I read everywhere that $N$ qubits correspond to $2^N$ bits. Let's start with 1 qubit, which is commonly represented by $\alpha |0\rangle + \beta |1\rangle$ where alpha and beta are complex numbers. This looks to me like infinite bits. How do they choose alpha and beta so that it becomes 2 bits?

Edit: Why is it commonly said that $N$ qubits correspond to $2^N$ bits?


2 Answers 2


Despite what badly written pop-science explanations of quantum computation may tell you, a qubit is not two classical bits and $N$ qubits is not $2^N$ bits. Qubits are fundamentally different from bits, pairs of bits or anything else (except for a representation of $\alpha$ and $\beta$ in binary to precision sufficient to classically simulate the computation to whatever precision is required).

As for why people talk about $N$ qubits being equivalent to $2^N$ bits, it's hard to say. To paraphrase Tolstoy, all correct explanations are alike but all incorrect explanations are incorrect in their own particular way. Perhaps it comes from the widespread mistaken belief that quantum computers somehow try all the options in parallel; perhaps it's a mistranslation of the fact that quantum computers offer exponential speedup over some classical algorithms. Scott Aaronson has some thoughts along these lines:

Fueling the belief that countless more quantum algorithms should exist (or that are not finding them is a failure), seems to be the idea that a quantum computer could just “try every possible answer in parallel”.

But we’ve understood since the early 90s that that’s not how quantum algorithms work! You need to choreograph an interference pattern, where the unwanted paths cancel

The miracle, I’d say, is that this trick yields a speedup for any classical problems, not that it doesn’t work for more of them

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    $\begingroup$ Thanks. I see everywhere that $N$ qubits correspond to $2^N$ bits. What kind of magic do they do with alpha and beta? $\endgroup$
    – asmani
    Commented Feb 10, 2018 at 22:10
  • $\begingroup$ Can you please explain what kind of precise representation of α and β do they assume? $\endgroup$
    – asmani
    Commented Feb 20, 2018 at 23:28
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    $\begingroup$ What precision does who assume, when? Any time you're simulating a physical system, you need to choose a level of precision that's appropriate to the results you're trying to get. For example, if you're calculating shortest routes between cities, you need to decide whether Berlin is 400km from Frankfurt, or 425km or 423km or 423.12347km or... $\endgroup$ Commented Feb 21, 2018 at 1:24
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    $\begingroup$ @Asmani If you're simulating a quantum computer on a classical computer then everything is represented in binary. There's no direct relationship between bits and qubits. They're fundamentally different objects. It's not productive to try to translate between them. It's like saying, "I've heard that cars are faster than horses. So how many horses would I need to go as fast as a car?" That's just not how it works. $\endgroup$ Commented Feb 21, 2018 at 14:37
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    $\begingroup$ @Asmani If you want to solve stuff on a quantum computer, you build a quantum computer. We're still in the very early stages of doing that so, at the moment, you probably can't build a quantum computer powerful enough to solve your problem. If you want to solve things on a classical computer, you use a classical algorithm. Simulating a quantum algorithm on a classical computer will be slower than just using a classical algorithm, because of the simulation overheads. $\endgroup$ Commented Feb 22, 2018 at 9:06

You can think of $ \lvert 0 \rangle $ or $ \lvert 1 \rangle $ as the standard basis of $ \mathbb{C}^2 $. In this sense, a single qubit $ \lvert \psi \rangle $ is nothing more than a unit vector in $ \mathbb{C}^2 $, that is, a linear combination of $ \lvert 0 \rangle $ and $ \lvert 1 \rangle $ with length $1$. Now, a $N$ qubit is an element of $ \mathbb{C}^{2^N} $ since it is defined via the tensor product of single qubits. Maybe that is where (the absurdity) "$N$ qubits correspond to $2^N$ bits" comes from.


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