What's the relation between Wilson Renormalization Group (RG) in Statistical Mechanics and QFT RG? What's the relation between Wilson Renormalization Group(RG) in Statistical Mechanics and QFT RG? For easier to compare, I choose scalar $\phi^4$ in both cases.
Wilson RG: 
Given $\phi^4$ model, 
$$Z=\int\mathcal{D}\phi(x)\exp[-\beta H[\phi(x)]]$$
$$\beta H[\phi(x)]=\int d^dx\left\{ \frac{k}{2}(\nabla\phi(x))^2 +\frac{t}{2}\phi^2(x)+\frac{u}{4!}\phi^4(x)\right\}$$
There is a natural UV-cutoff $\Lambda_0$ for statistical mechanics.
Integrating the momentum shell from $\Lambda=\Lambda_0 e^{-l}$ to $\Lambda_0$ and rescale, we can get the RG flow equation/Beta function:
$$\frac{du}{dl}=(4-d)u-\frac{3}{2}u^2\frac{K_d \Lambda_0^d}{(t+k\Lambda_0^2)^2} \tag 1$$
$$\frac{dt}{dl}=2t +\frac{u}{2}\frac{K_d\Lambda_0^d}{t+k\Lambda_0^2}\tag 2$$
with $K_d=S_d/(2\pi)^d$.
High energy QFT RG:
Given $\phi^4$ model,
$$\mathcal{L}=\frac{1}{2} (\partial_\nu \phi_R)^2+ \frac{1}{2}m_R^2\phi_R^2+\frac{\mu^\epsilon g_R}{4!} \phi^4_R + \frac{1}{2}(Z_\phi-1) (\partial_\nu \phi_R)^2+ \frac{1}{2}(Z_m Z_\phi -1)m_R^2\phi_R^2+(Z_g Z^2_\phi-1)\frac{\mu^\epsilon g_R}{4!} \phi^4_R$$
with $\epsilon=4-D$, $D$ the spacetime dimension, $\mu$ an arbitrary mass scale, $g_R, m_R$ dimensionless renormalized parameters. 
Using dimensional regurization, momentum substraction scheme, we can get the RG flow equation in $D \rightarrow 4$ like this:
$$\frac{\partial g_R}{\partial \ln\mu}=-\epsilon g_R+ \frac{3}{2}g_R^2+O(g_R^3) \tag 3$$
$$\frac{\partial m_R^2}{m_R^2 \partial \ln\mu}=\frac{K_4}{2}g_R + O(g_R^2)\tag 4$$
My questions:
1.Everyone says that Wilson RG is essentially same as QFT RG. No matter in technique or concept I can't understand this relation. 
Firstly in concept, Wilson RG describes how the effective parameter flow as you view the system in a larger and larger size.
In QFT, we can compute any observable $\sigma$
$$\sigma = \sigma(m_0,g_0,\Lambda)$$
with $m_0$, $g_0$ the bare parameters, $\Lambda$ the UV-cutoff. However if you fix the bare parameters and make $\Lambda\rightarrow \infty$ then every observable $\sigma$ must be divergent. The only way out is using experiment to fix the some observables $\sigma$, then RG tells us how bare parameters should grow with cutoff, that is $m_0(\Lambda), g_0(\Lambda)$, to keep observables  $\sigma(m_0(\Lambda),g_0(\Lambda),\Lambda)$ finite and independent with cutoff $\Lambda$. So what's the relation?
2.In technique, especially $(2),(4)$ are totally different. How to see explicitly the Wilson RG $(1),(2)$ and QFT RG $(3),(4)$ are essentially same? (or in weaker sense they can have same Wilson-Fisher fixed point? have same critical exponent?) 
3.Above calculation is only $1$-loop. Even though you can show in $1$-loop they are essentially same, how to prove that they are same in higher loop? Because I heard that in QFT RG, the coefficent of $g_R^3$ and higher will depend on the regularization. It seems impossible Wilson RG is same as QFT RG.
PS: There is a relating question Relation between Wilson approach to renormalization group and 'standard' RG But our questions are totally different.
 A: This is of course a very subtle problem, and I will only scratch the surface here. I think that the best reference for this question is the discussion by B. Delamotte in arxiv:0702.365, Section 2.6-  "Perturbative renormalizability,  RG  flows,  continuum limit,  asymptotic  freedom and all that..."
Although the two approaches seem completely different, they are in essence equivalent, in the sense that, as long as the perturbative expansion makes sense, they will give the same results : same fixed points, critical exponents, and so on.
Beyond one-loop, the two methods seem completely different : in Wilson's approach, one generates new interactions in the lagrangian ($\phi^6$ etc.), which are needed to get the fixed points at order $\epsilon^2$ for instance. In the QFT RG, one never generate new interactions in the lagrangian, everything in captured in $g_R$, which is used to generate all vertex functions (corresponding to $\langle\phi^6\rangle$ and so on). In fact, what the perturbative RG is doing in practice is to project the RG flow onto one specific RG trajectory (called L in the reference above), which can be parametrized by the flow of $m^2_R$ and $g_R$ only (thus describing the flow of all other interactions in terms of $m^2_R$ and $g_R$).
As I said in a previous answer (see below) :
In the Wilsonian approach, one starts from the microscopic scale $\Lambda$ and looks at what's going at smaller energy, whereas in the "standard" approach, one fixes than macroscopic scale and sends $\Lambda\to\infty$ in order to effectively probe smaller and smaller energy scales.
See also Divergent bare parameters/couplings: what is the physical meaning of it? Do this have any relation with wilson's renormalization group approach? and Why do we expect our theories to be independent of cutoffs?
