There've been a few things in the Wilsonian approach to renormalization that made me confused recently.
The partition function of the $\phi^{4}$-theory (in Euclidean signature) is given by
$$\mathcal{Z}\equiv\int[\mathcal{D}\phi]e^{-\frac{1}{\hbar}S[\phi]}, \tag{1}$$
where $S[\phi]$ is the action in Euclidean signature. i.e.
$$S[\phi]=\int d^{D}\mathbf{x}\left[\frac{1}{2}\phi(-\Delta+m^{2})\phi+\frac{\lambda}{4!}\phi^{4}\right]. \tag{2}$$
In the Euclidean momentum space, one can perform the Fourier transform
$$\phi(\mathbf{x})=\int\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}e^{i\mathbf{p}\cdot\mathbf{x}}\hat{\phi}(\mathbf{p}). \tag{3}$$
Then, the action in Euclidean signature can be written as
\begin{equation}
S[\phi]=\int d^{D}\mathbf{x}\left[\frac{1}{2}\phi(-\Delta+m^{2})\phi+\frac{\lambda}{4!}\phi^{4}\right] \\
=\frac{1}{2}\int\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\hat{\phi}(-\mathbf{p})(\mathbf{p}^{2}+m^{2})\hat{\phi}(\mathbf{p})+\frac{\lambda}{4!}\int\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\cdots\int\frac{d^{D}\mathbf{p}_{4}}{(2\pi)^{D}}\hat{\phi}(\mathbf{p}_{1})\cdots\hat{\phi}(\mathbf{p}_{4})(2\pi)^{4}\delta(\mathbf{p}_{1}+\cdots+\mathbf{p}_{4}). \tag{4}
\end{equation}
Since we are ignorant to the laws of physics in the UV, one must regularize the above functional integral by introducing a UV cut-off $\Lambda$ in the (Euclidean) momentum space. In the position space, introducing such a momentum cut-off means that the field is put on a lattice of size of the scale $L\sim 1/\Lambda$, and one has, instead of (4),
$$\phi_{\Lambda}(\mathbf{X}_{\mathbf{n}})=\underset{|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{p}}\hat{\phi}(\mathbf{p})\quad\mathrm{and}\quad\hat{\phi}(\mathbf{p})=\frac{1}{\Lambda}\sum_{\mathbf{n}}e^{-i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{p}}\phi_{\Lambda}(\mathbf{X}_{\mathbf{n}}), \tag{5}$$
where $\mathbf{X}_{\mathbf{n}}=2\pi\mathbf{n}/\Lambda$, and $\mathbf{n}\in\mathbb{Z}^{D}$ labels that lattice sites in $D$ dimensions.
Then, the first problem is that if we plug (5) back into equation (4), it seems to me that we cannot produce the Dirac-delta function for the kinetic Gaussian part, because it involves finite difference of second order. It seems that in QFT books one must pretend that we can differentiate wrt $\mathbf{X}_{\mathbf{n}}$, and we would have something like
$$\int d^{D}\mathbf{x}\phi(\mathbf{x})(-\Delta+m^{2})\phi(\mathbf{x})\approx\sum_{\mathbf{n}}\underset{|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\underset{|\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot(\mathbf{p}+\mathbf{k})}\hat{\phi}(\mathbf{p})(\mathbf{k}^{2}+m^{2})\hat{\phi}(\mathbf{k}). \tag{6}$$
Then, using the formula
$$\delta(\theta)=\frac{1}{2\pi}\sum_{n=-\infty}^{+\infty}e^{in\theta},$$
for $\theta\in(-\pi,+\pi]$, we obtain
\begin{align}
&\quad\quad\quad\quad\quad\quad\int d^{D}\mathbf{x}\phi(\mathbf{x})(-\Delta+m^{2})\phi(\mathbf{x}) \\
&\approx\underset{|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\underset{|\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\delta(\mathbf{p}+\mathbf{k})\hat{\phi}(\mathbf{p})(\mathbf{k}^{2}+m^{2})\hat{\phi}(\mathbf{k}) \\
&=\underset{|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\hat{\phi}(-\mathbf{p})(\mathbf{p}^{2}+m^{2})\hat{\phi}(\mathbf{p}).
\end{align}
1. If we had already put the field on a lattice, how could we still perform the second order derivative as opposed to the finite differentiation in (6)?
