How does a qubit represent the number two in quantum computing?

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A qubit can exist in a superposition of both 0 and 1 simultaneously, allowing it to represent multiple values, including the binary representation of two (10). When measured, a qubit collapses to a single state of either 0 or 1, but its potential to represent multiple states is what differentiates it from a classical bit. The probabilities of these states are determined by the coefficients in the qubit's representation, which also allows for quantum interference. Additionally, qubits can be entangled, enabling a system of qubits to represent a vast number of states, exponentially increasing their computational power. Understanding these principles is crucial for grasping the capabilities and applications of quantum computing.
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Qubit can be ZERO and ONE at the same time. Right?

1 and 0 can represent TWO (10) in a binary system. Right? Therefore one qubit can represent number 2. Right?

My question. When this qubit is used to give a result of a calculation (is measured/evaluated somehow at the end of a calculation) it collapses into only one defined state ONE or ZERO. Right? HOW IS IT MADE THAT THE ANSWER (from this qubit) IS THE NUMBER TWO (which it was supposed to be)?

Please excuse a layman for trying to get some idea of quantum computing and many thanks for attention.
 
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The difference between classic bit and qubit is that, while bit can take one of two different values ( 0 or 1) , qubit is a superposition (sum) of the two states simultaneously. When evaluated, qubit is presented in one of the two possible states ##|0\rangle## , ##|1\rangle## with a certain probability to be in one of them. The sum of the probabilities is 1 or 100%. One qubit is not equivalent to a classic bit - which has probabilities let's say p1, p2 to be 0 or 1 respectively, even though these may be equal to the probabilities of qubit, to be in the state ##|0\rangle## or ##|1\rangle##. The difference is that quantum superposition in qubit, encodes a phase between the two states - besides the probabilities, allowing interference of the two states. Also, the probability of each state is given by the square of the coefficient that define the specific set up of a qubit. As a formula: 1 qubit = ##a\cdot |1\rangle + b\cdot|0\rangle## where ##\left|{a}\right|^2 + \left|{b}\right|^2 = 1##
 
Well apart from all of that, one important distinction between qubits and bits is not only that the qubits can exist in superposition but they can also be entangled.
 
Not only can a qubit be in a superposition of 0 and 1, but a set of "N" qubits can be in a superposition of up to "2 to the N" states.
So, for example, 3 qubits can be in a state where they may code for 2 (010), 3 (011), 5(101), or 7(111), but not the other 4 codes. And 8 qubits could code for all prime numbers up to 251 or all of the composite numbers up to 255.

But simply coding for "2" may not be that useful. Most of the things you might want to do with "2" can be done with regular computers.
 
Hey Ut-Napishtim.

You should consider all possible states existing in super-position instead of having only one fixed value across the spectrum of states.

That is basically what the quantum aspect is - instead of having a definitive state you have everything happening at once and you have processes that decide how all of these things are entangled and eventually collapsed to some observable (single) state.
 
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