How is the set of states $Q$ logically replaced by a Hilbert space?

Question

Question is: How is the set of states $Q$ logically replaced by a Hilbert space if a classical Turing machine is described by a 7-tuple $M=\langle Q,\Gamma, b,\Sigma, \delta, q_ {0},F\rangle$?

I read here that the set of states Q is replaced by a Hilbert space.
But I want to understand how Hilbert Space is represented in quantum electronics, using Quantum gate circuits

I don't know if this is a physical question but I think that is important to understand what way we need to use if we want represent with (quantum) logical circuits this 'Hilbert space'.

I try to read also here but is not very clear how a Hilbert space is manipulated as a quantum logical circuit https://en.wikipedia.org/wiki/Quantum_finite_automaton

Answer 1

But I want to understand how Hilbert Space is represented in quantum electronics, using Quantum gate circuits

Here is a picture of a quantum gate circuit. Those input states --- $ |\psi\rangle , |0\rangle$ --- are elements of the Hilbert space that the computation takes place in.

How is the set of states Q logically replaced by a Hilbert space

This is harder to answer completely in short form. All I can say is that there is some function which takes as an input an element of $Q$ and maps it to an element of Hilbert space. For example, a trivial map would take any element of $Q$ and map it onto $|0\rangle$. That's not very useful, but it's the kind of thing you're asking about. More complicated maps exist.