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In addition to @D. Tran's answer and for anyone out there (just in case), this did it for me. The first mistake I made was that I kept on thinking superimposed qubits could only hold 0s and 1s as pairs (2 states) i.e. say for $n=3$ qubits I thought it was equivalent of ##-##-## (e.g. 01-11-01 which is just $2^{n+n}$ combinations at max) but it turned out a ...


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As other have said, you can see 4 qubits as having either 16 different combinations, or as continuum-many superpositions of these combinations. (Continuum-many means "the cardinality of the real numbers") However you can also ask about the dimension of the space, whcich is a different notion of size. A single qbit is described by a vector in a 2-dimensional ...


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This depends on precisely what you mean by "combinations". Let's go back to just one qubit to be clear. For one qubit, there are two classical states, $|0\rangle$ and $|1\rangle$. These are really the only things you can count. Of course, as a qubit, you are allowed any superposition of the form $$ \alpha|0\rangle+\beta|1\rangle,\qquad |\alpha|^2+|\beta|^2=1....


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Well, it's still $2^4 = 16$ I believe. The point of superposition is that these 16 combinations can encode 16 inputs, and with something called "quantum parallelism" they can be calculated simutaneously. Even though they don't give out all $16$ results simultaneouly, the result can be manipulated in various ways to give you what you need while drastically ...


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