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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:

  1. The Gaussian part:

$$S_{0}[\chi]\equiv\frac{1}{2}\underset{s\Lambda<|\mathbf{k}|\leq\Lambda}{\int}\frac{d^{D}\mathbf{k}}{(2\pi)^{D}}\hat{\chi}(-\mathbf{k})(\mathbf{k}^{2}+m^{2})\hat{\chi}(\mathbf{k}).$$

  1. 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?

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  • $\begingroup$ Made a few typo fixes and changed d'Alembertian to Laplacian since this is in Euclidean signature. Feel free to roll back edits if disagree. $\endgroup$ Commented Jan 5, 2023 at 23:37
  • $\begingroup$ There are several ways to put UV cutoffs, among those (A) the lattice regularization and (B) the (sharp) Fourier regularization. In (A), normally everything is discretized and $\Delta$ is by hand defined as a finite difference operator on the lattice, so question 1 would not even arise. In (B), one usually imposes a cap $|\mathbf{p}|\le\Lambda$ on momenta but the formulation is still in continuum (see the famous paper by Wilson and Kogut), namely the $\mathbf{x}$ are integrated over $\mathbb{R}^D$ instead of summed over a discrete lattice... $\endgroup$ Commented Jan 5, 2023 at 23:45
  • $\begingroup$ ...I think the source of confusion is due to mixing (A) and (B) instead of choosing one or the other. $\endgroup$ Commented Jan 5, 2023 at 23:46
  • $\begingroup$ @AbdelmalekAbdesselam Thank you very much for your comments. I thought choosing (A) would simultaneously leads to (B) and vice versa. $\endgroup$
    – Valac
    Commented Jan 6, 2023 at 6:56

1 Answer 1

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  1. If we had alread put the field on a lattice, how could we still perform the second order derivative as opposed to the finite differentiation in (6)?

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.

  1. How do I see the corrections 𝑍(Λ), 𝑚(Λ) and 𝜆(Λ) from (12)? Did I miss anything from the above calculations?

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.

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  • $\begingroup$ Thank you very much. $\endgroup$
    – Valac
    Commented Jan 5, 2023 at 18:16
  • $\begingroup$ Upvoted for that useful arxiv document link. $\endgroup$ Commented Jan 5, 2023 at 18:53

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