Next, we consider an energy scale $s\Lambda$, with $s\in(0,1)$, and integrate out the modes of $|\mathbf{p}|\in(s\Lambda,\Lambda]$. So the field splits into two parts:
\begin{align}
&\quad\quad\quad\quad\quad\quad\phi_{\Lambda}(\mathbf{X}_{\mathbf{n}})\equiv\varphi(\mathbf{X}_{\mathbf{n}})+\chi(\mathbf{X}_{\mathbf{n}}) \\
&=\underset{|\mathbf{p}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{p}}\hat{\phi}(\mathbf{p})+\underset{s\Lambda<|\mathbf{p}|\leq \Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{p}}\hat{\phi}(\mathbf{p}) \\
&\equiv\underset{|\mathbf{p}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{p}}\hat{\varphi}(\mathbf{p})+\underset{s\Lambda<|\mathbf{k}|\leq \Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot\mathbf{k}}\hat{\chi}(\mathbf{k}) \tag{7}
\end{align}
Then, the regularized action in Euclidean signature can be written as
$$S[\phi;\Lambda]=\frac{1}{2}\underset{|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\hat{\phi}(-\mathbf{p})(\mathbf{p}^{2}+m^{2})\hat{\phi}(\mathbf{p})+\frac{\lambda}{4!}\underset{|\mathbf{p}_{i}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\cdots\frac{d^{D}\mathbf{p}_{4}}{(2\pi)^{D}}\hat{\phi}(\mathbf{p}_{1})\cdots\hat{\phi}(\mathbf{p}_{4})(2\pi)^{4}\delta(\sum_{j=1}^{4}\mathbf{p}_{j}).$$
The quadratic cross terms vanish because
\begin{align}
&\quad\,\,\int d^{D}\mathbf{x}\chi(-\Delta+m^{2})\varphi\approx\sum_{n}\underset{|\mathbf{k}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\underset{s\Lambda<|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}e^{i\mathbf{X}_{\mathbf{n}}\,\cdot(\mathbf{p}+\mathbf{k})}\hat{\phi}(\mathbf{p})(\mathbf{k}^{2}+m^{2})\hat{\phi}(\mathbf{k}) \\
&=\underset{|\mathbf{k}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\underset{s\Lambda<|\mathbf{p}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\delta(\mathbf{p}+\mathbf{k})\hat{\phi}(\mathbf{p})(\mathbf{k}^{2}+m^{2})\hat{\phi}(\mathbf{k})=0.
\end{align}
Therefore, the regularized action takes the form
\begin{align}
&\quad\,\,S[\phi;\Lambda]=S[\varphi+\chi;\Lambda] \\
&=\left(\frac{1}{2}\underset{|\mathbf{p}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\hat{\varphi}(-\mathbf{p})(\mathbf{p}^{2}+m^{2})\hat{\varphi}(\mathbf{p})+\frac{\lambda}{4!}\underset{|\mathbf{p}_{i}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\cdots\frac{d^{D}\mathbf{p}_{4}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p}_{1})\cdots\hat{\varphi}(\mathbf{p}_{4})(2\pi)^{4}\delta(\sum_{j=1}^{4}\mathbf{p}_{j})\right) \\
&+\left(\frac{1}{2}\underset{s\Lambda|<\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\hat{\chi}(-\mathbf{k})(\mathbf{k}^{2}+m^{2})\hat{\chi}(\mathbf{k})+\frac{\lambda}{4!}\underset{s\Lambda<|\mathbf{k}_{i}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\cdots\frac{d^{D}\mathbf{k}_{4}}{(2\pi)^{D}}\hat{\chi}(\mathbf{k}_{1})\cdots\hat{\chi}(\mathbf{k}_{4})(2\pi)^{4}\delta(\sum_{j=1}^{4}\mathbf{k}_{j})\right) \\
&+\left(\frac{\lambda}{4!}\cdot 4\underset{\begin{align}&|\mathbf{p}|\leq s\Lambda \\ s\Lambda &<|\mathbf{k}_{i}|\leq\Lambda \end{align}}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{3}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p})\hat{\chi}(\mathbf{k}_{1})\hat{\chi}(\mathbf{k}_{2})\hat{\chi}(\mathbf{k}_{3})(2\pi)^{4}\delta(\mathbf{p}+\mathbf{k}_{1}+\mathbf{k}_{2}+\mathbf{k}_{3})\right. \\
