3
$\begingroup$

I've been looking into renormalization lately and would like to know 2 things about the Polchinski Exact Renormalization Group Equation (PERGE): According to Wikipedia, the PERGE is defined by $$\frac{dZ_{\Lambda}}{d\Lambda}=0 \qquad \Rightarrow $$ $$\frac{dS_{Int,\Lambda}}{d\Lambda}=\frac{1}{2}\int \frac{d^4p}{(2\pi)^4}[(\frac{\delta S_{Int,\Lambda}(p)}{\delta \phi(p)})\frac{dR_{\Lambda}^{-1}(p)}{d\Lambda}\frac{\delta S_{Int,\Lambda}(p)}{\delta \phi^{*}(p)}-Tr[R_{\Lambda}^{-1}(p)\frac{\delta^2S_{Int,\Lambda}(p)}{\delta \phi(p)\delta \phi^{*}(p)}]],$$ with $Z_{\Lambda}$ the partition function for the momentum scale, $\Lambda$, and $S_{Int,\Lambda}$ the Euclideanized interaction action corresponding to $\Lambda$. Supposedly, $\phi(p)$ are the momentum-space field configurations.

  1. My 1st question is: Is this formulation of the PERGE correct when considering a complex scalar field?

  2. My second question is: what are $R_{\Lambda}$ and $\phi(p)$? Specifically, if $\phi(p)$ can be taken as a momentum-based complex scalar field, is it then basically a superposition of the 4-momentum matter annihilation and anti-matter creation operators?

$\endgroup$

1 Answer 1

4
$\begingroup$
  1. The Polchinski exact renormalization group flow equation (ERGE) [1] $$\begin{align} &\frac{\partial W_{\rm int}[J]}{\partial\Lambda}\cr ~=~&\color{red}{\frac{1}{2}}\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{\partial K}{\partial\Lambda}\cr &\left(\frac{\delta W_{\rm int}[J]}{\delta\phi(k)}\frac{1}{k^2+m^2}\frac{\delta W_{\rm int}[J]}{\delta\phi(-k)}-\frac{\hbar}{k^2+m^2}\frac{\delta^2 W_{\rm int}[J]}{\delta\phi(k)\delta\phi(-k)}+\hbar\delta^d(0)\right).\end{align}\tag{1}$$ for a real scalar field can easily be adapted to a complex scalar field by replacing the field $\phi(-k)$ with $\phi^{\ast}(k)$ and recalling that there is conventionally no longer a symmetry factor $\color{red}{\frac{1}{2}}$, cf. the kinetic action term (3).

  2. $\phi(k)$ is the Fourier transformed of the field $\phi(x)$. And $\frac{1}{R_{\Lambda}(k)}=K(k)$ is a smooth regulator that is 1 for $|k|\ll \Lambda$ and 0 for $|k|\gg \Lambda$.

Let us for completeness sketch Polchinski's original derivation [1] of the ERGE (1) for a real scalar:

  1. The partition function is $$Z[J]~:=~ \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\},\tag{2}$$ where the Euclidean action consists of two parts: $$S_2~:=~\color{red}{\frac{1}{2}}\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}~\phi(-k)\frac{k^2+m^2}{K}\phi(k),\tag{3} $$ and the Wilsonian effective action $$W_{\rm int}[J].\tag{4}$$

  2. The partition function should not depend on the scale $\Lambda$: $$\begin{align} 0~\stackrel{?}{=}~&\hbar\frac{d Z[J]}{d\Lambda}\cr ~=~&-\int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}\left(\frac{\partial W_{\rm int}[J]}{\partial\Lambda}+\color{red}{\frac{1}{2}}\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}~\phi(-k)(k^2+m^2)\frac{\partial K^{-1}}{\partial\Lambda}\phi(k)\right)\cr &\qquad\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\}.\end{align}\tag{5}$$

  3. If we assume that boundary terms in the path integral vanish, we get that $$\begin{align} 0~\sim~&\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{\partial\ln K}{\partial\Lambda} \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}\frac{\hbar\delta}{\delta\phi(k)}\phi(k)\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\}\cr ~=~&\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{\partial\ln K}{\partial\Lambda} \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}\cr &\left(\hbar\delta^d(0)-\phi(-k)\frac{k^2+m^2}{K}\phi(k)-\phi(k)\frac{\delta W_{\rm int}[J]}{\delta\phi(k)}\right)\cr &\qquad\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\},\end{align}\tag{6}$$ and $$\begin{align} 0~\sim~&\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{\partial K}{\partial\Lambda} \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}\cr &\qquad\frac{\hbar\delta}{\delta\phi(k)}\frac{1}{k^2+m^2}\frac{\hbar\delta}{\delta\phi(-k)}\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\}\cr ~=~&\int_{\mathbb{R}^d}\!\frac{d^dk}{(2\pi)^d}\frac{\partial K}{\partial\Lambda} \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}\cr &\left(-\frac{\hbar}{K}\delta^d(0)-\frac{\hbar}{k^2+m^2}\frac{\delta^2 W_{\rm int}[J]}{\delta\phi(k)\delta\phi(-k)} +\phi(-k)\frac{k^2+m^2}{K^2}\phi(k)+\frac{2\phi(k)}{K}\frac{\delta W_{\rm int}[J]}{\delta\phi(k)}+\frac{\delta W_{\rm int}[J]}{\delta\phi(k)}\frac{1}{k^2+m^2}\frac{\delta W_{\rm int}[J]}{\delta\phi(-k)}\right)\cr &\qquad\exp\left\{-\frac{1}{\hbar}\left(S_2+W_{\rm int}[J]\right)\right\}.\end{align}\tag{7}$$

  4. It is straightforward to check that Polchinski ERGE (1) together with eqs. (6)+(7) implies eq. (5). $\Box$

For a more modern proof, see my Phys.SE answer here.

References:

  1. J. Polchinski, Renormalization and effective lagrangians, Nucl. Phys. B231 (1984) 269; eqs. (18).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.