Jacobian in path integral equal to one?

[geoduck]

Consider:

\int d\phi e^{iS[\phi]}=\int d\phi' J e^{iS'[\phi']}

where J is the Jacobian. If the transformation of variables to phi' is a symmetry of the action [i.e., S'=S], then this becomes:

\int d\phi e^{iS[\phi]}=\int d\phi' J e^{iS[\phi']}

But doesn't this imply that the Jacobian has to equal one?

But surely that doesn't have to be true in general? If the action has a symmetry, and you perform the change of coordinates corresponding to the symmetry transformation, then does the Jacobian of that transformation have to equal one?

 

[Hypersphere]

I think you miss that, in general, \mathrm{d} \phi \neq \mathrm{d} \phi'.

For a simple example, consider S=\frac{1}{2}\phi^2. Obviously, this is invariant under the inversion \phi \rightarrow \phi'=-\phi, for which \mathrm{d}\phi'=-\mathrm{d}\phi and J=\frac{\partial \phi'}{\partial \phi}=-1.

 

[geoduck]

I think you miss that, in general, \mathrm{d} \phi \neq \mathrm{d} \phi'.

For a simple example, consider S=\frac{1}{2}\phi^2. Obviously, this is invariant under the inversion \phi \rightarrow \phi'=-\phi, for which \mathrm{d}\phi'=-\mathrm{d}\phi and J=\frac{\partial \phi'}{\partial \phi}=-1.

The Jacobian should take care of differences in measure, so what happens to

∫ e^{-x^2} dx from -∞ to ∞

under y=-x is:

∫ e^{-y^2} dx/dy dy from ∞ to -∞

which equals ∫ e^{-y^2} (-1) dy from ∞ to -∞

so it's the change of the order of the limits in the integration that allows the Jacobian to not have to be equal to 1.

Maybe a stronger statement is true: If the limits in an integration are unchanged by a transformation, then the Jacobian must equal one?

 

[Hypersphere]

so it's the change of the order of the limits in the integration that allows the Jacobian to not have to be equal to 1.

I guess that is the more common way of putting it, yeah.

Maybe a stronger statement is true: If the limits in an integration are unchanged by a transformation, then the Jacobian must equal one?

I don't have an explicit example, but couldn't we have a local transformation y=f(x) that doesn't change the endpoints (i.e. limits) and J=\frac{\partial f}{\partial x} \not\equiv 1?

I think your statement should hold for linear functions f(x) though.