Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.3 4.3bsd-beta 6/6/85; site ucbvax.ARPA Path: utzoo!linus!philabs!cmcl2!seismo!harvard!talcott!panda!genrad!decvax!decwrl!ucbvax!ucbmiro!rimey From: rimey@ucbmiro.ARPA (Ken Rimey) Newsgroups: net.physics Subject: Re: Heisenberg Uncertainty Principle Message-ID: <9761@ucbvax.ARPA> Date: Thu, 8-Aug-85 09:34:23 EDT Article-I.D.: ucbvax.9761 Posted: Thu Aug 8 09:34:23 1985 Date-Received: Mon, 12-Aug-85 00:42:05 EDT References: <3506@decwrl.UUCP> Sender: nobody@ucbvax.ARPA Reply-To: rimey@ucbmiro.UUCP (Ken rimey) Organization: U.C. Berkeley Lines: 99 >In over-simplified terms Heisenberg's Uncertainty Principle says that we >cannot know the simultaneous position and momentum of an individual >elementary particle with unlimited accuracy. Yet, we are able to >determine the simultaneous position and momentum of conglomerations of >these elementary particles. No, you cannot simultaneously determine the exact position and momentum of a conglomeration either. >What is different about the individual particles and groups of particles? Your intuition about groups of particles is really an intuition about relatively massive objects in the relatively large-scale everyday world. The uncertainty principle is (unc. in pos) (unc. in momentum) >= h = 6.6E-34 Js If you want to insure that some object is in its proper place to an accuracy of 1 micrometer, the uncertainty in its momentum will be 6.6E-28 kg m / s. To get the uncertainty in velocity, we divide by the mass. Let's say the thing weighs a milligram. Then it will likely have a velocity of the order of 6.6E-22 m/s - very small. The problem is that everyday distances are large, and that everyday masses are huge, by atomic standards. The mass of an electron is 9.1E-31 kg; experiment with that number. >Is it strictly a case of the measurement process itself disturbing the >individual particle, or is something else going on here? > > Mike Augeri, DEC, Maynard MA USA Something else is going on here. In classical mechanics, a particle has a position and a momentum. You can be uncertain about the position and momentum, but so what? Lack of knowledge doesn't account for the discrete energy levels of a hydrogen atom. In quantum mechanics, what you don't know is also irrelevant. However, the state of particle simply isn't described by a position and a momentum as in classical mechanics. It makes sense to ask, given a particle in some state what might I get if I measure its position or its momentum? Given the state, I could calculate probability distributions describing what I might get. If the distribution for the position measurement is very tightly peaked around a single position, I say that the state has a well-defined position. If I measure the position, I almost always get values very close to that particular one. Similarly, a state might have a well-defined momentum. Can I have a state that has both a well-defined position and a well-defined momentum? No, there is no such thing. Say you have a single particle. It is described by wavefunction, a complex number for every point in space. The Schrodinger equation determines how this wavefunction changes with time. If at some time the wavefunction is zero someplace, looking for the particle there then is guaranteed fruitless. The probability of finding the particle in a given place is given by the magnitude of the complex number there squared. (I should say "The probability of finding it in a given region ..." but you know what I mean.) Take the fourier transform of this wavefunction. This gives you the momentum space wavefunction, a complex number for each momentum. This determines the probability of finding the particle with a given momentum in the same way as the position wavefunction determines the probability of finding the particle with a given position. Note that the momentum wavefunction cannot be chosen independently of the position wavefunction - it is the fourier transform of the latter. Is it possible to have a function that is nonzero only in a very small region, such that its fourier transform is nonzero only in a very small region? No, that is not possible. The electrical engineers among you may recall that the fourier transform of a narrow gaussian is a wide gaussian and vice versa, that the fourier transform of a spike has infinite extent, and so on. This is the uncertainty principle. I'll mention two questions that my explanation above doesn't address. First, does this momentum wavefunction that I have defined really relate to the p=mv we all know and love? Sure. Take a wavefunction with a moderately peaked position distribution, and a moderately peaked momentum distribution. If you calculate how this wavefunction evolves in time according to the Schrodinger equation, you will find that the peak in the position wavefunction moves with a velocity corresponding to the momentum at which the peak in the momentum wavefunction lies. Second, if it is impossible to determine exact position and momentum simultaneously, then what happens if I try? Does lightning strike me or what? The only way I can think of to measure the position of a free particle is to bounce another particle off of it. If that other particle is to have a well-defined position, there will be a large uncertainty in its momentum, and it will transfer some of that momentum to the particle whose position is being determined. Some people's reaction to this thought experiment is "Aha! So that's what's really going on." They are fooling themselves if they think this is a non-mathematical explanation of why the uncertainty principle is true. The argument is circular. The probe particle is a large disturbance only because it itself must satisfy the uncertainty principle. Ken Rimey rimey@berkeley