# Computing the Wilsonian Action

$$Z = \int\left[\mathcal{D}{\phi}\right] e ^{-\int d^dx \, \frac{1}{2} (\partial \phi)^2 + \frac{m^2}{2}\phi^2 + \frac{\lambda}{4!}\phi^4} \cdot \underbrace{\int\left[\mathcal{D}\hat{\phi}\right] e^{-S_{\text{kin}}[\hat{\phi}]} \, e^{-\int d^dx \, \frac{\lambda}{4}\phi(x)^2\hat{\phi}(x)^2 + \text{ other }\hat{\phi} \text{ interactions}}}_{\text{compute perturbatively to provide effective correction to first factor}}$$

$$S_{\text{kin}}\left[\hat{\phi}\right] = \int d^dx \, \frac{1}{2} \left( \partial \hat{\phi} \right)^2$$ gives the propagator (Euclidean space)

$$D(x-y) = \langle\hat{\phi}(x)\hat{\phi}(y)\rangle = \int_{b\Lambda\leq |p| \leq \Lambda}\frac{d^dp}{(2\pi)^d} \, e^{-ip(x-y)} \frac{1}{p^2}$$

### Zeroth order

This gives $$Z = \int\left[\mathcal{D}\hat{\phi}\right] e^{-S_{\text{kin}}[\hat{\phi}]} = \langle\Omega|\Omega\rangle=1$$

### First order

$$$$\begin{split} Z & = \int\left[\mathcal{D}\hat{\phi}\right] e^{-S_{\text{kin}}[\hat{\phi}]} \left( -\int d^dx \, \frac{\lambda}{4}\phi(x)^2\hat{\phi}(x)^2 \right)\\ & = -\int d^dx \, \frac{\lambda}{4}\phi(x)^2 \, \langle\hat{\phi}(x)^2\rangle\\ & = -\int d^dx \, \frac{\lambda}{4}\phi(x)^2 \, \int_{b\Lambda\leq |p| \leq \Lambda}\frac{d^dp}{(2\pi)^d} \, e^{-ip(x-x)} \frac{1}{p^2}\\ & = -\int d^dx \, \frac{\mu}{2}\phi(x)^2 \end{split}$$$$

Where $$$$\begin{split} \mu &= \frac{\lambda}{2} \, D(0) = \frac{\lambda}{2} \int_{b\Lambda\leq |p| \leq \Lambda}\frac{d^dp}{(2\pi)^d} \, \frac{1}{p^2}\\ & = \frac{\lambda}{2} \int \frac{d\Omega_d}{(2\pi)^d} \, \int_{b\Lambda}^{\Lambda} dr \frac{r^{d-1}}{r^2}\\ & = \frac{\lambda}{(4\pi)^{d/2}} \frac{1-b^{d-2}}{\Gamma{(d/2)}} \frac{\Lambda^{d-2}}{d-2} \end{split}$$$$

### Second order

$$$$\begin{split} Z & = \int\left[\mathcal{D}\hat{\phi}\right] e^{-S_{\text{kin}}[\hat{\phi}]} \frac{1}{2} \left( -\int d^dx \, \frac{\lambda}{4}\phi(x)^2\hat{\phi}(x)^2 \right)^2\\ & = \frac{1}{2} \left( \frac{\lambda}{4}\right)^2 \int d^dx \, d^dy \, \phi^2(x) \phi^2(y) \, \langle\hat{\phi}^2(x) \, \hat{\phi}^2(y)\rangle \end{split}$$$$

From Wicks' theorem $$\langle\hat{\phi}^2(x) \, \hat{\phi}^2(y)\rangle = D^2(0) + 2D^2(x-y)$$

The first term gives $$$$\begin{split} & = \frac{1}{2} \left( \frac{\lambda}{4}\right)^2 \int d^dx \, d^dy \, \phi^2(x) \phi^2(y) \, D^2(0) \\ & = \frac{1}{2} \left[ \frac{\lambda}{2} \int d^dx \, \frac{1}{2}\phi^2(x) \, D(0) \right]^2\\ & = \frac{1}{2} \left[ \int d^dx \, \frac{\mu}{2}\phi^2(x)\right]^2\\ \end{split}$$$$

Combining 0th, 1st, 2nd and succesive orders we get $$1 + \int d^dx \, \frac{-\mu}{2}\phi(x)^2 + \frac{1}{2} \left[ \int d^dx \, \frac{-\mu}{2}\phi(x)^2\right]^2 + \dots = e^{\int d^dx \, \frac{-\mu}{2}\phi(x)^2}$$

effectively providing a correction $$m^2+\mu$$.

## The second term - where I get stuck

Proceeding as above I would write

$$$$\frac{1}{2} \left( \frac{\lambda}{4}\right)^2 \int d^dx \, d^dy \, \phi^2(x) \phi^2(y) \, 2D^2(x-y)$$$$

I don't understand the expansion in Fourier space Peskin&Schroeder do here, with respect to external momenta:

$$\int d^dx \, d^dy \, \phi^2(x) \, \phi^2(y) \approx \int d^dx \, \phi^4(x) + \phi^2(x) \, \left(\partial \phi(x) \right)^2 + \dots$$

Also, I don't understand how they get in eq. 12.15 a power of -4 in the integration momentum, instead of the whole propagator squared, and why no $$e^{-ip(x-y)}$$ factor appears:

$$[\dots] \int_{b\Lambda\leq |p| \leq \Lambda}\frac{d^dp}{(2\pi)^d} \,\frac{1}{p^4}$$

Can somebody help me out with this?

• I don't have the book on me, but what happens is that D is short range in real space (because it involves only high momenta) thus y is close to x, so you can use this Taylor expansion. – Adam Nov 27 '18 at 23:10