# Geometric interpretation of perturbation theory in quantum field theory

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I am studying GR right now, and one interesting thing I learned about vectors is that they are defined to have the same properties as derivatives.

With this in mind, can I make a differential geometric interpretation of ordinary perturbation theory used in quantum field theory in the following way:?

1. The set of all field configurations makes a (very complicated) manifold.
2. We do perturbation theory about a background field configuration corresponding to a point on this manifold.
3. The perturbations are like expanding functions out to first order -- can be viewed like tangent vectors

Am I on some track to enlightenment, or is this a dead-end train of thought?

This post imported from StackExchange Physics at 2014-03-28 17:12 (UCT), posted by SE-user QuantumDot

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At the non-rigorous/intuitive level, OP's observations are spot on. To facilitate such thinking, physicists often use DeWitt's condensed notation, where a field $\phi^{\alpha}(x)$ is written as $\phi^{i}$, while pretending that $i=(\alpha,x)$ is an index of a local coordinate $\phi^{i}$ for some differential manifold.

The problem is that the space of all field configurations is typically an infinite-dimensional space, while ordinary differential geometry is usually only discussed in the context of finite-dimensional manifolds.

Thus strictly speaking, one would have to master/study/develop an infinite-dimensional mathematical version of differential geometry to make OP's picture precise/rigorous.

This post imported from StackExchange Physics at 2014-03-28 17:12 (UCT), posted by SE-user Qmechanic
answered Nov 16, 2012 by (2,860 points)
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$k$th order perturbation theory is just the approximation in terms of a $k$th order Taylor expansion in the perturbed variable. So if you have a problem depending on a single parameter $g$, perturbation theory essentially works out a low order polynomial approximation to a path (parameterized by $g$) in an infinite-dimensional space of fields.  While this give some geometric interpretation, it does not simplify anything - computing the approximation you still need all the standard machinery.

answered Mar 30, 2014 by (13,637 points)

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