4
$\begingroup$

In the path integral approach one defines in some heuristic way the functional path integral \begin{equation} Z=\int{\cal{D}}\phi e^{iS(\phi)} \end{equation} and the one claims that one must integrate over all paths.

I understand that the domain of the integral is the configuration space of the theory.

My question is:

How does the integral depend on our initial choice of configuration space?

EDIT:

For example, 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 the Sobolev Spaces $H^{1}(\Sigma)\times H^{0}(\Sigma)$. However one can also prove that the problem is well-posed for initial data in $H^{k}(\Sigma)\times H^{k-1}(\Sigma)$.

These two well-possessedness results gives two different configuration spaces $H^{1}$ in the first case and $H^{k}$ in the second.

How does the path integral change in this case?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ For gauge theories, you need to be careful and mod out the gauge transformations. That is, integrate over physically distinct states. See, e.g., Sergey V. Shabanov's "Phase space structure and the path integral for gauge theories on a cylinder" arXiv:hep-th/9308002 for a study of 2d examples. $\endgroup$ – Alex Nelson Apr 2 '14 at 20:17
  • $\begingroup$ To add to A. Nelson's comment, one can systematically fix the overcounting of states due to the gauge symmetry by employing the Faddeev-Popov method, as described in Peskin and Schroeder. In some cases, this may introduce additional fields known as 'ghost fields.' $\endgroup$ – JamalS Apr 2 '14 at 20:52
  • $\begingroup$ @Alex Nelson: Thank you. If I understand correctly in the case of one interacting scalar field there is no need to fix the gauge and the classical configuration space is exactly the function space you use to integrate; and if you have gauge freedom you must quotient the space by the action of the gauge. Is this correct? $\endgroup$ – yess Apr 3 '14 at 14:46
  • $\begingroup$ @yess, that's correct. $\endgroup$ – Alex Nelson Apr 3 '14 at 16:57
2
$\begingroup$

You do integrate over all paths in configuration space, but beware : differentiable paths contribute to a measure of 0 in the integral. The real contribution comes from fractal paths of dimension 2 (cf "The Dimension of a Quantum-Mechanical Path" by Abbott and Wise).

This "spreading" of the path is the equivalent in path integrals of the Heisenberg uncertainty principle, something of the form

$\langle m \frac{x_{k+1} - x_k}{\varepsilon} x_k \rangle - \langle x_k m \frac{x_{k} - x_{k-1}}{\varepsilon} \rangle = \frac{\hbar}{i} \langle 1 \rangle$

(cf Feynman and Hibbs)

the angle brackets indicating a path integration of some functional with some action. It means that there is no real speed, but only an average one, since the paths are all non-differentiable at every points. The speed has a standard deviation linked to the standard deviation of the position of your measurement (this is also expressed in the informal relation you sometimes see in path integral book : $dx^2 \propto dt$)

In phase space, things get a bit more complicated, and only discontinuous paths in phase space contribute ("Feynman Path Integrals in a Phase Space" by Berezin).

The same logic applies for fields, and indeed you have to be careful to not integrate over the same field configuration twice if it's a gauge field.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.