Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxn!ihnp4!qantel!hplabs!sri-unix!AI.Mayank@MCC.ARPA From: AI.Mayank@MCC.ARPA Newsgroups: net.physics Subject: Re: Quantum Field Theory Message-ID: <403@sri-arpa.ARPA> Date: Mon, 15-Jul-85 20:22:21 EDT Article-I.D.: sri-arpa.403 Posted: Mon Jul 15 20:22:21 1985 Date-Received: Thu, 18-Jul-85 06:32:09 EDT Lines: 133 From: Mayank Prakash> >> 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. Is that a joke? Now who needs to seriously study QFT? >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. WRONG. 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 position measurement does not leave you in an eigenstate of the position operator (for such eigenstates do not exist), but in a state which is localized in an arbitrarily small (depending on the accuracy of the measurement) region of space (but not of zero volume). Only such states are well defined. The beginning texts on QM talk about position operators etc. only as a convenient shorthand that allows them to explain basic concepts without going into the details of Hilbert spaces and the like. >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?) WRONG AGAIN. It doesn't matter what representation you use for the state of the system, the same things are still happening in them. What you achieve by using the momentum or the number representantion is an excellent disguise which makes it difficult to see what is going on in a position measurement. In other words, the state of the system does not depend on the representation used, the state before the measurement corresponded to a distributed system (the "particle" in this case), and the one after to a localized one. >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 the system is not disturbed, then the expectation value of N will not change with time, and (2) If a measurement is made for the value of N, thus forcing it to enter into an eigenstate of N, it will stay in that state unless disturbed. To see that the measurement does disturb the system, consider the folloawing - first make a measurement of the position, thus forcing the system to be localized in a small region of space, that is, if the position is measured again immediately, the system will be found within the same region again. Now, immediately measure N. If, as you claim, the system was not disturbed by this measurement, then a measurement of the position will find the system in the same region as before, which would be the case if the measurement of N wasn't made. If you still can't see what's wrong with your claim, then don't ask me, please go and read some books on QM (perhaps start with Feynman's lectures). >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), >. >. >. >. >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. Now is when you really start blowing through your head. Can you tell me why one cannot measure the field operators of the electrons or anything else? What is so special about the EM field that only its field operators are directly measurable? The answer of course is that no;ne of them are measurable. In the case of the EM field, however, an even stronger case can be made why it cannot be measured. One reason the quantum fields are not observable is roughly the same as the reason that X is not - they are unbounded operators, only worse in this case since we now have uncountably infinite degrees of freedom. In the case of the EM field, even the classical field corresponding to the quantum field is not observable. For in case you are not aware, (and judging from your message, you certainly are not), one cannot quantize the electric and magnetic fields directly - one quantizes the (even classically unobservable) electromagnetic potential. If the quantized fields are not obsrevable, then what are the classical fields? To answer that question, one must be rather tricky. One cannot even assume that the classical fields are the expectation values of the quantum fields, for they don't obey the Maxwell's equations. In order to get the classical limit of a theory, one does a so-called "loop expansion" of the generating functional of the Green's functions. The loop expansion is a power series expansion, in which all Feynman digrams with the same number of loops are lumped together in one term. The reason this is useful is that each loop is Feynman diagram carries with it a factor of the planck's constant, h. This means that the tree digrams, which have the lowest power of h, are the only ones that will survive when we take the limit h --> 0, and are the ones that describe the classical theory. This turns out to be give the right thing in the sense that thae classical limit of QED can be understood in terms of Maxwell's equations etc. 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. My reason for subscribing to this bboard is to have useful and interesting discussions on open questions in physics with people who know physics. - mayank. ========================================================================== II Mayank Prakash AI.Mayank@MCC.ARPA (512) 834-3441 II II 9430 Research Blvd., Echelon 1, Austin, TX 78759. II ========================================================================== -------