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Explore Stack InternalWell, 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.
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 integrals 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.
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.
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 integrals 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.
