Share this thread: 
#19
Feb312, 09:12 AM

Emeritus
Sci Advisor
PF Gold
P: 9,540




#20
Feb312, 09:20 AM

P: 883




#21
Feb312, 10:31 AM

P: 381




#22
Feb312, 11:01 AM

Emeritus
Sci Advisor
PF Gold
P: 9,540

However, in nonrigorous QFT, I think the idea is just to ignore that the interacting Hilbert space is really a different Hilbert space, and just introduce operators that can take nparticle states to (n+1)particle states for example. In this context, Fock space is, as you put it, "pretending to have interaction when it doesn't really". I really suck at QFT beyond the most basic stuff, so I can't explain it better, and I might even be wrong (about the stuff in this paragraph). Anyway, "particle" doesn't mean "classical particle", so you can't assume that something has properties like position just because a quantum theory calls it a "particle". 


#23
Feb312, 11:18 AM

Sci Advisor
P: 8,795

http://www.phys.psu.edu/~collins/563/LSZ.pdf: "Note that in both formulae, the vacuum state 0> is very definitely and strictly the true vacuum. This is just the same as in the definition of the coefficient c, (3), where the vacuum and oneparticle states are definitely the true vacuum and oneparticle states, i.e., the true physical states. In contrast, many textbook treatments appear to suggest that the state 0> should be the freefield unperturbed vacuum; if that approach is tried, very delicate limits involving adiabatic switching of the interaction are called for." Edit: Here's another presentation by Srednicki that starts off with the more common point of view, but he goes on to discuss that it's wrong, and says that renormalization computes the corrections to having started the calculation with the wrong ground state. http://web.physics.ucsb.edu/~mark/msqftDRAFT.pdf, p51: "However, our derivation of the LSZ formula relied on the supposition that the creation operators of free field theory would work comparably in the interacting theory. This is a rather suspect assumption, and so we must review it." 


#24
Feb312, 11:30 AM

Emeritus
Sci Advisor
PF Gold
P: 9,540




#25
Feb312, 05:33 PM

P: 381




#26
Feb312, 06:53 PM

Emeritus
Sci Advisor
PF Gold
P: 9,540

I suspect that even some QFT experts don't know rigorous QFT. It's like an entirely different field of physics. A typical student at an "introduction to QFT" course would probably need two more years of math before he can really begin to learn rigorous QFT. 


#27
Feb312, 07:10 PM

Sci Advisor
P: 8,795

QCD is believed to be completely consistent. It's still a Clay problem, but you can see that they do make use of axiomatic field theory. For example, Gupta's notes (p23) say that QCD has OsterwalderSchrader reflection positivity. This is a condition for the analytic continuation of a Euclidean theory to meet the Wightman axioms, which is constructive field theory. 


#28
Feb312, 08:30 PM

P: 381

http://en.wikipedia.org/wiki/Quantum_field_theory Wiki:"Quantum field theory is thought by many[who?] to be the unique and correct outcome of combining the rules of quantum mechanics with special relativity." Fact: it is not exactly correct as you emphasized. Wiki:"In perturbative quantum field theory, the forces between particles are mediated by other particles. The electromagnetic force between two electrons is caused by an exchange of photons. Intermediate vector bosons mediate the weak force and gluons mediate the strong force. " Fact: Fock space doesnt handle interactions so those pertubative approach are just temporary and is fundamentally invalid" Wiki:"In QFT, photons are not thought of as "little billiard balls" but are rather viewed as field quanta – necessarily chunked ripples in a field, or "excitations", that "look like" particles." Fact: Particles dont have positions so they are not really excitations of the field. One must not visualize it that way. Agree with everything? Maybe its time to correct Wiki and state things are not that rosy and indeed bleak. 


