# Are qubits just analog, continuous classical bits?

Topologically, classical bits (cbits) are essentially special cases of qubits restricted to the poles of the Bloch sphere. However, this restriction doesn't seem to be classical per se, but is simply inherited from the historical fact that transistors are either on or off. One could very well conceive of a classical bit which lives in a "classical" Bloch sphere of sorts, i.e., imagine some tiny, fully classical sphere with a needle that can point in all directions. Obviously, one of these dimensions won't be imaginary, but topologically speaking, it would be equivalent to the quantum Bloch sphere. E.g., if the needle were pointing along the equator, then we'd have a perfect example of classical superposition, which one could argue is "coherent" since it has a well defined phase and isn't the result of a statistical mixture.

I'll try to expand on the above with an example. Let's say our classical measurement basis is $$NS$$ (North $$1$$, South $$0$$) vs. $$EW$$ (East $$1$$, West $$0$$). To answer the question "is the wind blowing in the $$NW$$ direction?", we would need at least two measurements: One along the NS basis vector and the other along the $$EW$$ vector. If the results is $$10$$, then the answer is affirmative. However, would may as well apply a "classical Hadamard gate" which rotates the classical basis so as to answer the same question with a single measurement---as opposed to two---hence exhibiting the same "speedup" as what is typically purported to be unique to quantum superposition.

Ultimately, I'm looking for the "secret sauce" that qubits have that cbits don't. Clearly, there has to be more to it than what could be possible with my example of classical, analog Bloch spheres.

Cross-posted on qc.SE

Are qubits just analog, continuous classical bits?

No. The continuous (vs. discrete) nature of qubit states is not the secret sauce that makes them so powerful. The secret sauce that is missing from a classical model is entanglement.

One minor failure of your proposed classical model of qubits is that (as far as I can tell) it doesn't have the update rule that measurements necessary force the system into the observed state. But that isn't the end of the world; you could add that in manually, and while it might not be terribly natural, that would reproduce that aspect of QM.

But the more important aspect that your model fails to capture is entanglement between qubits - more specifically, the exponential scaling of the Hilbert space dimension with the number of qubits. You're actually correct that a single qubit is essentially classical: it's a little-appreciated fact that the Kochen-Spekker theorem happens to fail for a single qubit, and there does exist a classical model that fully describes one qubit. But once you start combining multiple qubits together, the topology of the full system is not just the $$n$$-fold product of the 2-sphere, as it would be if qubits were fundamentally classical. Instead, it's a much higher-dimensional manifold $$\mathbb{C}P^n$$.

If qubits were classical spheres, then you could still describe correlations between them by simply defining a probability distribution over $$(S^2)^{\times n}$$. But such a classical probabilistic model wouldn't be capable of violating the Bell inequality, efficient quantum computing (as far as we know), and all of the other strange things that qubits can do. There just aren't enough degrees of freedom in your proposed classical model to make all of that stuff happen.

• I see how you pointed out the crux of the problem when you said that "multiple qubits are not just the n-fold product of the 2-sphere". That said, I kind of lost you after that. Is there a very basic example you could give to illustrate this? Apr 18, 2022 at 12:34
• @Tfovid, although we could encourage tparker to edit this answer to lengthen it into an even more lucid tutorial/textbook on the topic, an alternative approach is for us students of this matter to dig into each tentacle that tparker mentions, then submit fresh questions on Physics.SE for anything perplexing about that tentacle. There is much to unpack in tparker's answer here that is filled with depth of wisdom; the follow-up question digging into each tentacle might be the best way to illuminate all the nooks, crannies, & niches of his wise answer. Apr 18, 2022 at 12:55
• @Tfovid I'm inclined to agree with Andreas ZUERCHER; this is a huge topic, and basically the entire field of quantum information theory is dedicated to exploring your question, so I don't think it's worth going into more detail in this forum. I might suggest that you look into either an intro quantum computing textbook or a discussion of Bell's theorem. But to (very minimally) address your question, I'd just point out that if you pick any two-qubit Bell state, then there's no obvious way to represent that state with one point on each of two different balls. (Try it!) Apr 18, 2022 at 14:13

Your proposed qubit of an arrow on a sphere is a fair model of a single qubit, but it does not scale to multiple qubits. Your qubit's state can be described by a single complex number. If you had $$N$$ Of your classical spheres you could describe the total state with $$N$$ complex numbers, but $$N$$ qubits require $$2^N$$ complex numbers to describe the state. As explained in the other answer, this is because your classical model doesn't capture entanglement.

This is the exponential scaling of quantum resources. With your classical spheres you would be able to model $$N$$ qubits if you had $$2^N$$ spheres, but that's not really better than using a classical computer to model a quantum system, which doesn't scale that well.

It's true that the analog nature of quantum computers is part of the secret sauce that makes them powerful (it's also their achilles heel in that it makes them error-prone), but it's more than that. Another analog classical model that captures complex amplitudes would be discrete acoustic modes of a vibrating string, but again, what you miss out on is the exponential scaling due to entanglement. In quantum mechanics if you add one "thing" (a qubit) you double the dimension of the Hilbert space you model. As far as I can tell, if you try to make a classical analog linear algebra solver, you will need to add $$2^{N-1}$$ "things" to scale your model from $$N-1$$ dimensional to $$N$$ dimensional.

• I am a little bit confused by the phrasing where you said "If you had $N$ qubits you could describe the total state with $N$ complex numbers, but $N$ qubits require $2^N$ complex numbers to describe the state." Isn't a qubit described by two complex numbers (one for the $0$ and the other for $1$)? How does the $2^N$ arise? Apr 18, 2022 at 13:15
• @Tfovid oops that was a typo, I edited Apr 18, 2022 at 14:05
• @Tfovid but anyways, you seem to be misunderstanding the very basic $2^N$ scaling of quantum Hilbert space. You should read some about this. But basically if you have 2 qubits you have 4 basis states in your Hilbert space now: $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$. A generic quantum state is a superposition of these 4 states with each state getting a weighting by a complex number. Apr 18, 2022 at 14:08
• I understand that 2 qubits live in a superposition of 4 basis states, but I could say the same of 2 cbits. They also live in 4-D, hence my confusion as where the "quantum advantage" lies---especially if one were to go with that classical Bloch sphere idea I proposed. Apr 18, 2022 at 14:16
• No. Two cbits live in a discrete space with $2^N$ discrete states. This is very different than a $2^N$ dimensional Hilbert space which has an infinite number of states. Your Bloch sphere idea has states which live in an N dimensional complex space (because you need N complex numbers to describe the state of N of your spheres). To wit: your spheres can’t be entangled Apr 18, 2022 at 14:18