&+\frac{\lambda}{4!}\cdot 6\underset{\begin{align}&|\mathbf{p}_{i}|\leq s\Lambda \\ s\Lambda &<|\mathbf{k}_{i}|\leq\Lambda \end{align}}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{2}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p}_{1})\hat{\varphi}(\mathbf{p}_{2})\hat{\chi}(\mathbf{k}_{1})\hat{\chi}(\mathbf{k}_{2})(2\pi)^{4}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{k}_{1}+\mathbf{k}_{2}) \\
&\left.+\frac{\lambda}{4!}\cdot 4\underset{\begin{align}&|\mathbf{p}_{i}|\leq s\Lambda \\ s\Lambda &<|\mathbf{k}|\leq\Lambda \end{align}}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{3}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p}_{1})\hat{\varphi}(\mathbf{p}_{2})\hat{\varphi}(\mathbf{p}_{3})\hat{\chi}(\mathbf{k})(2\pi)^{4}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}+\mathbf{k})\right) \\
&=S[\varphi;s\Lambda]+S_{_{(s\Lambda,\Lambda]}}\,\,[\varphi,\chi]. \tag{8}
\end{align}
Likewise, the functional integral measure factorizes as
$$[\mathcal{D}\hat{\phi}]\equiv\prod_{|\mathbf{p}|\leq\Lambda}d\hat{\phi}(\mathbf{p})=\prod_{|\mathbf{p}|\leq s\Lambda}d\hat{\varphi}(\mathbf{p})\prod_{s\Lambda<|\mathbf{p}|\leq\Lambda}d\hat{\chi}(\mathbf{p})=[\mathcal{D}\varphi][\mathcal{D}\chi],$$
where I have used the fact that Fourier transforms are unitary.
Performing the functional integral over $\chi$, one has
$$e^{-\frac{1}{\hbar}S_{\mathrm{eff}}\,[\varphi;s\Lambda]}=\underset{C^{\infty}\,((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S[\varphi+\chi;\Lambda]}\quad\mathrm{or}\quad S_{\mathrm{eff}}[\varphi;s\Lambda]\equiv-\hbar\log\left[\,\underset{C^{\infty}\,((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]\exp\left(-\frac{1}{\hbar}S[\varphi+\chi;\Lambda]\right)\right]. \tag{9}$$
From (8), the regularized action contains a part $S[\varphi;s\Lambda]$ which is independent of the field $\chi$, and so it can be factored out of the integral over $\chi$. Then,
$$e^{-\frac{1}{\hbar}S_{\mathrm{eff}}\,[\varphi;s\Lambda]}=e^{-\frac{1}{\hbar}S[\varphi;s\Lambda]}\underset{C^{\infty}\,((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{_{(s\Lambda,\Lambda]}}\,\,\,[\varphi,\chi]}. \tag{10}$$
The interaction action $S_{(s\Lambda,\Lambda]}[\varphi,\chi]$ consists of two parts:
- The Gaussian part:
$$S_{0}[\chi]\equiv\frac{1}{2}\underset{s\Lambda<|\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\hat{\chi}(-\mathbf{k})(\mathbf{k}^{2}+m^{2})\hat{\chi}(\mathbf{k}).$$
- The perturbation part, which we symbolically denote as
$$\frac{\lambda}{4!}\left(\int\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}+6\int\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\right).$$
To the leading order of $\lambda$, we have
\begin{align}
\mathcal{Z}[\varphi]&\equiv\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]\exp\left\{-\frac{1}{\hbar}S_{(s\Lambda,\Lambda]}[\varphi,\chi]\right\} \\
&=\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\left[1-\frac{1}{\hbar}\frac{\lambda}{4!}\left(\int\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}+6\int\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\right)+\mathcal{O}(\lambda^{2})\right] \\
&=\mathcal{N}-\frac{1}{\hbar}\frac{\lambda}{4!}\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\left(\int\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}+6\int\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}+4\int\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\right)+\mathcal{O}(\lambda^{2}) \tag{11}
\end{align}
So at the leading order, we obtain five terms. The first term $\mathcal{N}$ is a constant resulted from integrating out $\chi$ in the Gaussian part.
