brooks@lll-crg.ARPA (Eugene D. Brooks III) (07/11/85)
> 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. :-)
DAM%MIT-OZ@MIT-MC.ARPA (07/15/85)
I appreciated your message on quantum field theory; I found it enlightening and will read it more carefully later. However certain aspects of the message bothered me. In particular consider the following statement: Notice that I avoid the use of the term "wave function collapse" here as it creates a misunderstanding of what is going on ... What IS going on? I would be perfectly willing to listen to a mathematical description of the measurement process. Are the laws of physics governing "measurement" different from the laws of physics which govern "non-measurement" interactions such as scattering? If we assume that some events are measurements and others aren't and that we know the difference between measurement events and non-measurement events then the mathematics is clear: measurements result in eigenstates. But I find this assumed distinction between measurement and non-measurement quite bothersome. 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. But isn't a phototube a physical object? Can't we assume that a phototube is made of atoms and that our theories of atoms apply to the phototube? Or would you say that the phototube itself is just a computational device for predicting perception? Is awareness the only reality? I would like to believe in external physical reality and I would like to have a good mathematical model of what that reality is. What do you think? Is there an external reality? What is it? I'm willing to listen to the mathematics. I even think I understand it to some extent. But I don't think that the mathematics has yet answered these basic questions.
DAM%MIT-OZ@MIT-MC.ARPA (07/15/85)
It is interesting that you pointed out that the wave function is not distributed in space. I find this to be one of the most disturbing properties of quantum mechanics. The wave function is distributed in configuration space, i.e. the space of all possible CONFIGURATIONS of the system. For classical n-particaal systems this configuration space has 3n dimensions. (Footnote: Actually, as was pointed out, many representations of the wave function are possible. The wave function can be thought of as a point in a Hilbert space and the Hilbert space can have different spectral representations corrosponding to different sets of operators. But I like to think in terms of configuration space because it allows me to switch between classical and quantum-mechanical thinking.) What bothers me is that I don't fully understand the relationship between a wave function distributed over configuration space and real-live space-time. Given a wave function distributed over configuration space how can one talk about "events" which occur at particular points of "space-time"? It seems to me that the physical theory should account for events which accur in a SINGLE FOUR DIMENSIONAL SPACE-TIME MANIFOLD. But I don't see how to define such a manifold in terms of quantum field theory (but then again I don't really understand quantum field theory). As a somewhat sophisticated non-physicist it seems to me that this question would be important for understanding the relationship between quantum field theory and gravitation. If one could define a single four-dimensional "causal manifold" in terms of quantum field theory one might see why such a manifold is curved.
AI.Mayank@MCC.ARPA (07/16/85)
From: Mayank Prakash <AI.Mayank@MCC.ARPA> > >> 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 ========================================================================== -------
brooks@lll-crg.ARPA (Eugene D. Brooks III) (07/18/85)
> It is interesting that you pointed out that the wave > function is not distributed in space. I find this to be one > of the most disturbing properties of quantum mechanics. The > wave function is distributed in configuration space, i.e. the > space of all possible CONFIGURATIONS of the system. For classical > n-particaal systems this configuration space has 3n dimensions. As an added twist you might consider isospin. It behaves like spin or angular momentum except that it lives in isospin space. Its not the 3 dimensional space that we live in!
brooks@lll-crg.ARPA (Eugene D. Brooks III) (07/20/85)
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.
DAM%MIT-OZ@MIT-MC.ARPA (07/24/85)
I would be very interested to hear an explanation of how the four dimensional space we live in is defined in terms of quantum field theory. Simple quantum mechanics would do since I am more familiar with that. Furthermore we can ignore gravity; I am willing to assume that space-time is flat. Eventually I would like to understand the "last principle" derivations. I would be thankful for any time you take to explain this to me.
gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>) (07/28/85)
> I would be very interested to hear an explanation of how the four > dimensional space we live in is defined in terms of quantum field theory. Excellent question; perhaps the folks who've been saying that their pet theory explains everything will explain this (ha!). The only halfway sensible answers I've seen to this were contained in some of the more speculative work of Heisenberg and of Eddington. Unfortunately, the latter died before finishing his explanation (which was algebraic) ... Some schools of epistemology would say that it is possible that no better answer is available than "the three dimensions of space and one of time are tied directly to our perceptual abilities and are unlikely to be reducible to simpler concepts".
