Talk:Propagator

Feynman Propagator non-zero for space-like separations

Is this actually true? Need citation here. Yes, the momentum space propagator is off-shell in that k^2+m^2 is not equal to E^2 since these are really dummy variables in the virtual propagator. Need a source here that says for certain that the Feynman propagator NEEDS to be non-zero for space-like separations. What I mean is, although the Feynman propagator is non-zero outside the lightcone, wouldn't you get exactly the same results if you set this to zero? Is it non-zero just for mathematical simplicity rather than anything physical.

Suspect formulas

Formulas in this article are suspect in many ways.

(1) Position-space KG propagator, m>0 case, does not reduce to m=0 case for $m\to 0$ . (For small x, $J_{1}$ (x) ~ x, so it becomes constant)

(2) There's propagation outside the light cone in both cases.

(3) See http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/31/02/ and http://www.physicsforums.com/showthread.php?t=161235 for other definitions of position-space KG propagator.

(4) There is a well known formula relating propagators for Dirac and Klein-Gordon equations: $S_{F}(x-x')=(i\gamma ^{\mu }\partial _{\mu }-m)\Delta _{F}(x-x')$ and it does not appear to hold for Dirac propagator in this article (whichever KG propagator is used) because differentiation of $J_{1}$ will necessarily produce either $J_{0}$ , $J_{2}$ or both.

4 Answer:) Yes this is true but J_0, J_1, and J_2 are not all independent functions. Look at the definition of J1 and see it is related to J1' and J1 by an equation. — Preceding unsigned comment added by 109.145.158.194 (talk) 19:38, 26 January 2014 (UTC) [reply]

(5) What happens to fermions if $m\to 0$ ? Position-space Dirac propagator seems to vanish!

<off topic> I've checked several books on QFT; not one of them gives the expression for a position-space Dirac propagator. Not P&S, not Weinberg, not even Zee. Unbelievable.</off topic> --Itinerant1 08:55, 3 August 2007 (UTC)[reply]

I'm not certain the last comment is really off topic. Not being able to verify information in the article is of some concern.

Regarding concern 1, a constant is indeed a solution to the Klein-gordon equation, it doesn't have the necesary singularity at x=x' however. (For a propagator or Green function we want the differential equation to equal a delta function not 0.) You should be able to combine these two solutions to get the boundary conditions you want.

When m is not equal to 0, my guess is that we should use the Bessel function of the second type Y_1(m|x-x'|) which does have the proper limit. (Or to be more accurate, use a linear combination of the two which matches the desired boundary conditions.) Choosing boundary conditions is roughly (or at least partially) equivalent to chosing a Feynman propagator, a retarded Green Function, or an advanced Green function.

Lee Loveridge Ph.D. —Preceding unsigned comment added by 157.185.96.158 (talk) 17:38, 3 October 2007 (UTC)[reply]

(6) I think there is a sign problem in the section on the causual (retarded) propagator. It should either read x0<y0 or the theta function needs to be \theta(y0-x0). Am I correct? —Preceding unsigned comment added by 132.199.100.237 (talk) 09:01, 20 November 2008 (UTC)[reply]

Just perused this entry when looking for a brief way to introduce propagators in a discussion. The formulas are not "well formed" Yes, the Heaviside with argument x-y needs to be only the time components. The term with Bessel function should also be deleted since Heaviside factor is zero for spacelike intervals. I'm not adept at editing. Someone should edit the equations. As is this section is not up to academic standards. And apparently it has been in error for nearly a decade at least.

Theophilus71 (talk) 12:59, 24 June 2023 (UTC)[reply]

The feynman propagator images

The Feynman propagator pictures in the article seem a little confused. The propagator only depends on the distance between the spacetime points x and y, so a 2d graph over the whole range of x and y seems a complete waste, as it sacrifices the ability to represent the actual value of G_F(x-y) in anything other than a color shade, for a degree of freedom which G_F does not depend on, that is, x+y.

