I known that in the Wilson renormalization process fast modes are integrated out in order to define an effective action for the low modes field. Considering phi to the fourth theory it's easy to see how the quadratic and quartic terms are corrected at 1-loop. These corrections lead to couplings renormalization but I don't understand what term in the fast modes path integral leads to the field strength renormalization Z. In other words where are the derivative interactions?

More precisely. If I define $\bar{\phi_0} =\phi_0 + \hat{\phi}_0$, in which $\hat{\phi}_0$ is the fast modes field, the path integral over $\hat{\phi}_0$ is: $$\int D\hat{\phi}_0 \, e^{-\int d^Dx [\frac{1}{2}(\partial\hat{\phi}_0)^2 +\frac{1}{2} m_0^2 \hat{\phi}_0^2+\frac{\lambda_0}{4!}( \hat{\phi}_0^4 + 4\phi_0^3\hat{\phi}_0 + 4 \phi_0\hat{\phi}_0^3 + 6 \phi_0^2\hat{\phi}_0^2)]}$$

Considering the free action as $S_0=\int d^Dx \, \frac{1}{2}(\partial\hat{\phi}_0)^2$ I can expand the exponential in this way: $$\int D\hat{\phi}_0 \,e^{-S_0}(1-\frac{1}{2} m_0^2 \hat{\phi}_0^2 -\frac{\lambda_0}{4!}( \hat{\phi}_0^4 + 4\phi_0^3\hat{\phi}_0 + 4 \phi_0\hat{\phi}_0^3 + 6 \phi_0^2\hat{\phi}_0^2) + \dots) $$ Now for example from $\phi_0^2\hat{\phi}_0^2$ I obtain the first correction to the 2-point function and so on. In order to calculate $1+\Delta Z$ and write the effective action as: $$S_{eff}[\phi_0]=\int d^Dx\,[\frac{1}{2}(1+\Delta Z)(\partial \phi_0)^2 +\frac{1}{2} (m_0^2+\Delta m^2)\phi_0^2+\frac{1}{4!}(\lambda_0 + \Delta \lambda)\phi_0^4 + ...]$$

I need a term in the expansion which contains $(\partial\phi_0)^2$, right? Where is it?


1 Answer 1


Derivative corrections appear in just the same way as usual. For example, quadratic terms with derivatives can be written as $\phi f(\partial) \phi$ where $f(\partial)$ is some combination of partial derivatives, such as $\partial^2 + m^2$ in free field theory. The corresponding propagator is $1/f(ip)$.

So derivative terms just correspond to nontrivial momentum-dependence of the propagator. You can directly compute the propagator in the Wilsonian picture (i.e. taking an expectation over fast modes). Any nontrivial dependence on the external momentum whatsoever corresponds to derivative terms.

In $\phi^4$ theory there are no quadratic derivative terms created at one-loop, because the one diagram you can draw doesn't actually have the external momentum flowing through the loop at all. But derivative terms do appear at two loops; they're totally generic.


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