# Jacobian in path integral equal to one?

1. Nov 18, 2012

### 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?

Last edited: Nov 18, 2012
2. Nov 18, 2012

### 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 $J=\frac{\partial \phi'}{\partial \phi}=-1$.

3. Nov 18, 2012

### geoduck

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?

4. Nov 19, 2012

### Hypersphere

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

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.