brooks@lll-crg.ARPA (Eugene D. Brooks III) (07/28/85)
> I would be very interested to hear an explanation of how the > four dimensional space we live in is defined in terms of quantum field > theory. Simple quantum mechanics would do since I am more familiar > with that. Furthermore we can ignore gravity; I am willing to assume > that space-time is flat. Eventually I would like to understand the > "last principle" derivations. In quantum field theory, the four dimensional space we live in (x,y,z,t) are scalar indeces of the field operators. The position in space-time is not an operator as it is in simple quantum mechanics. One expresses a Hamiltonian (energy) operator in terms of the field operators. There are also corresponding momentum operators Px, Py and Pz which are expressed in terms of the field operators. If one defines the initial conditions on the field operators at time t=0 then the Hamiltonian is used to propagate the initial state of the operators using the standard Heisenberg picture Schroodinger equation for the field operators. The specification of the field operators on the t=0 surface of 4D space time is a non-Lorentz invariant element of the theory. This can also be gotten rid of by specifying the initial conditions on any space like surface. If you do this the evolution of the system runs along a normal to this surface. The generalized Hamiltonian (Px, Py, Px, H) is used to do the propagation of the initial specification of the field operators. See: "Relativistic Quantum Fields", Bjorken and Drell "Quantum Field Theory", Itzykson and Zuber
brooks@lll-crg.ARPA (Eugene D. Brooks III) (07/28/85)
> Eventually I would like to understand the > "last principle" derivations. No one to my knowledge has ever constructed the field operators for a baseball so classical physics will be with us for some time yet. :-)
DAM%MIT-OZ@MIT-MC.ARPA (07/30/85)
Thanks for the response. I appreciate the time you spend writing the messages. We seem to be dealing with a "configuration space" in which each configuration is a state of the 3-dimensional field. A particular configuration is an eigenfunction of all (continuum many) field operators. > If one defines the initial conditions on the field operators at time t=0 then > the Hamiltonian is used to propagate the initial state of the operators > using the standard Heisenberg picture Schroodinger equation for the field > operators. I assume that QFT is similar to ordinary quantum mechanics in that position eigenfunctions are highly unstable over time (if I know position I don't know momentum). In QFT this should mean that if I have an eigenfunction of all field operators then very soon I have a wave function which is spead out over all field configurations. In ordinary QM a solution of Schodingers equation is a function of the form psi(c, t) where c is a variable ranging over configurations of the system. Is the same true in QFT? In QFT each configuration c is a distribution of the field in THREE-space? Why does time have such a distinguished status? Why shouldn't solutions of (whatever) physical equations be functions of the form psi(c') where c' ranges over configurations of the field in FOUR-space? I'll have to think about this further.
mikes@AMES-NAS.ARPA (07/31/85)
From: mikes@AMES-NAS.ARPA (Peter Mikes) If you consider functions f(xyxt) - adding time to xyz, you are talking about the trajectories or orbits and you should not call them 'configurations' which imply a snapshot in time. In case of fields I would say you deal with 'histories' - until somebody suggests a better word. Also - if u want to be really relativistic, it may not be enough to think about psi as scalar. Even Dirac eq. has more.
brooks@lll-crg.ARPA (Eugene D. Brooks III) (08/05/85)
> In ordinary QM a solution of Schodingers equation is a function > of the form psi(c, t) where c is a variable ranging over configurations > of the system. Is the same true in QFT? In QFT each configuration c > is a distribution of the field in THREE-space? Why does time have such > a distinguished status? Why shouldn't solutions of (whatever) physical > equations be functions of the form psi(c') where c' ranges over > configurations of the field in FOUR-space? I'll have to think about > this further. Time does not have distinguished status, the t=0 surface just happens to be the standard space like surface that is used to establish the initial conditions of the field operators. You can pick any space like surface you like to set up the initial field operators and commutation relations. If you pick the t=0 surface the equations of motion only involves the H operator. If you pick some other surface then the equations of motion involves the full (H, Px, Py, Pz) operator and the t variable in the equations of motion are replaced by a variable s, the distance along a normal to the surface.