#29
Feb312, 09:41 PM

P: 381

I read the Mysearch shared site in http://www.quantumfieldtheory.info/Chap01.pdf and need to ask a critical question:
"1.8 Points to Keep in Mind When the word “field” is used classically, it refers to an entity, like fluid wave amplitude, E, or B, that is spread out in space, i.e., has different values at different places. By that definition, the wave function of ordinary QM, or even the particle state in QFT, is a field. But, it is important to realize that in quantum terminology, the word “field” means an operator field, which is the solution to the wave equations, and which creates and destroys particle states. States (= particles = wave functions = kets) are not considered fields in that context. " Why not call it Quantum Operator Theory instead of Quantum Field Theory as the above fact showed that the Field in QFT was not related to the classical field. I thought QFT was just about performing canonical quantization on the classical field. Or could be this true only to QED? Isnt QED about performing quantization on the electromagnetic field? 


#30
Feb312, 09:59 PM

P: 1,583




#31
Feb312, 10:34 PM

P: 381




#32
Feb412, 03:47 AM

P: 66

knowledge of QFT (mostly the nonrelativistic kind) as part of their training in QM, but needn't be experts in the mathematical foundations of QM. necessarily have a position, what we mean, in layman's language, is that those 'basic vibrations' aren't confined to a single point in space. Note however that they may (but don't NEED to be) confined to a very tiny region from our macroscopic point of view. This is completely analogous to the case of nonrelativistic ordinary QM. As to how to visualise a quantum field... well, quantum operators behave a lot like stochastic variables. They have an expectation value and a complete set of moments which give you the indeterminacy of said expectation value. So in principle, any such operator can be visualised as a 'fuzzy' quantity, centered around the expectation value and with the fuzziness being proportional to the indeterminacy. So for the case of a field, it's a 'fuzzy' field. As a visualisation technique, this is probably only useful for bosonic fields in states such that the indeterminacy is much smaller than the expectation value. This is the case for instance for the electromagnetic field in most ordinary cases. Fermionic fields OTOH don't have a classic limit and are thus much harder to visualise. 


#33
Feb412, 03:59 AM

P: 66

ideal solution, but are necessary for those cases where the full solution to the problem isn't available. Note such techniques are /extensively/ used across both pure and applied physics (including engineering). For instance, we don't have a general solution for the Nbody problem, so we need to resort to approximations like numerical and/or perturbative methods. 


#34
Feb412, 04:19 AM

P: 66

operators (or POVMs, which are a related but more complicated object). The 'actual' field IS the 'operator' field. I'll give you two examples: the total momentum of a system, P, and the electromagnetic field, A. In CLASSICAL physics, these, or their components in some reference frame, are numbers. P = {Px, Py, Pz}; A = {phi, Ax, Ay, Az}. In QUANTUM physics, these are operators. That's a more complicated kind of object. An important difference with the above case is, operators don't have a value by themselves. This is where the state comes in in the theory. Quantum states give operators their values (and their indeterminacy). So, while in classical physics you have A=A(x,y,z,t) as a vector with a definite value assigned to every point (x,y,z), in quantum physics you have A=A(x,y,z,t) as an operator field, that is, an operator assigned to every point of space (and time). Once you're given a state you can assign a value (actually, an expectation value and an indeterminacy) to those operators. If the indeterminacy is sufficiently small, it can be ignored and you recover the classical field (this can only happen for fields which do possess a classical limit, of course. The em field does.) 


#35
Feb412, 04:51 AM

PF Gold
P: 522

I do not want to be accused of blatant scepticism, although it is said that a certain amount is healthy. Equally, I do not want be accused of just cherrypicking comments by other people out of context just because they might appear to question some aspect of QFT. However, from the perspective of somebody simply interested in the subject, I am beginning to wonder just how many years of maths is now required to even come close to understanding QFT, let alone questioning any of its fundamental premises. As such, it seems that QFT may now extend beyond the reach of most people to quantify for themselves and therefore they must “stand on the shoulders of giants” or, at least, on the shoulders of somebody taller than themselves. However, it seems that any conclusions drawn will still depend on whose shoulders you decide to pick, e.g. see article “The search for a quantum field theory” for a somewhat pessimistic, and possibly outdated, take on the current state of play. Of course, this author, although apparently well qualified, may have simply lost his way and been left behind by leading edge thinking. Therefore, I am assuming that his concerns can now be dismissed? 


#36
Feb412, 06:44 AM

P: 381




Register to reply 