7}{2}= \frac{4}{2} \equiv\ket{2}\\
\ket{2}+\ket{7}-\ket{2}-\ket{3}=\frac{15}{2}-\frac{7}{2}= \frac{8}{2} \equiv\ket{4}
\end{align}

Then instead of cumbersome matrices. Represent $\ket{4}$ as the difference of two primes!

\begin{align}
2 \equiv \ket{2}\\
3 \equiv \ket{3}\\
4 \equiv \ket{7}-\ket{3}\\
5 \equiv \ket{5}\\
6 \equiv \ket{11}-\ket{5}\\
\end{align}

Further \begin{align}
\frac{16}{3}=\ket{3}+\ket{5}\\
\frac{7}{3}=\ket{2}+\ket{3}
\end{align}

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