Timeline for QFT and its notations
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2020 at 15:10 | comment | added | gmvh | @AbdelmalekAbdesselam: Indeed, one needs to consider contact terms. Otherwise, the Dyson-Schwinger equations would be rather different, not to mention that the tree-level two-point function wouldn't be the Green function of the free field equation anymore. | |
| Apr 17, 2020 at 14:02 | comment | added | Abdelmalek Abdesselam | @gmvh: Also, another comment, physicists use evaluation at points instead of the point of view of Schwartz distributions. The reason is not the presence of a regulator which makes sample fields $\varphi(x)$ smooth. It is because correlations $\langle \varphi(x_1)\cdots\varphi(x_n)\rangle$ have singular support on the diagonal. Namely, one can evaluate them pointwise at noncoincident points. Sometimes, however, one needs to consider contact terms. | |
| Apr 17, 2020 at 2:50 | comment | added | user1504 | To be fair to gmvh: calling the renormalized classical lattice interaction a 'normally ordered product' is itself an abuse of notation. My apologies. :) | |
| Apr 16, 2020 at 20:31 | comment | added | Abdelmalek Abdesselam | @gmvh: As you know there are two points of view: 1) operators, 2) path integrals. In 1) normal ordering means putting creation ops on the left and annihilation ops on the right. In 2), it is not like there is no normal ordering. There is one but it is expressed differently, i.e., changing monomials like $\phi^4$ to Hermite polynomials $:\phi^4:$. That's what user1504 is referring to. | |
| Apr 16, 2020 at 19:47 | comment | added | gmvh | In the path integral formalism, the fields aren't operators, so there is nothing to normal order at that level. Renormalization is accounted for by writing the lattice action in terms of bare parameters (typically indicated by a subscript 0), which of course aren't the physical renormalized quantities. QFT on the lattice is really pretty rigorous at finite lattice spacing (the devil is in the continuum limit). | |
| Apr 16, 2020 at 15:14 | comment | added | user1504 | Perhaps it's worth adding that using exactly the classical action is simply abuse of notation. Even on the lattice, it's better to use an appropriately renormalized interaction (in this case, the normally-ordered product $:\phi^4:$). Leaving out the renormalization is also an abuse of notation. I think physicists do it because it works in the d=1 case (quantum mechanics), which was the first case where the path integral was really understood. Not really recommended, as doing this confused their thinking about QFT for decades. | |
| Apr 16, 2020 at 11:58 | history | answered | gmvh | CC BY-SA 4.0 |