- The second term:
$$-\frac{1}{\hbar}\frac{\lambda}{4!}\underset{s\Lambda<|\mathbf{k}_{i}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{3}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{4}}{(2\pi)^{D}}\delta(\mathbf{k}_{1}+\mathbf{k}_{2}+\mathbf{k}_{3}+\mathbf{k}_{4})\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\hat{\chi}(\mathbf{k}_{1})\hat{\chi}(\mathbf{k}_{2})\hat{\chi}(\mathbf{k}_{3})\hat{\chi}(\mathbf{k}_{4})$$
- The third term:
$$-\frac{4}{\hbar}\frac{\lambda}{4!}\underset{|\mathbf{p}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p})\underset{s\Lambda<|\mathbf{k}_{i}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{3}}{(2\pi)^{D}}\delta(\mathbf{p}+\mathbf{k}_{1}+\mathbf{k}_{2}+\mathbf{k}_{3})\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\hat{\chi}(\mathbf{k}_{1})\hat{\chi}(\mathbf{k}_{2})\hat{\chi}(\mathbf{k}_{3})$$
- The fourth term
$$-\frac{6}{\hbar}\frac{\lambda}{4!}\underset{|\mathbf{p}_{i}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{2}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p}_{1})\hat{\varphi}(\mathbf{p}_{2})\underset{s\Lambda<|\mathbf{k}_{i}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{k}_{2}}{(2\pi)^{D}}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{k}_{1}+\mathbf{k}_{2})\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\hat{\chi}(\mathbf{k}_{1})\hat{\chi}(\mathbf{k}_{2})$$
- The fifth term:
$$-\frac{4}{\hbar}\frac{\lambda}{4!}\underset{|\mathbf{p}_{i}|\leq s\Lambda}{\int}\frac{d^{D}\mathbf{p}_{1}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{2}}{(2\pi)^{D}}\frac{d^{D}\mathbf{p}_{3}}{(2\pi)^{D}}\hat{\varphi}(\mathbf{p}_{1})\hat{\varphi}(\mathbf{p}_{2})\hat{\varphi}(\mathbf{p}_{3})\underset{s\Lambda<|\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}+\mathbf{k})\underset{C^{\infty}((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{0}[\chi]}\hat{\chi}(\mathbf{k}_{1})$$
To express the effective Lagrangian in the form of an asymptotic series in $\lambda$, we factor out the infinite constant $\mathcal{N}$ in (11):
\begin{align}
&\,\,\quad e^{-\frac{1}{\hbar}S_{\mathrm{eff}}\,[\varphi;s\Lambda]}=e^{-\frac{1}{\hbar}S[\varphi;s\Lambda]}\underset{C^{\infty}\,((s\Lambda,\Lambda])}{\int}[\mathcal{D}\chi]e^{-\frac{1}{\hbar}S_{_{(s\Lambda,\Lambda]}}\,\,\,[\varphi,\chi]}=e^{-\frac{1}{\hbar}S[\varphi;s\Lambda]}\mathcal{Z}[\varphi] \\
&=e^{-\frac{1}{\hbar}S[\varphi;s\Lambda]}\cdot\mathcal{N}\frac{\mathcal{Z}[\varphi]}{\mathcal{N}} \\
&=\mathcal{N}e^{-\frac{1}{\hbar}S[\varphi;s\Lambda]}\left(1-\frac{1}{\hbar}\frac{\lambda}{4!}\frac{\langle\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+6\langle\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\rangle}{\mathcal{N}}+\mathcal{O}(\lambda^{2})\right),
\end{align}
and take logarithm on both sides:
\begin{align}
&\quad\,\,S_{\mathrm{eff}}[\varphi;s\Lambda]=S[\varphi;s\Lambda]-\hbar\log\mathcal{Z}[\varphi] \\
&=-\hbar\log\mathcal{N}+S[\varphi;s\Lambda]-\hbar\log\left(1-\frac{\lambda}{4!\hbar}\frac{\langle\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+6\langle\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\rangle}{\mathcal{N}}+\mathcal{O}(\lambda^{2})\right) \\
&=-\hbar\log\mathcal{N}+S[\varphi;s\Lambda]+\frac{\lambda}{4!}\frac{\langle\hat{\chi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\chi}\hat{\chi}\hat{\chi}\rangle+6\langle\hat{\varphi}\hat{\varphi}\hat{\chi}\hat{\chi}\rangle+4\langle\hat{\varphi}\hat{\varphi}\hat{\varphi}\hat{\chi}\rangle}{\mathcal{N}}+\mathcal{O}(\lambda^{2}). \tag{12}
\end{align}
Then, it is said that the kinetic term and the interaction vertex in $S[\varphi;s\Lambda]$ will receive quantum corrections from the functional integral in $-\hbar\log\mathcal{Z}[\varphi]$, and the effective action looks like
$$S_{\mathrm{eff}}[\varphi,s\Lambda]=\frac{1}{2}Z(\Lambda)\varphi(-\Delta+m(\Lambda)^{2})\varphi+\frac{\lambda(\Lambda)}{4!}\varphi^{4}+\sum_{I}c_{I}(\Lambda)\mathcal{O}_{I}[\varphi,\nabla\varphi,\cdots], \tag{13}$$
where the in last term $\mathcal{O}_{I}$ represents more complicated interactions of $\varphi$ in higher orders, and $c_{I}(\Lambda)$ are the corresponding coupling constants.
