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A classical bit is represented by either 0 or 1. However a superposition state is combination of 0 and 1 both with some probability ($|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$). Does it mean that creating superposition actually quantizes the classical bit ? And let us assume we applied an quantum mechanical gate to $|0\rangle$ or $|1\rangle$ to go to superposition state. Does this gate now be called as a quantized operator or quantization gate ?

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  • $\begingroup$ It is hard to assess what "be thought of as" means here. Sure a qubit can "be thought of as" as quantum version of a bit. $\endgroup$
    – Mauricio
    Aug 25, 2023 at 10:10

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Adding a little bit of background/context to @banercat's answer:

The comparison between classical and quantum bits can be a useful analog to understand why the (reversible) operations on a universal gate quantum computer preserves an exponential set of information that can be used in clever ways for specific parts of interesting algorithms (https://en.wikipedia.org/wiki/Quantum_algorithm).

That said, the quantum nature of superposition and entanglement is provably non-classical. Many scientists were thinking there might be a more classical way to interpret microscopic behavior (such as using hidden variables) but following Bell's work in the mid-sixties (https://en.wikipedia.org/wiki/Bell_test) experimentalists was able to prove there is no underlying classical explanation.

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Does it mean that creating superposition actually quantizes the classical bit ?

Creating a superposition state goes beyond the classical notion of bits. In the pure theoretical side of quantum computing, it is sufficient to think of $|\psi\rangle$ as a classical bit displaying quantum properties (effectively what a qubit is). However, it's not accurate to say that this "quantizes" the classical bit, as quantization usually refers to the process of converting continuous values (like in classical physics) to discrete values (like in quantum physics).

Does this gate now be called as a quantized operator or quantization gate?

This is a bit of a strange question ask, as these gates are not applied to classical bits, but applied to qubits, which are already quantum objects. Due to this, gates are not really referred to as 'quantization gates', but just quantum gates.

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"...A classical bit is represented by either 0 or 1. However a superposition state is combination of 0 and 1 both with some probability $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. Does it mean that creating superposition actually quantizes the classical bit...?

A "classical" bit has a qubit equivalent that is a "superposition" (no "creating superposition" is needed) - just represent it as:

$1.0|0\rangle+0.0|1\rangle$ for "0",

$0.0|0\rangle+1.0|1\rangle$ for "1".

"...And let us assume we applied an quantum mechanical gate to $|0\rangle$ or $|1\rangle$ to go to superposition state. Does this gate now be called as a quantized operator or quantization gate ?..."

An example of a gate that does this is the Hadamard gate.

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