Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 8/28/84; site lll-crg.ARPA Path: utzoo!linus!philabs!cmcl2!seismo!umcp-cs!gymble!lll-crg!brooks From: brooks@lll-crg.ARPA (Eugene D. Brooks III) Newsgroups: net.physics Subject: Re: Re: Quantum Field Theory Message-ID: <713@lll-crg.ARPA> Date: Sat, 20-Jul-85 02:44:43 EDT Article-I.D.: lll-crg.713 Posted: Sat Jul 20 02:44:43 1985 Date-Received: Sun, 21-Jul-85 03:44:44 EDT References: <403@sri-arpa.ARPA> Organization: Lawrence Livermore Labs, CRG group Lines: 103 Myank, I really don't wish to trade insults with someone who is uninformed about my background but I will provide rebuttal to some of your incorrect points. > >I think you need to seriously study quantum field theory before you make this > >statement. The field operator in a quantized electromagnetic field is an > >observable. When you measure it you get a value of the "classical" field. > > Is that a joke? Now who needs to seriously study QFT? No, it was not a joke. I have very seriously studied QFT. My thesis title was "Non-pertubative analysis of some simple field theories on a momentum space lattice", Phd Theoretical Physics, California Institute of Technology, 1984. > Strictly speaking, there is no operator whose eigenvalues are the > positions. The mathematical reason for this is that the position operator is > unboundeda and has number of undesirable properties as a consequence. A The Dirac delta function does indeed have a lot of undesirable properties. To a physicist its quite acceptable as a tool for solving problems. This is not relevant to my previous points however, pick another operator that is bounded if you want to replace with X with in those discussions. If we physicists worried about filling in the epsilons and deltas we would never get anywhere as we would not be physicists. We would be mathematicians. Perhaps you are one of THEM! They are known to have trouble seeing beyond all those epsilons and deltas. You will find even the most advanced texts in quantum field theory making use of the Dirac delta function as if its a perfectly respectable wave function. When needed it will be discussed in terms of limits of gaussians. Net.physics, where physics if often confused with astrology, is hardly the place to be worrying about epsilons and deltas. > >Now suppose that you instead measure N, the number operator, and get the > >eigenvalue n. This is a positive integer. N commutes with the Hamiltonian > >H and if you measure N at a later time you will still get n. Voila!, we are > >measuring without disturbing a darn thing. Stick that under your quantum > > You are simply incredible!!! Without disturbing a darn thing? Really? N > commutes with H simply means that N is a constant of motion, that is, (1) If No, I am not that incredible, and you are correct about N being a constant of the motion. The nice behavior, lack of disturbance on measurements after the first (I did not say the first would not disturb) comes from the fact that N is a constant of the motion. That was not, however, the central point. There are other operators to measure, these include the position operator X (modulo the unimportant boundedness arguments which do not affect the outcome) or momentum operator P (which has the same unimportant problems), and ideal times to measure them for certain states in a harmonic oscillator. Consider something the QND folks at Caltech call a squeezed state. Read their papers as the stuff gets too technical for net.physics. You can, just like the N operator case above, measure the position operator for a squeezed state and do it without disturbing it if its done at the right times. These times are not infinitesimally close to each other (but are periodic in time) and X is not a constant of the motion. I'll let you figure it out, you seem have some mathematical facility. I did not claim that measuring N the first time did not disturb the system. Read my statement carefully, the SUCCEEDING measurements of N do not disturb the system. Suppose you want to detect gravity waves by making measurements on a harmonic oscillator. You want to measure without disturbing as you want to detect when the gravity wave has disturbed the oscillator. The cleanest way to do this is to measure N repeatedly. Of course, its hard to do. > Can you tell me why one cannot measure the field operators of the electrons? > What is so special about the EM field that only its field operators are > directly measurable? Yes I can, The electron field operators (ie the electron field) is not measurable basically because it is a Fermi field. Bose field operators represent measurable fields. The EM field is a Bose vector field. It is measureable and people have been measuring it since long before QM came along. The situation is similar for any other Bose field. For the Fermions you can only measure their presense in the sense of particles. Understanding these connections does require quite a bit of work in QFT. The loop expansion in h!, give me a break Mayan, of course I am acquainted with it. Like perturbation expansions, it is suspected not to converge and be only useful as an asymptotic expansion. Please be careful what you do with it. With your attitude on eplisons and deltas the use of it for anything is very POOR form. > NOTE. This is the last time I am responding to a message on this topic which > is written by someone who doesn't know what he is talking about. I am not > interested in teaching fundamentals of physics to the various people on the > net. Perhaps a good idea Mayan but certainly for the wrong reasons. You would not take me up on the blindness tests but perhaps we can help you prove to yourself that the EM field is measureable. I suggest the following experiment: Procure from a hardware store a few hundred feet of very fine wire and a helium balloon. One day during a thunderstorm inflate the balloon and attach the wire to it letting it out as high as it will go. I can't guarantee that you will prove to yourself the measureability of the EM field but perhaps you will get lucky. I remember Franklin was rumored to have done something similar so the experience may indeed be enlightening. You asked for a cup of EM field, you can consider it an oscillating vector potential if you wish, I delivered. You asked for the reason for electron field operators not being observable. I again delivered. Please leave your foot in your mouth for a while and think a little before pulling it out.