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Let's say I have a square matrix of size $2^n\times 2^n$ with entries being 8 bit integers, where $2^n\times 2^n=b\times b\times b=2^l\times 2^l\times 2^l$, then if I want to represent that matrix in the form of a cube, is the following representation correct? $$|A\rangle=\dfrac{1}{2^{3k/2}}\sum_{i=0}^{2^l-1}\sum_{j=0}^{2^l-1}\sum_{k=0}^{2^l-1}|A(i,j,k)\rangle\otimes |i\rangle|j\rangle|k\rangle,$$ where $A(i,j,k)$ is the value at the location $(i,j,k)$ and $|A(i,j,k)\rangle$ is the binary representation of the decimal value, and $|i\rangle,|j\rangle,|k\rangle$ are the position coordinates each of length $b$ bits. Can we represent the cube like this?

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    $\begingroup$ $2^9\times 2^9=2^6\times 2^6\times 2^6$, so here $b=64$, yes binary representation means the binary of the number, for example, $7=|111\rangle $, associating a $ket$ with the binary representation, yes $A$ is the matrix with integer entries with each entry of $8$ bits. $\endgroup$ – Upstart Apr 24 at 20:54
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To be clear, you essentially have a cube consisting of $2^l$ points along each direction, and associated with each point $(i,j,k)$ is an 8 bit integer $A(i,j,k)$?

In that case, a state of $3l+8$ qubits, $$ |A\rangle=\frac{1}{2^{3l/2}}\sum_{i,j,k=0}^{2^l-1}|A(i,j,k)\rangle|i\rangle|j\rangle|k\rangle $$ is one possible way of representing the data. It is very reminiscent of quantum fingerprinting schemes.

The real question is what you want to use such a representation for? That will determine if the representation is any good. What you cannot do is use single copies to deterministically extract information about the values of $A(i,j,k)$. This should be obvious - you cannot use $3l+8$ qubits to give you arbitrary access to $24l$ bits of data.

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  • $\begingroup$ Yes that is the representation that i also wrote, except for the typo in the normalizing part. I didn't understand the last part of your answer? $\endgroup$ – Upstart Apr 25 at 7:37
  • $\begingroup$ Well, you question "Can I represent it like this?" is answered with a qualified "yes" (it contains all the data, but it depends on what you need to do with the data). I'm trying to check why you want to represent that data as a quantum state in order to determine if your chosen representation is any good for that purpose. $\endgroup$ – DaftWullie Apr 25 at 7:46
  • $\begingroup$ actually, i need this for information scrambling purpose. $\endgroup$ – Upstart Apr 25 at 7:48

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