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Classical bits can be in one of two states 0 or 1, but a qubit can be in a superposition of those. Thus the number of states increase exponentially so does the computational power.

I understand that the number of avalable states obviously plays a major role in the computational power but I don't understand what is the effect of the continuous range of the values of a qubit on that computational power.

Contrary to the classical bit, a qubit can have an infinite number of values, those on the surface of the Bloch sphere (at least until it is measured), so:

  1. How do the availability of infinite values affect the computational power? In a classical computer a bit is bound to use 0 and 1 but when a qubit can have an infinite amount of values how is the performance affected?
  2. What aspects of the quantum computer are affected by the availability of those infinite values?
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It isn't important that they have values on a real scale.

The point isn't just that qubits superimpose 2 states.

The point is that 2 qubits superimpose 4 states. 3 qubits superimpose 8 states.

1024 qubits can superimpose around 10^308 states.

Each of those states has a complex amplitude, which is the square root of its probability. The computer acts like a SIMD (single instruction multiple data) machine. As it runs, the amplitude (and probability) concentrates in the answer state(s) so they can be observed.

The fact that it's doing all those computations in parallel is where the power comes from.

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  • $\begingroup$ Yes I know where the power comes from. I was just wondering if there was any point in having a continuous range of values. Thank you though. $\endgroup$ – Adam Nov 18 '17 at 20:14
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We don't want the continuous range of values to have an effect. Controlling their effect for the input (preparation) and output (measurement) is easy. The accuracy of the intermediate values and results can be controlled by allowing noise. There is an error-correction theory for quantum computation, which is believed to cope with that part. (Gil Kalai is a skeptic who believes that experiments will show that this noise modelling and error-correction theory was too optimistic.)

But let me explain the part relevant to your question: If a quantum algorithm would only work if intermediate results of exponential precision were available, and no more robust reformulation of that quantum algorithm were possible, then we would not regard that algorithm as physically realizable. But the error-correction theory hopefully tells us that such a reformulation will always be possible, by only using polynomially many more qubits.

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