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Let's say I have a vector $\vec{v}=(1,0)$ and a state $|10\rangle$, and a decimal number $3$. I know that I can associate the decimal $3$ with the vector $\vec{v}$, but can I also associate the state $|10\rangle$ with the decimal $3$? Is there some kind of isomorphism between these two, which I don't think there is because the state $|10\rangle$ corresponds to a 4 dimensional vector. Can someone clarify these doubts?

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  • $\begingroup$ I'm confused about what you are calling the isomorphism. Do you mean to equate your vector to $2$ (binary $(1,0)$)? $\endgroup$ – Mark S Apr 21 at 23:04
  • $\begingroup$ yes that is what i mean $\endgroup$ – Upstart Apr 22 at 2:12
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Let's use four bits/qubits for exposition.

With four bits, you want there to be an isomorphism between the classical vectors $(d,c,b,a)$ (with $d,c,b,a\in \{0,1\}$) and decimal numbers $D$. There is the obvious mapping $D=8d+4c+2b+a$. If you have $(d,c,b,a)=(0,0,1,1)$ then you map to $3$.

When you have four qubits that you know are in $\vert dcba\rangle$ with $(d,c,b,a)=(0,0,1,1)$ (because, say, you measured them) then it's OK to have that same mapping. The qubits are, in sense, classical bits after measurement.

It sounds like you're nervous about whether this mapping is still valid even when the system is in a superposition. But why wouldn't it be? After Hadamarding $4$ qubits that are initially $\vert 0000 \rangle$ then you can think of it as a uniform superpositon of $\vert 0000\rangle, \vert 0001\rangle, \cdots, \vert 1111\rangle$ or as a uniform superposition of $0,1,\cdots ,15$ as you see fit.

EDIT

Classically, with modern electronic computers, we have an "isomorphism" between bits and voltages. For example, we can equate +0V to $0$, and +1.1V to $1$. If we have a plurality of inputs, then we can equate the $n$ input voltages to an $n$-bit binary number.

Quantum mechanically, we have an "isomorphism" between, say, qubits and spin states. For example, we can equate spin down to $\vert 0\rangle$, and spin up to $\vert 1\rangle$. If we have a plurality of spin states, then we can equate a superposition of the $n$ spin states to a superposition of an $n$-bit binary number.

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  • $\begingroup$ If I hadamard say $2$ zero states i.e $|0\rangle$, $|0\rangle$, I get (ignoring the normalizing factor) states $|00\rangle, |01\rangle,|10\rangle, 11\rangle$, this is before measuring the 2 qubits, now how do i think of them as $0,1,2,3$? This is still unclear $\endgroup$ – Upstart Apr 22 at 9:35

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