## References

The primary reference for this post is the classic paper by Wilson and Kogut (the renormalization group and the $\epsilon$ expansion). Also we used the article by Fisher (renormalization group theory: its basis and formulation in statistical physics). Also the book by Huang is as always extremely helpful.

## The set up

As an example we consider the Landau theory given by the Hamiltonian
\begin{eqnarray}
E[m]=\int d^dx {\cal E}(m(x))~,~{\cal E}=\frac{1}{2}(\vec{\partial}m(x))^2+\sum_{n=1}^{\infty}K_nm^n(x)+....
\end{eqnarray}
The field $m(x)$ is the order parameter (magnetization) and the dot stands for higher derivative terms. The coupling constants $K_n$ depend on the cutoff $\Lambda$. The most important example for us is the Landau-Wilson model given by
\begin{eqnarray}
{\cal E}=\frac{1}{2}(\vec{\partial}m(x))^2+\frac{1}{2}r_0 m^2(x)+u_0 m^4(x).
\end{eqnarray}

The renormalization group transformation consists of three steps:

Coarse-Graining: We integrate in the partition function over the high momentum modes.

Rescaling: We rescale the momenta back to the original domain in order to restore the original coarse-grained “picture”.

Normalization: We rescale the fields appropriately which is necessary since the relative size of the fluctuations of the rescaled magnetization is in general different from the original.

These three steps define a renormalization group transformation which  in general is not invertible  and thus we are really dealing with a semigroup structure. By repeating these steps we get then highly non-linear equations which are the renormalization group equations.

The partition function of the Landau theory in momentum space is of the form
\begin{eqnarray}
Z=\int \prod^{\prime}_{|\vec{k}|<\Lambda}[d\tilde{m}(\vec{k})][d\tilde{m}^{\dagger}(\vec{k})]\exp(-E[\tilde{m}(\vec{k})]).\label{Lambda}
\end{eqnarray}
The prime indicates that we integrate only over half of the momentum space. The energy functional in momentum space reads (with $K_2=r_0/2$)
\begin{eqnarray}
E[\tilde{m}]=\frac{1}{2}\int d^dk (\vec{k}^2+r_0)|\tilde m(\vec{k})|^2+....
\end{eqnarray}
We now apply the above stated three steps of the renormalization group transformation.

### Coarse-Graining (Integration):

We integrate in the partition function over all momentum modes which satisfy $\Lambda/b\leq |\vec{k}|\leq \Lambda$ with $b>1$. Thus the new energy functional should be defined by the expression
\begin{eqnarray}
\exp(-E^{\prime}[\tilde{m}(\vec{k})])=\int \prod^{\prime}_{\frac{\Lambda}{b}|<\vec{k}|<\Lambda}[d\tilde{m}(\vec{k})][d\tilde{m}^{\dagger}(\vec{k})]\exp(-E[\tilde{m}(\vec{k})]).
\end{eqnarray}
Clearly, the new Hamiltonian $E^{\prime}$ depends only on the fields $\tilde{m}(\vec{k})$ for which $|\vec{k}|<\Lambda/b$. This new Hamiltonian is generally of the same form as the original one but with different coupling constants, viz
\begin{eqnarray}
E^{\prime}[\tilde{m}]=\frac{1}{2}\int_{|\vec{k}|<\frac{\Lambda}{b}} d^dk (A\vec{k}^2+\tilde{r}_0)|\tilde m(\vec{k})|^2+....
\end{eqnarray}
In fact, $E^{\prime}$ will also contain additional terms which did not exist in the original action and which will be dropped (Wilson truncation). The partition function becomes
\begin{eqnarray}
Z=\int \prod^{\prime}_{|\vec{k}|<\frac{\Lambda}{b}}[d\tilde{m}(\vec{k})][d\tilde{m}^{\dagger}(\vec{k})]\exp(-E^{\prime}[\tilde{m}(\vec{k})]).
\end{eqnarray}

