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  Is it possible to have transformations that transform the action and the measure while leaving the functional integral invariant?

+ 2 like - 0 dislike

Anomalous symmetries are those for which the Lagrangian stays invariant but the measure of the functional integral does not. I wonder if there are transformations that change both the action and the measure, such that the functional integral is left invariant, is this possible?

asked Jan 31, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]

Any arbitrary change of variables does this, no?

@RyanThorngren: Not quite; see my answer.

1 Answer

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On a formal level, the functional integral is an infinite-dimensional generalization of a scaled Lebesgue integral in $R^n$. For the latter, any diffeomorphism (bijective $C^1$ transformation) of the vector $x$ over which the integration extends can be compensated by changing the measure by the corresponding absolute value of the Jacobian determinant. In infinite dimensions, $x$ becomes a field, but the transformation law should still be valid.

However, since there is no proper definition of the functional integral (and an infinite-dimensional Lebesue integral doesn't exist) it is not clear to which extent the above reasoning is valid. People use it (just like the functional integral itself) on a heuristic basis.

Note that the property of renormalizability is not preserved under nonlinear field transformations, but renormalizability is needed to give the functional integral a well-defined and unique perturbative meaning. Thus performing a nonlinear tranformation of fields in the context of functional integrals is a hazard that easily leads to inconsistent results.

A proper answer to your question must wait for the day where functional integrals have a meaningful nonperturbative definition.

answered Jan 31, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

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