On Tue, Jul 31, 2018 at 12:01 AM Bruce Kellett <bhkell...@optusnet.com.au> wrote:
> From: Jason Resch <jasonre...@gmail.com> > > On Mon, Jul 30, 2018 at 11:21 PM Bruce Kellett <bhkell...@optusnet.com.au> > wrote: > >> From: Jason Resch <jasonre...@gmail.com> >> >> On Mon, Jul 30, 2018 at 8:33 PM Bruce Kellett <bhkell...@optusnet.com.au> >> wrote: >> >>> From: Jason Resch <jasonre...@gmail.com> >>> >>> >>> You can use "itself" only if this "it" can be in multiple locations and >>> heading in different directions. >>> >>> >>> That is a property of waves. But you will only ever observe a single >>> photon from this wave..... >>> >> >> Waves/Photons, doesn't matter what you call them. >> >> Within the quantum computer this wave/photon is simultaneously in many >> different locations/doing many different things, performing computations >> and doing useful work using all of its separate superposed instances of >> itself. Once it's done doing all this work it settles down on a final >> value which we can read. And it will be correct, and may have finished an >> enormous computation in a short period of time, if and only if, it did in >> fact split up and do all these independent things simultaneously. >> >> >> Or you can view the action of a quantum computer as a simple interference >> effect. Incorrect solutions to the algorithm destructively interfere. You >> don't have to introduce ideas such as 'being in different locations and >> doing different things.' It is just simple interference in a wave. (And it >> is all in one world, because interference can only occur within the one >> world.) >> >> >> > To add some clarity, I would say interference effects of a superposed > system can only be seen from the vantage point of another system which has > not interacted with that superposed/interfering system. > > > Hmmm! I'm not sure what you mean here. Certainly, the interference from > the two-slit set up is only observable when the photons interact with a > screen. I am not an expert in quantum computers, but I understand that the > output from the computation is generally read off from a set of qubits that > are separate from, but interact with the qubits that are actually used for > the computation. Is that what you mean? > I think it depends on the computation. It may be for practical reasons that separate qubits are used to store the output, but I don't see any conceptual reason why they would need to. It is difficult enough a problem to scale a QC with more qubits that I would assume it is more practical to reuse any qubit whenever the computation allows it to be reused. > In general, though, observing something means interacting with it. > > What I was referring to with the "can only be seen" were the *effects* of the interference, be they the final results of the quantum computation or the light and dark bands in the two slit experiment. If you entangle yourself (measure/observe/interact with) the system in the superposition, then you yourself become part of that superposition, and no longer will see interference effects from that system. > > On that we agree. But where did those other photons come from? How did >> they get to be in different positions going in different directions? >> >> >> They aren't. >> > > How do do you explain the experiment with beam splitters and recombining > light at a half silvered mirror to interfere and only be reflected one way? > > > Photons have both wave-like and particle-like properties. That is quantum > physics. > > So do you accept or reject that this "wave" can be in different places > simultaneously? > > > A wave is not a localized object, so the same wave can extend to different > locations. > > So then "a photon is not a localized object, so the same photon can extend > to different locations." -- is this right or wrong? > > > A photon is what causes a localized spot on a screen. A wave is what > interferes with itself. If you want to equivocate on the meanings of words, > you can say that a photon is extended. But I prefer to keep the terms > distinct, and apply them as appropriate in different contexts. > But really there is only a photon here. There is not a wave that is separate from the photon. When a photon strikes a semisilvered mirror and splits along two paths, do you say the wave is created at that point? It seems more natural to me to simply say the photon was split by the splitter. Jason -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
