# Quantum representation of cube

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?

• $2^9\times 2^9=2^6\times 2^6\times 2^6$, 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. Apr 24 '19 at 20:54

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.