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  Dependence of the Path Integral in the Function Space

+ 2 like - 0 dislike

In a globally hyperbolic spacetime with compact initial cauchy surface $\Sigma$ one can have well-posed problems for the scalar field, $\phi$ with initial data in \(H ^1 \Sigma \times H ^0 \Sigma\). However one can also prove that the problem is well-posed for initial data in \(H^{3}\Sigma\times H^{3}\Sigma\). These two spaces define two different sets of functions and two different configuration spaces.

Does the path integral calculations give the same results for both cases?

asked Apr 16, 2014 in Theoretical Physics by y (10 points) [ revision history ]
edited Apr 16, 2014 by dimension10

1 Answer

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The path integral is not defined over any kind of Sobolev space, so neither restriction is important. The typical paths you select in a function-space path integral are extremely wild, they are distributions, the space of solutions to the differential equation is only important for finding extremal points of the path integral, not for the path integral itself. So there is no difference between the two spaces, neither is remotely related to the actual space on which the path integral is performed, which is the space of distributions. For a proper account, see recent articles of Hairer.

You should imagine the manifold is somehow replaced by a discrete grid or mesh, or else, as Hairer does (and Wilson did), that the fields are smeared by a test function, and you are taking the limit as the test function gets narrow. As the test function gets narrower, the random-pick field values at any given point diverge, while the averages are still convergent, so the convergence is in distribution. Neither Sobolev restriction on initial conditions makes any sense in a path integral.

answered Apr 18, 2014 by Ron Maimon (7,720 points) [ no revision ]

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