### Rescaling:

We rescale the momenta back to the original domain, i.e. we restore the cutoff back to $\Lambda$, as $k_i\longrightarrow k_i^{\prime}=b k_i$. The action becomes
\begin{eqnarray}
E^{\prime}[\tilde{m}]=\frac{b^{-d}}{2}\int_{|\vec{k}|<\Lambda} d^dk (\frac{A}{b^2}\vec{k}^2+\tilde{r}_0)|\tilde m(\frac{\vec{k}}{b})|^2+....
\end{eqnarray}

### Normalization:

In order to restore the canonical normalization of the kinetic term we rescale the fields as
\begin{eqnarray}
\tilde{m}^{\prime}(\vec{k})&=&\sqrt{\frac{A}{b^{d+2}}}m(\frac{\vec{k}}{b}).
\end{eqnarray}
The energy becomes
\begin{eqnarray}
E^{\prime}[\tilde{m}^{\prime}(\vec{k})]=\frac{1}{2}\int_{|\vec{k}|<\Lambda} d^dk (\vec{k}^2+{r}_0^{\prime})|\tilde m^{\prime}(k)|^2+...,
\end{eqnarray}
where the renormalized mass is given by
\begin{eqnarray}
{r}_0^{\prime}=\frac{b^2}{A}\tilde{r}_0.
\end{eqnarray}
The partition function becomes
\begin{eqnarray}
Z^{\prime}=\int \prod^{\prime}_{|\vec{k}|<\Lambda}[d\tilde{m}^{\prime}(\vec{k})][d\tilde{m}^{\prime\dagger}(\vec{k})]\exp(-E^{\prime}[\tilde{m}^{\prime}(\vec{k})]).\label{Lambda1}
\end{eqnarray}

In summary, the coupling constants $K=(K_2,K_3,...)$ in the original Hamiltonian $E$ will be mapped under the renormalization group transformation to new coupling constant $K^{\prime}=(K_2^{\prime},K_3^{\prime},...)$ in the renormalized Hamiltonian $E^{\prime}$. The coupling constants $K^{\prime}$ are called renormalized coupling constants. We also say that the coupling constants of the theory "flow" under a change of the energy scale. If we denote the renormalization group transformation by $R$ then this mapping can be written as ($K=K^{(0)}$, $K^{\prime}=K^{(1)}$)
\begin{eqnarray}
K^{(1)}=R(K^{(0)}).
\end{eqnarray}
This step can be iterated an arbitrary number of times. The renormalization group transformation is the same for all iterations and as a consequence we can write after $i+1$ iterations the renormalization group transformation
\begin{eqnarray}
K^{(i+1)}=R(K^{(i)}).\label{fundamental}
\end{eqnarray}
By inspection, since the partition functions (\ref{Lambda}) and (\ref{Lambda1}) are given by the same function except for the replacement $K\longrightarrow K^{\prime}$, the free energy should satisfy the relation
\begin{eqnarray}
f(K)=f_0(K)+f(K^{\prime}).
\end{eqnarray}
Next, since the volumes (numbers of degrees of freedom) before and after the renormalization group transformation are related by $(\Lambda^{\prime}/\Lambda)^d=b^{-d}$, the free energy per degrees of freedom must satisfy the relation
\begin{eqnarray}
F(K)=F_0(K)+b^{-d}F(K^{\prime}).\label{fundamental1}
\end{eqnarray}

## Formalism

We are interested in the most general properties of the following renormalization group transformation
\begin{eqnarray}
K^{(i+1)}=R(K^{(i)}).
\end{eqnarray}
A fixed point  $K^*$ in the coupling-constant space is a point which does not move, i.e. it is invariant, under the successive action of the above renormalization group transformation. Thus we must have
\begin{eqnarray}
K^{*}=R(K^{*}).
\end{eqnarray}
Many points $K$ in the coupling-constant space will approach the fixed point under a large number of iterations of the renormalization group transformations. We write this as
\begin{eqnarray}
K^{(i)}\longrightarrow K^*~,~i\longrightarrow\infty.
\end{eqnarray}
The fixed point $K^*$ describes the system at the critical temperature $T=T_c$. At this point the system becomes scale invariant, i.e. it becomes invariant under a change of length scale. This can also be expressed by the fact that correlation length $\xi$ (inverse mass) diverges at the fixed point.

