3
$\begingroup$

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 physics.SE

$\endgroup$
2
  • 2
    $\begingroup$ Qubits can be entangled and there is no classical way to do this. If it weren't for this, quantum computing is extremely similar to classical analog computing but with complex amplitudes. $\endgroup$
    – Eelvex
    Commented Apr 18, 2022 at 13:05
  • 1
    $\begingroup$ Have you read about "rebits"? These seem to be a lot like the model you're talking about (if I've understood correctly). They are universal for quantum computing (i.e. can compute with the same efficiency as any other quantum computer) and therefore presumably faster than our standard notion of classical computers. $\endgroup$
    – DaftWullie
    Commented Apr 20, 2022 at 10:24

2 Answers 2

1
$\begingroup$

I consider only one qubit in this explanation.

A qubit does not contain as much information as a "continuous classical needle" living in a sphere, in the following sense.

If I assume that it is possible to perfectly measure this classical needle, I could access information about its orientation with only one copy of this needle. Conceptually I could imagine measurement that do not perturb its state and with enough repetition on this unique needle I can characterize its orientation.

For a qubit, the situation is different. If it can be in an arbitrary initial state, it is conceptually impossible to design an experiment that will tell you which state it is in with certainty. This is because when you will do the first measurement, you will modify the quantum state that is encoded ( * ). Hence you will possibly extract a little bit of information after the first measurement but not enough to characterize the quantum state it was in.

With one qubit, in some sense, we cannot store more than one bit of information. A way to do it is to encode from the start the state in either $|0\rangle$ or $|1\rangle$ and to perform a measurement in the $|0\rangle, |1\rangle$ basis. Doing so you will find with certainty the state the qubit is in. But this behave exactly as a classical bit. Another way to see it is that the Von Neumann entropy of a qubit is upper bounded by $\log(2)$ (same maximum entropy than for a classical bit).

( * ) Apart from if "by chance" the initial state of the qubit was an eigenstate of the observable you measured. But if you want to fully characterize the qubit state you will need "at some point" to measure an observable for which the qubit state is not an eigenstate. And the same issue will arize.

$\endgroup$
1
$\begingroup$

Another point of view for this could be a complexity one. Namely, we know how to efficiently simulate analog computation classically. Whereas we still haven’t figured out how to do so for quantum computation.

This suggests (not proves) that quantum computation and analog computation are completely different.

$\endgroup$
2
  • 1
    $\begingroup$ Hello. Maybe to be a little bit more accurate: we can simulate quantum computers on classical (digital) computers, but not in polynomial time. The paper you refer to seems to suggest that we can simulate analog classical computers in polynomial time with a classical digital computer, under some (reasonable) hypothesis from complexity theory. $\endgroup$ Commented Apr 20, 2022 at 15:05
  • $\begingroup$ Thanks, I forgot to add that we can’t simulate QC efficiently on a classical device. $\endgroup$ Commented Apr 20, 2022 at 15:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.