Timeline for QFT and its notations
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2020 at 3:27 | vote | accept | JustWannaKnow | ||
| Apr 16, 2020 at 18:50 | comment | added | Abdelmalek Abdesselam | The statement $\int_{\mathscr{S}'(\mathbb{R}^d)}\varphi(f)^n d\mu_G(\varphi)=\infty$ is false. | |
| Apr 16, 2020 at 18:33 | answer | added | Abdelmalek Abdesselam | timeline score: 5 | |
| Apr 16, 2020 at 14:51 | comment | added | Aaron Bergman | Now, if all you want to do is get the formal perturbation expansion, you could be asking about how to rigorously calculate the n-point functions you bet by expanding the exponential. I’m sure that’s been done on the math side, but from the physics side, as Igor Khavkine said, the buzzwords are normal ordering and Wick’s theorem. | |
| Apr 16, 2020 at 14:30 | comment | added | Aaron Bergman | I think the issue is that you are taking the method of making the Gaussian integral rigorous and trying to extend that to other QFTs. This hasn’t worked, and, as Michael said above, it’s far from obvious that it’s the right direction to go to. The action is a classical quantity, with honest functions that can be point wise multiplied. The path integral is a not rigorous thing based on that classical action that has a plausible path to definition via the lattice using exactly that classical action. | |
| Apr 16, 2020 at 11:58 | answer | added | gmvh | timeline score: 4 | |
| Apr 16, 2020 at 6:57 | comment | added | Michael Engelhardt | The $\phi^{4} $ is motivated by the physics. The real world is not described by a quadratic action. The infinities you note associated with the Gaussian case are quite innocuous compared to what you have to contend with once you introduce interactions such as the $\phi^{4} $. | |
| Apr 16, 2020 at 4:48 | comment | added | JustWannaKnow | apart from that, I've seen some books where even thins like $\int_{\mathcal{S}'(\mathbb{R}^{d})}\varphi(f)^{n}d\mu_{G}(\varphi)$ were written as $\int \varphi(x)^{n}d\mu_{G}(\varphi)$, so the pointwise notation was used again but just as a matter of notation for something that has meaning. This led me to believe that the $\varphi(x)^{4}$ was also a notation convention or something, as I said, to idk see field variables as random variables or something. But from the content of the comments it seems that this is a formal convention which we cannot avoid, maybe because of the physics of it (?). | |
| Apr 16, 2020 at 4:43 | comment | added | JustWannaKnow | (cont) the addition of a term such as $\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx$ which shall be treated as a perturbation of the former Gaussian measure is, in this case, quite strange to me since you worked hard to give precise meaning to the Gaussian measure etc and then you add some formal object to the theory. I mean, to my understanding, this has two possible explanations: (a) this is motivated by the physics behind it and there's not much we can do or (b) this is just a matter of notation | |
| Apr 16, 2020 at 4:37 | comment | added | JustWannaKnow | As far as I understand (and I might be wrong) if you take $g=0$ in the action (\ref{2}), you can derive a rigorous (Gaussian) measure $\mu_{G}$ on $\mathcal{S}'(\mathbb{R}^{d}$ whose covariance is $-\Delta+m^{2}$, so this is a Gaussian measure associated to the action $S$. Of course, this has a problem which is that moments such $\int_{\mathcal{S}'(\mathbb{R}^{d})}\varphi(f)^{n}d\mu_{G}(\varphi) = \infty$, so we need to regularize the theory. (continues) | |
| Apr 16, 2020 at 4:30 | comment | added | JustWannaKnow | I'm happy my question gathered such a nice set of comments. I'm learning a lot already. Let me say some words above to adress you all at once. | |
| Apr 16, 2020 at 2:48 | comment | added | Michael Engelhardt | Recognizing that most configurations in the path integral are "rough", or are distributions, if you wish, doesn't solve the problem of defining field theory. Pretty much everything you calculate will still be infinite. You have to implement the renormalization program, i.e., regularize the theory and scale the couplings such as to produce a well-defined limit. However, once you accept that the theory needs to be regulated anyway, the notion of thinking of the field configurations as distributions becomes much less compelling. | |
| Apr 16, 2020 at 2:13 | comment | added | Igor Khavkine | The obvious answer to your question is that the common conventions and notations in field theory are historical. In the naive approach to path integrals, one only considers continuous (even smooth) field configurations, for which $S(\varphi)$ is well-defined. AFAIK, distributional configurations are forced on you for technical reasons, like the ability to properly define Gaussian integrals, but I don't know much about that. Anyway, that's why people invented normal ordering, to extend local field polynomials to some distributional configurations. | |
| Apr 16, 2020 at 1:49 | comment | added | Aaron Bergman | If you’re trying to think of the path integral rigorously, you’re going to have a lot of trouble. The case above, for example, likely does not exist as a theory. Unless you want to pursue it as a field of research, you’ll probably have better luck of thinking of it as an integral over some inchoate ‘space’ of ‘functions’ and assuming that the lattice has a continuum limit when the theory makes sense, even if it’s been too hard to prove in most cases as yet. | |
| Apr 16, 2020 at 1:27 | comment | added | JustWannaKnow | @AaronBergman my point here is basically the pointwise dependence of tempered distributions. It is a very common language (I've encountered it in many notes, papers etc) in QFT and I don't know the real benefit of it, since it is clear formal. Also, you're correct, I'm seeing it as a perturbative theory. About the lattice regularization, I don't know how rigorous those continuum limits can be (as I said, I'm not an expert) but the formulation on the continuum causes me trouble, as I pointed out, when tempered distributions are represented as 'real valued functions'. | |
| Apr 16, 2020 at 1:19 | comment | added | Aaron Bergman | I’m having trouble understanding the question, but the separation of the $\phi^4$ is not natural from the physics and really is just the statement that you’re doing perturbation theory. The whole thing is really a measure on ... something. Maybe looking at lattice regularizations will help your intuition? | |
| Apr 16, 2020 at 0:53 | comment | added | Michael Engelhardt | QFT per se doesn't reference random variables. However, one observes that, formally, QFT path integrals, continued to the Euclidean, look like partition functions in Statistical Mechanics, which does reference random variables. This mapping both allows one to use intuition from Statistical Mechanics to think about QFT, as well as provides paths towards computational methods. To do physics, it's enough to be quite informal about this mapping, but by all means flesh it out for yourself in whatever way seems most natural to you. | |
| Apr 15, 2020 at 22:46 | history | edited | JustWannaKnow | CC BY-SA 4.0 |
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| Apr 15, 2020 at 22:45 | comment | added | JustWannaKnow | @DmitriPavlov you are right! I didn't even noticed it. Thanks. Well, this reinforces the statements in my post, right? | |
| Apr 15, 2020 at 22:32 | comment | added | Dmitri Pavlov | Your formula for φ does not depend linearly on f. Distributions are by definition linear functionals on a space of test functions, so your formula cannot define a distribution. | |
| Apr 15, 2020 at 20:37 | history | edited | JustWannaKnow | CC BY-SA 4.0 |
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| S Apr 15, 2020 at 20:20 | history | suggested | RobPratt |
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| Apr 15, 2020 at 20:13 | review | Suggested edits | |||
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| Apr 15, 2020 at 20:05 | history | asked | JustWannaKnow | CC BY-SA 4.0 |