I can kinda expect that after integrating out the $\chi$ field, there should be some new types of interactions in the low-energy effective field theory, but I don't know where the corrections in the kinetic term and $\varphi^{4}$-vertex come from.
2. How do I see the corrections $Z(\Lambda)$, $m(\Lambda)$ and $\lambda(\Lambda)$ from (12)? Did I miss anything from the above calculations?
Best Answer
The difference between the second derivative and a finite difference will scale like a negative power of the cutoff, so corresponds to an irrelevant operator. You should include these operators if you carry out a rigorous calculation (eg numerically), but to leading order they aren't important. In particle physics, these operators would break Lorentz invariance, corresponding to the fact that a lattice regulator breaks Lorentz invariance. So it's more common in analytic calculations in particle physics to use a different regulator, like dimensional regularization, that respects Lorentz invariance so that this issue doesn't come up.
I think you've actually written down a path integral that computes two-loop corrections, which is fine but I think it would be easier to answer your conceptual question by understanding what happens at 1 loop.
To compute the one-loop corrections, you should first expand the field as a background (large wavelength) part and fluctuation (short wavelength) part to quadratic order
\begin{equation} \phi = \Phi + \varphi \end{equation} so \begin{equation} S[\Phi + \varphi] = S[\Phi] + \frac{\delta S}{\delta \Phi} \varphi + \frac{\delta^2 S}{\delta \Phi^2} \varphi^2 + \cdots \end{equation} We can treat the first term $S[\Phi]$ as a constant in the path integral over $\varphi$; it corresponds to the classical action for the long wavelength fluctuations $\Phi$. The second term vanishes to this order (see eg https://arxiv.org/abs/hep-th/0701053). The third term gives the first loop result after integrating over $\varphi$.
\begin{eqnarray} e^{i S_{\rm Wilson}[\Phi]} &\approx& e^{i S[\Phi]} \int D \varphi e^{i \varphi \mathcal{O} \varphi} \\ &=& e^{i S[\Phi]} \left(\det (2\pi \mathcal{O}) \right)^{-1/2} \\ &\propto& e^{i \left(S[\Phi] + \frac{1}{2}{\rm tr} \log \mathcal{O} \right) } \end{eqnarray} where \begin{equation} \mathcal{O}(x,y) = \frac{\delta^2 S}{\delta \Phi(x) \delta \Phi(y) } \end{equation}
Generally $\mathcal{O}$ will depend on $\Phi$ (for example, consider a $\phi^4$ theory; $\mathcal{O}$ will involve an integral of the log of a function of $\Phi^2$), so the one-loop correction to the action $\sim \frac{1}{2} {\rm tr} \log \mathcal{O}$ will correct the $\Phi$ dependence coming from the classical action, including corrections to the quadratic part of the action.
A classic calculation that is well worth working through along these lines is the Coleman-Weinberg potential, which is a 1-loop correction to the potential, including the mass term and quartic interaction term.
I strongly recommend working through https://arxiv.org/abs/hep-th/0701053, which has some explicit one loop calculations of the Wilson action in a toy model where you can see where the various corrections come from.