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In Folland's Quantum Field Theory he mentions that we can apply Feynman's formula (Feynman parameterization) to either the Wick rotated integrals or the non-Wick rotated integrals corresponding to Feynman diagrams. However, later in the section he outlines a regularization procedure, in which he says we must first Wick rotating the integrals. Why is this Wick rotation necessary? Is it to make use of certain properties of Euclidean space that are not shared by Minkowski space? If so, which ones?

Qmechanic
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CBBAM
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    I don't have Folland's book at hand, so I will not post an answer as I don't have the context. The idea is to use the fact that the inner product in Euclidean space is non-degenerate, as opposed to the Lorentzian one. This means we can safely use the "ball parametrization" of the space, where the radius variable we integrate on is zero only at the origin. – Jeanbaptiste Roux Jan 07 '23 at 09:11
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    @JeanbaptisteRoux By a non-degenerate inner product I assume you mean $(u,v) = 0$ for all $v$ implies $u = 0$? – CBBAM Jan 07 '23 at 09:20
  • Good general question, it may be useful to list here some related questions for future users: https://physics.stackexchange.com/q/457591/226902 https://physics.stackexchange.com/q/455387/226902 https://physics.stackexchange.com/q/525092/226902 https://physics.stackexchange.com/q/110360/226902 https://physics.stackexchange.com/q/374900/226902 – Quillo Jan 07 '23 at 09:39

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Well, as long as the singularities of the propagators in Minkowski signature are regularized with a Feynman $i\epsilon$ prescription, Wick rotation is in principle not necessary. However in practice, the loop momentum integrals of the Feynman diagram are simpler to evaluate in Euclidean signature, e.g. because $SO(d)$ symmetry is simpler to handle than $SO(d-1,1)$ symmetry.

See also this related Phys.SE post.

Qmechanic
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    So it is a matter of making calculations easier due to symmetry? – CBBAM Jan 07 '23 at 09:21
  • @CBBAM Wick rotation also allows to make better sense of the path integral (https://physics.stackexchange.com/a/110410/226902), but then regularization is still needed: https://physics.stackexchange.com/a/110410/226902 (Minkowski path integral with the Feynman "epsilon prescription" is equivalent to the Euclidean path integral). See also this good answer: https://physics.stackexchange.com/a/96106/226902 (Euclidean spacetime is "more convenient"). – Quillo Jan 07 '23 at 09:48