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!decvax!tektronix!uw-beaver!cornell!vax135!petsd!pesnta!amd!vecpyr!lll-crg!brooks From: brooks@lll-crg.ARPA (Eugene D. Brooks III) Newsgroups: net.physics Subject: Quantum Field Theory Message-ID: <693@lll-crg.ARPA> Date: Thu, 11-Jul-85 00:39:33 EDT Article-I.D.: lll-crg.693 Posted: Thu Jul 11 00:39:33 1985 Date-Received: Tue, 16-Jul-85 08:02:12 EDT References: <368@sri-arpa.ARPA> Organization: Lawrence Livermore Labs, CRG group Lines: 161 > Not true. Let us not confuse between the classical and quantum electromagnetic > fields. The quantum EM fields are as much unobservable as the wave function. 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. The measurement begets eigenvalues phi(x) of the field operator PHI(x) just as the position operator X of the harmonic oscillator can be measured to obtain the eigenvalue x. What you have to be careful not to confuse is the operator X of the harmonic oscillator with the scalar index x of the quantum field. To understand this clearly consider the harmonic oscillator. You can measure the operator X (the position operator) and get a value x (one of the eigenvalues of the operator X) as a result. The state becomes an eigenstate of the operator X, with eigenvalue x, as a (I think, desireable) side effect of the measurement. This is natural, what else should happen to a state when you measure it? Notice that I avoid the use of the term "wave function collapse" here as it creates a misunderstanding of what is going on and makes you want to worry about relativity and all that. It does not apply for harmonic oscillator anyway. The classical physics underlying the harmonic oscillator is Newtonian. The wavefunction is not something that is distributed in space. If you use the momentum (P) or number (N) representation to represent the state, the notion of a wavefunction that is spread in space does not even arise. The wavefunction is now spread in momentum space or in a discrete but infinitely long vector respectively. You can have it anyway you want it. (Signals going out to the wavefunction that is distributed in space to tell it to collapse??? Gag me with a spoon! Try transforming to the P or N representation and working there for a while to get rid of the tendancy for such thinking. Where are your signals propagating now?) Note that the operator X does not commute with H and as a result if you measure X at a later time you will get a new and potentially different eigenvalue "x prime", you can compute the probability of various values of x prime using QM. The particle has not moved in the classical sense that you could have followed it along a path. First is was here and then it was there. Thats all there is to it. Using the Newtonian equations of motion which the the harmonic oscillator is based on the mass can be found rather far away rather quickly. There just isn't much chance of it. 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 mechanical hat for a while. It will lead you down the road to the QND work at Caltech. Now what does all this have to do with Quantum Field theory? In Quantum Field theory the classical field (the value of the field at a given point in space) is the thing that corresponds to the the position of the mass in a harmonic oscillator. There is an operator PHI that you can measure at each point in space x, which is no longer an operator as was the position X of the mass in the harmonic oscillator. x is simply the index that determines which field operator PHI(x) you are going to measure. When you measure PHI(x), and you might indeed only measure the field PHI at only one point in space x say 0 for instance, you toss the state into an eigenstate of the operator PHI(0) with eigen value phi(0). The "wave function" is the probability amplitude associated with each possible eigenvalue of the field operator PHI(0). You see, the wave function is not spread out in space at all! Its spread out over the possible eigenvalues of the field operator PHI(0). Now of course there is an operator associated with each value of the index x which turns out to be continous in this case. Thats a whole lot of operators! The complete wavefunction for the whole field everwhere (in x) is the outer product of the wavefunctions associated with the field at each value of the index x. The wavefunctions are functions of phi(x) and not of x. You can measure all of the field operators PHI(x) as the same time and get a whole set of eigenvalues phi(x). Note that you are obtaining a value for the classical field everywhere in this case. Its perfectly measureable. It turns out that the operators PHI(x) do not commute with the Hamiltonian H and if you later measure all of the operators PHI(x) again you will get different values phi(x) prime just as for the harmonic oscillator. Now then, where are those darned photons in all of this. Suppose you transform your description of the classical field phi(x) to momentum space p. This is a transformation on the classical description BEFORE you quantize. Instead of talking about phi(x) you talk about phi~(p) which are the expansion coefficients of the classical function phi(x) into a set of continous functions corresponding to differnt wavelengths p, sines and cosines of (x) if you will, to over simplify it a bit. Now you quantize this description of the field. You can think of the operator PHI~(p) and its associated momentum, in the quantum mechanical sense, operator XHI~(p) which corresponds to the operator P for the harmonic oscillator. You can also transform in the quantum mechanical domain to the number representation N(p). When you measure N(p) you get an eigenvalue n(p) which corresponds to the number of photons of momentum p. In the case of a free field, this operator commutes with the Hamiltonian H and if you measure it again you get the same eigenvalue n(p). Again you are measuring without disturbing a darn thing, except perhaps during the first measurement. n(p) is them photons and they correspond exactly to the number of quanta n in the simple old harmonic oscillator. Its just that there is a whole lot of oscillators each denoted by the value of the index p. You can also talk of the number (photon) representation in x space instead of p space. This set of operators N(x) does not commute with the hamiltonian and if you measure the number of photons at a given position x and get the eigenvalue n(x) and then try measuring later on you will NOT get the same eigen value n(x). This is in exact analogy with the harmonic oscillator measurements of the operator X. You are measuring an operator which does not commute with Hamiltonian, and therefore if you remeasure later you won't get the same results. Suppose you measure N(x) everywhere and get 1 at x=0 and 0 everywhere else. It turns out that the sum of the operator N(x) over x does commute with the Hamiltonian H and if you measure N(x) everywhere at a later time you will get 1 for some x prime, perhaps not equal to 0 and 0 everywhere else. The total number of photons will not change. One might be tempted to say that the photon moved from x to x prime and you might even hazard a discussion on what path was taken. This is not what happed, nothing "moved" in the sense that you think of a ball that has been thrown by a baseball pitcher. The photon simply fired a phototube at position x at time t and then fired another phototube at position x prime at time t prime. If you didn't have your phototubes on inbetween you don't have any information about where the photon was inbetween t and t prime and there is no sense in even talking about it. Of course there are the domain limits that would be set by relativity, and enforce by the correct relativistic equations of motion for the FIELD. If the phototubes had been on during the interval t to t prime the "path" would have not gone out of this domain. This would be a uniquely different experiment which "kicks" the hell out of the photon before it gets to x prime and the probability of getting there as a result will be quite different. Put a phototube at each slit of an interference experiment and see what happens to the interference pattern. I assume that you have a mythical phototube that fires without absorbing the energy of the photon. All of the quantum mechanical equations of motion, wavefunctions and stuff are "in your head" just as the newtonian equations of motion are "in your head". The only reality is the firing of the phototubes. The quantum mechanical wave function is a computational entity that allows you to make some probabilistic predictions about when and where. I hope that I have case some more light on this point above but if you still don't understand then I suggest a decade (or more in most cases) of serious study. I think that the above discussion should appear in a book sometime somewhere but who really knows when the phototube will fire! After all, GOD really does play dice with universe. Copyright(c) 1985, all rights reserved. Any Unix computers on the net may copy freely, hard copies are limited to one per user. VMS couputers will be sued. :-)