Now we linearize the above  renormalization group equation around the fixed point  $K^*$ by writing
\begin{eqnarray}
K^{(i+1)}_{\alpha}-K^{*}_{\alpha}=(K^{(i)}-K^{*})_{\alpha}=W_{\alpha\beta}(K^{(i)}_{\beta}-K^{*}_{\beta}).
\end{eqnarray}
The linearized renormalization group transformation is given by
\begin{eqnarray}
W_{\alpha\beta}=\frac{\partial R_{\alpha}}{\partial K_{\beta}}|_{K=K^{*}}.
\end{eqnarray}
The origin of the coupling-constant space is chosen to be $K^{*}$. The left eigenvectors $\phi$ of the renormalization group matrix $W$ are defined by
\begin{eqnarray}
\phi_{\beta}W_{\beta\alpha}=\lambda\phi_{\alpha}.
\end{eqnarray}
For each left eigenvector $\phi$ we define a scaling field $v$ by the relation
\begin{eqnarray}
v^{(i)}=\sum_{\alpha}\phi_{\alpha}(K^{(i)}-K^{*})_{\alpha}.
\end{eqnarray}
It satisfies under the renormalization group transformation the crucial property
\begin{eqnarray}
v^{(i+1)}=\lambda v^{(i)}.
\end{eqnarray}
We say that $v$ scales under the renormalization group transformation with a factor of $\lambda$. Thus the scaling field $v$ increases if $\lambda>1$ and decreases if $\lambda<1$. In other words, the eigenvalue $\lambda$ associated with the left eigenvector $\phi$ is what controls the flow behavior of the corresponding scaling field. Recall that under the renormalization group transformation the length scales as $L\longrightarrow bL$, i.e. it increases since $b>1$. Thus, if the eigenvalue $\lambda$ increases under the renormalization group transformation then we may write $\lambda=b^{y}$ where $y>0$ whereas for $y<0$ the eigenvalue $\lambda$ decreases. The exponent $y$ is called the dimension of the scaling field $v$. The scaling fields are in some sense the proper coupling constants whereas their exponents are their quantum mass dimensions.

Hence we will choose in the coupling-constant space the origin at  $K^{*}$ whereas the preferred directions are given by the scaling fields $v$. The fundamental equation (\ref{fundamental1}) reads in terms of the scaling fields
\begin{eqnarray}
F(v_1,v_2,...)=F_0(v_1,v_2,...)+b^{-d}F(\lambda_1 v_1,\lambda_2 v_2,...).
\end{eqnarray}
The function $F_0$ is regular at the fixed point (critical temperature) whereas the second term gives the singular part of the free energy.

Some definitions are now in order.

Irrelevant, relevant and marginal fields: For $\lambda<1$ the scaling field is called irrelevant (decreases in the RG transformation) whereas for $\lambda>1$ it is called relevant (increases in the RG transformation). Thus if we want to remain on a fixed point we must set the relevant scaling fields to zero. The case $\lambda=1$  corresponds to marginal fields.

Critical surface and fixed point: The fixed point lies on a hypersurface called the critical surface in the coupling-constant space. The critical surface is defined by setting all the relevant scaling fields to zero. Thus starting from any point on the critical surface we will approach the fixed point under a large number of iterations of the renormalization group transformation whereas starting from any point off the critical surface we will diverge from the fixed point. All actions (physical systems) located on the critical surface define collectively a so-called universality class. See figure.

