This is probably a dumb layman question but as I understand it quantum computing makes use of the superposition principle in which a qbit can be in more states at once until it’s observed. A wavefunction has been described in a metaphor as if a person is missing in New York, the wavefunction describes him to be anywhere in that area, and when the police find him in Central Park the wavefunction collapsed. But if a wavefunction is not a physical object and a superposition only describes our knowledge (according to the Copenhagen interpretation) of the system how can it help with computing?
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$\begingroup$ The missing-person metaphor is not valid. It suggests that if $X$ is something we can measure (an observable, like "where's the missing person") and if $S(X)$ is the set of possible outcomes of such a measurement (such as "in Central Park"), then every physical state always has one of the properties in $S(X)$, whether or not we know which property it has. That is wrong. Most physical states don't have any of the properties in $S(X)$, and this can be shown experimentally (say, by violations of the CHSH inequality). In technical terms: most observables don't commute with each other. $\endgroup$– Chiral AnomalyCommented Sep 20, 2019 at 18:21
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$\begingroup$ ... I posted that as a comment, not as an answer, because it's really just setting the stage for this request for clarification: Is this meant to be a question about quantum computing? Or is it meant to be a question about the fundamentals of quantum theory, using computing as an example to illustrate the issue? $\endgroup$– Chiral AnomalyCommented Sep 20, 2019 at 18:21
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$\begingroup$ Both but more the last one. I would like to know what superposition means if the wavefunctiom is only describinh knowledge of the system. As it is been described to mem although the metaphor can be wrong. $\endgroup$– gerrit uitdenhagenCommented Sep 20, 2019 at 19:30
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1$\begingroup$ By the way, I wouldn't call this a dumb question at all. Learning quantum physics is difficult. You can rest assured that it has been difficult for everybody. It would be difficult even if everything ever written about it were perfectly accurate, and that's a very hypothetical "if"! $\endgroup$– Chiral AnomalyCommented Sep 20, 2019 at 19:32
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$\begingroup$ The same question could be ask about vectors $\endgroup$– BioPhysicistCommented Sep 24, 2019 at 0:30
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5 Answers
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Even if a wave function is not a physical element of reality, that is no sufficient reason for it not being useful for quantum computing.
Imagine for example you have a coin described by a classical probability distribution. I.e. its probability of heads is $p_1$ and that of tails is $p_2 = 1-p_1$. Now suppose additionally you have some kind of yes-no problem and some way of manipulating the probability distribution of your coin such that if the answer to your question is yes, the probability of the coin to be heads after your procedure is $p_1 = 75\%$. Conversely if the answer to your problem is no, the probability for tails becomes $p _2 = 75\% $. Then to get the answer to your problem you just get a bunch of coins, apply your procedure and after sufficiently many results (suppose 73 heads, 27 tails) you can be reasonably sure that your answer is yes.
So what you have done is manipulated the probability distribution of a coin to gain insight into some problem. But nobody would argue that the probability distribution of the coin is an element of reality. At any point there exists only the physical state of the coin, the probability distribution is just there because of our ignorance of the true state of the coin. In your words the probability distribution is not a physical object, but still it can in principle be used for computation.
The truly interesting question is whether we can devise ways of manipulating the probability distribution in the way we want. For quantum systems, which are described by a slightly fancier version of a probability distribution (a wave function) this turns out to be possible for some interesting problems.
In Summary: Even if probability (or wave functions) are not physical elements of reality, they can still be manipulated in principle in such a way as to perform computations.
answered Sep 20, 2019 at 14:20
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$\begingroup$ Are you basically saying that the strength of a quantum computer is just basically parallelisation? Rather than flipping a coin sequentially and calculate the probability, I can get a bunch of coins and get the result at once rather than wait for the many flipping to happen in a sequence. Or I am misunderstanding you? $\endgroup$– userCommented Sep 24, 2019 at 12:26
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The wave function (or more generally the state vector) may be an abstract function but the prediction that we make using this function have definite reality. After all, from the wavefunction we extract probability of outcomes of measurements, and we can certainly perform these measurements in real life.
Indeed quantum mechanics is just a set of rules to tell us how to manipulate wave functions and compute probabilities. We believe in those rules because they have been extensively tested by experiments.
answered Sep 20, 2019 at 15:31
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The characterization of quantum mechanics that you give only refers to probability, and never to phases or interference. That's an incomplete picture of quantum mechanics. If the incomplete picture were accurate, then it would make sense that quantum computing could not be useful. But the incomplete picture leaves out essential elements. For example, in the incomplete picture, we could never do double-slit interference with electrons.
On a more technical note, what you're talking about seems to be similar to a common way of trying to describe quantum mechanics in which Copenhagen-style collapse is nothing more than updating one's estimates of probability based on new information. This interpretation has been advocated by some people who advocate the "shut up and calculate" approach, but it doesn't really work. See this question: The Pusey, Barrett and Rudolph (PBR) theorem and "shut up and calculate"
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A metaphor can never give you the full picture because metaphors rely on our understanding of the macroscopic world, and nothing in the macroscopic world (as we see it) behaves exactly like the wavefunction. Your metaphor seems to suggest that the wave function is merely a probabilistic discription of a classical object, but this is not the case. The wavefunction is known to describe atoms better than any classical probabilistic description can provide.
In the Copenhagen interpretation the wavefunction is treated simply as an unphysical tool which is used to model the physical in a way that is usually consistent with experiment, but that is just one possible interpretation. The Copenhagen interpretation is one interpretation of many, and it is known to be inherently flawed because it interprets quantum mechanics in terms of classical notions; whereas we know that quantum mechanics is more fundamental than classical physics. The question of how to best interpret the wavefunction is still up for debate and is an active area of research in the philosophy of physics.
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$\begingroup$ The Copenhagen interpretation as I understand it typically relies on the notion of a classical observer. It's possible that has changed over the years since it was ill defined. $\endgroup$– DanielCommented Sep 20, 2019 at 15:11
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$\begingroup$ Still, there is no such thing as a classical measuring apparatus. $\endgroup$– DanielCommented Sep 20, 2019 at 15:17
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$\begingroup$ It nerds a classical measurement apparatus. No observer. A quantum object needs to interact with a macroscopic object (that can also be described with quantum mechanics, but because of the ammount of atoms can be described classicaly). Copenhagen is the most popular, imost phycisists think it’s correct. Mwi is extremely unpopular (I’ve done a poll). It’s Proponents are very loud. $\endgroup$ Commented Sep 20, 2019 at 15:20
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$\begingroup$ Yes there is a classical measurement apparatus, dspace.library.uu.nl/bitstream/handle/1874/24086/… $\endgroup$ Commented Sep 20, 2019 at 15:21
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$\begingroup$ That reference treats the measuring apparatus as a quantum object, and it has a different interpretation of wavefunction collapse than the traditional Copenhagen interpretation, so I would consider it a separate interpretation altogether. $\endgroup$– DanielCommented Sep 20, 2019 at 15:33
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First of all, it is important to note that a "wavefunction" is really just a way to represent a quantum state. The question is thus equivalent to the following: "how does quantum computing work if quantum states have no reality?".
The answer is that it's not clear what exactly you mean by saying that a state/wavefunction "has no reality". It does "have reality" in so far as it correctly describes nature, I don't know why you would believe that it doesn't.
If what you meant by "has no reality" is instead that some correlations possible within a quantum mechanical framework are not possible classically (that is, that quantum mechanics cannot be explained by a local hidden variable theory), then this is true. But then it's not clear why this should somehow imply that quantum computing is not possible.
