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Continuing to Euclidean Space Justified in Path Integral?

  1. Feb 25, 2015 #1
    It seems to me that in a path integral, since you are integrating over all field configurations, that going into Euclidean space is not valid because some field configurations will give poles in the integrand of your action, and when the integrand has poles you can't make the rotations required to make the integration limits take on real values. In other words, while you can always make the substitution [itex]t=-i\tau [/itex], in order to pretend [itex]\tau [/itex] is a real number and have the limits of integration over [itex]d\tau [/itex] be over real numbers, you need to be able to Wick rotate and that requires the field configurations in your action be well-behaved, but in the path integral the field configurations go over all values so aren't always well behaved.

    So how is that you can justify going into Euclidean space in the path integral?
     
  2. jcsd
  3. Feb 27, 2015 #2

    Orodruin

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    Poles develop in the propagators and if you insert the ##+i\epsilon## pole prescription, you will always be able to make the Wick rotation without crossing the poles.
     
  4. Feb 28, 2015 #3

    atyy

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    There is also the question of whether one can continue from Euclidean space back to Minkowski space. In general, one way of ensuring one gets a relativistic QFT from the Euclidean path integral is to make sure the Osterwalder-Schrader conditions http://ncatlab.org/nlab/show/Osterwalder-Schrader+theorem are satisfied. I think it is OS0 and OS1 that ensure that one can rotate from Euclidean space back to Minkowski space.

    (If I understand correctly, from point of view from the OS axioms, the question is not so much whether one can go from Minkowski to Euclidean, but whether one can go from Euclidean to Minkowski, since if one postulates a Euclidean path integral that satisfies the OS axioms, that directly defines a relativistic QFT.)
     
    Last edited: Feb 28, 2015
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