Eudoxie (talk) 09:05, 6 August 2010 (UTC)[reply]

Agreed... also x and y are four-component vectors, so a plot treating them as single variables makes no sense. Even if it did make sense, plots with no explanation of what the axes are or what the shadings represent are pretty pointless anyway. Removed. 92.11.114.194 (talk) 10:06, 19 September 2012 (UTC)[reply]

undefined symbol

In the section on non-relativistic propagators, the symbol D[q(t)] is introduced. Anyone who can clarify the meaning? Thanks, 99.93.47.26 (talk) 14:04, 15 August 2010 (UTC)[reply]

Detail and usefulness

I am unconvinced by the level of detail and the explicitness of the derivations provided in this article. While what exists is fairly detailed, the derivations are difficult to follow (For reference, I hold a Masters' degree in physics with a heavy dose of theory). For example, comparing the derivation of the position-space scalar field propagator to my textbook (the recommended one), I can't match the starting point in the book to the implied jump in the article - it simply mentions 'by Fourier transformation' (a Fourier transform of what?) with no elaboration, whereas the book (Quantum Field Theory, Mandl F and Shaw G, Wiley, 1993) starts with the scalar field commutators and proceeds through a long-winded sequence of (admittedly poorly-explained) steps. Yet the derivation takes several lines to define the 'box operator' and the Dirac delta - which anyone with any education in field theory should recognise and be able to remind themselves of in seconds, even if they haven't memorised them implicitly.

While I realise that demanding extensive derivation is not entirely reasonable, neither is it reasonable to assume that this article's quite limited audience - those of the population positioned in between 'total incomprehension' and 'comfortable professional familiarity' - would be able to gain much useful insight from derivations which assume too much prior understanding. Most modestly-knowledgeable people in the field (read: students) simply aren't capable of parsing this much jargon and extracting detailed understanding from it, myself included. If anyone sufficiently-knowledgeable feels like expanding this, I would suggest that at least a couple of intermediate steps and/or a couple of lines of prose explaining same in words would be beneficial. Sojourner001 (talk) 18:27, 11 October 2010 (UTC)[reply]

Question

So what do you call the propagator that results from going over the left pole and under the right pole? 173.165.239.237 (talk) 06:04, 27 February 2012 (UTC)[reply]

Dyson? YohanN7 (talk) 19:55, 25 January 2013 (UTC)[reply]

Is the propagator infinite?

The propagator D(x-y) is defined as an integral in p-space, in 4 dimensions. The integrand is exp(i(k*(x-y)) times a function that, for p-->infinity, goes like 1/p^2. Doesn't this imply that D=infinity?!?!? Indeed, the integrated function should vanish faster than 1/p^5 to get a finite integral in d=4.

I probably missed a point or two. — Preceding unsigned comment added by 94.161.219.216 (talk) 21:47, 21 August 2012 (UTC)[reply]

The propagator is a distribution. From its Fourier transform one can split off a suitable polynomial such that the remainder has an absolutely convergent Fourier transform. The Fourier transform of the polynomial yields distributions. I prefer the precise determination of the propagator in Bogoliubov's book. Norbert Dragon (talk) 17:38, 5 March 2020 (UTC)[reply]

Inline Referencing of Textbooks

This article refers readers to textbooks for certain calculations, e.g. This choice of contour is equivalent to calculating the limit (see Huang p. 30) in the section Relativistic propagators.

In my opinion this is not at all helpful to an unfamiliar reader. Either some explanation of the calculation should be given or the reference should be cited in the canonical way. That is to say, the references should be properly formatted at the bottom of the page, with enough information for a novice to understand them. Simply writing Huang may not be sufficient for a reader who doesn't know anything about the books Huang has authored. PhysicsSean (talk) 04:27, 9 February 2018 (UTC)[reply]

OK, relegated to ref in refute, a superior format. You may improve by formatting in ref template, if you consider it necessary. Pls review WP referencing rules: Author in parenthesis suffices, unless there is a year ambiguity involved, and it always refers to the eponymous text in the reflist. So, consideration of Huang's collected works is a non-sequitur. You appear confused about the reason for these refs. Firstly they are "defensive", i.e., they block obstreperous editors from peremptorily junking material as "unsourced", or "OR", an acrimonious procedural problem in WP editing. Secondly they send the patient aggrieved readers to the original text, if a moment's reflection failed them. There is no problem with dropping in useful explanatory works in the footnote, which you are welcome to, but "unfunded mandate" exhortations to what others should do are unconventional. Cuzkatzimhut (talk) 13:41, 9 February 2018 (UTC)[reply]

No derivation or explanation of the propagator AT ALL!

Just very poor. Another academic failing to explain something! This is how you do it: https://www.youtube.com/watch?v=uPsTGIkWKjA&t=1912s

@User:Koitus~nlwiki you should sign your posts with 4 tildes, being a register user. This page is not a forum. You fail to propose concrete improvements to the article, which is not a tutorial, in any case! It is unclear what your sour hectorings are meant to achieve, or how subpar video links elicit improvements. Cuzkatzimhut (talk) 17:03, 15 November 2020 (UTC)[reply]