X-Message-Number: 8570 From: Andre Robatino <> Subject: Re: CryoNet #8561 - #8567 Date: Tue, 9 Sep 97 11:59:15 EDT > Message #8567 > Date: Mon, 8 Sep 1997 22:58:14 -0700 (PDT) > From: John K Clark <> > Subject: Digital Shakespeare > <snip> > In #8556 Andre Robatino <> On Sun, 7 Sep 97 Wrote: > > >because of nonzero decoherence, the quantum state ends up being > >changed slightly which means that for a given input there are more > >than one possible output with nonzero probability. > > > OK, I'll buy that, what stuck in my craw is when you said this means it has > an infinite number of internal states, in fact if there in no relationship > between input and output it becomes pretty pointless to talk about internal > states at all. > I think I see your problem - sure, there are a finite number of input states, so there are a finite number of _intended_ internal states, but decoherence will smear each of them out after the computation starts, so the set of possible internal states at any point in the computation is what one gets by taking a finite number of points and smearing each of them out. This means that when one tries to read the output, more than one output will have nonzero probability. If the smearing is small enough, the correct output will still have significant probability which means that if one has independent means of checking the answer, the QC is useful. significant probability. > >>Me: > >>we can't measure the quantum wave function F(x) of a particle, we > >>can only measure the intensity of the wave function [F(x)]^2 > >>because that's a probability and probability we can measure. > > >True if you're doing a measurement of position, but a measurement of > >momentum involves taking partial derivatives of the wave function, > >so any change in the wave function other than a global phase factor > >leads to changes in the probabilities of measurements of various > >quantities. > > > I don't understand your distinction between position and momentum because it > makes no difference, the more certain you become about one the less certain > about the other. If you want to find the momentum of an individual photon, If two wave functions psi1(x,t) and psi2(x,t) satisfy psi2(x,0) = c(x)*psi1(x,0), where c(x) is a complex-valued function such that |c(x)| = 1 everywhere, the two have different probabilities for measurements of various quantities other than position at time 0, unless c(x) is a constant function. Also, even if one is only interested in measuring position, letting c be a function of position will cause psi1 and psi2 to have different time evolution, so for t > 0, psi1(x,t) and psi2(x,t) will no longer have the same intensity at each point. > hf, you can find it by using the formula hN/X , h is the Planck constant, > and N is the number of wave crests the photon makes over a distance X. > If we were to actually perform this experiment when we got near the end of > distance X it would not be entirely clear if we should include the last wave > crest or not. We could minimize this problem and get a more accurate > measurement of momentum by making X bigger, but then you'd know less about > the position of the photon, X, thus momentum and position have an inverse > relationship. > > >Anyway, the change in the wave function due to decoherence _does_ in > >general affect the intensity, along with the phase. > > > I never said it didn't, what I said was that we can't measure the wave > function but we can measure it's intensity. You can only measure the intensity at a given point by creating the same wave function a large number of times and measuring the likelihood of finding the particle at that point. You can do the same with any other physical quantity, so in the sense you're talking about the entire wave function (except for a global phase factor), including its phase at each point, is just as measurable as its intensity. Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=8570