Critical exponents: The eigenvalues $\lambda$ determine the critical exponents $y$. All systems in the same universality class have the same critical exponents.
\end{itemize}

## Critical exponents

The correlation length $\xi$ behaves around the critical temperature as
\begin{eqnarray}
\xi\sim |t|^{-1/D_t}~,~D_t=\frac{1}{\nu},
\end{eqnarray}
where $\nu$ is the mass critical exponent and $D_t$ is the temperature critical exponent. The gap (order parameter) critical exponent $\beta$ and the magnetic field critical exponent $D_h$ are given by
\begin{eqnarray}
M|_{H=0}\sim |t|^{\beta}~,~\beta=\frac{d-D_h}{D_t}.
\end{eqnarray}
The anomalous exponent $\eta$ is defined by the behavior of the $2-$point function at the critical temperature given by
\begin{eqnarray}
\Gamma(x)\sim \frac{1}{r^p}\exp(-r/\xi)~,~p=d-2+\eta=2d-2D_h.
\end{eqnarray}
There are three more critical exponents which can be determined solely in terms of $D_t$ and $D_h$ (scaling relations). These are $\alpha$ (specific heat), $\gamma$ (susceptibility), $\delta$ (magnetization at non-zero magnetic field).

## Exercise

Consider the Landau-Wilson energy functional in a constant magnetic field
\begin{eqnarray}
{E}[m]=\int d^dx\bigg[\frac{1}{2}(\vec{\partial}m(x))^2+\frac{1}{2}r_0 m^2(x)+u_0 m^4(x)\bigg]+\int d^dx h m(x).
\end{eqnarray}
We will integrate over a small momentum shell characterized by $b\simeq 1$, i.e. over the spherical shell of radius $\Lambda$ and thickness $\delta k=\tau.\Lambda$ where $\tau=\ln b$. We expand the field as
\begin{eqnarray}
m(x)=\bar{m}(x)+\delta m(x)~,~\delta m(x)=\sum_{i=1}^Nc_i\phi_i(x).
\end{eqnarray}
The $N$ wave packets (plane waves!!) $\phi_i(x)$ are of spatial size $\delta x= 1/\delta k$ and they are localized in the momentum shell $\delta k$. They are orthonormal and they also satisfy
\begin{eqnarray}
\int d^dx (\phi_i(x))^{2n+1}=0.
\end{eqnarray}
The background $\bar{m}$ is assumed to be constant over wave packets.

Question 1: Show that the RG-transformed energy functional (by integrating out the background at one-loop) is given by
\begin{eqnarray}
{E}^{\prime}[\bar{m}]={E}[\bar{m}]+\frac{1}{2}\sum_{i=1}^N\log\bigg(1+\frac{r_0}{\Lambda^2}+\frac{12u_0}{\Lambda^2}\bar{m}^2(x_i)\bigg),
\end{eqnarray}
where $x_i$ is the center of mass of the wave packet $i$.

Question 2: We approximate the sum over $x_i$ by an integral $\int dx$ and then we expand the logarithm. Show that we obtain
\begin{eqnarray}
E^{\prime}[\bar{m}]=\int d^dx\bigg[\frac{1}{2}(\vec{\partial}\bar{m}(x))^2+\frac{1}{2}\tilde{r}_0 \bar{m}^2(x)+\tilde{u}_0 \bar{m}^4(x)\bigg]+\int d^dx h \bar{m}(x).
\end{eqnarray}
Determine the values of $\tilde{r}_0$ and $\tilde{u}_0$ in terms of $r_0$ and $u_0$.

Question 3: By performing the second step (rescaling) and the third step (normalization) of the renormalization group transformation show that we can express the RG transformation as
\begin{eqnarray}
r_0^{\prime}-r_0=\bigg[2r_0+12C_d(\Lambda^{d-2}u_0-\Lambda^{d-4}r_0u_0)\bigg]\ln b.
\end{eqnarray}
\begin{eqnarray}
u_0^{\prime}-u_0=\bigg[(4-d)u_0-36C_d\Lambda^{d-4}u_0^2)\bigg]\ln b.
\end{eqnarray}
\begin{eqnarray}
h^{\prime}=b^{\frac{d+2}{2}}h.
\end{eqnarray}

Question 4: Determine the magnetic field critical exponent $D_h$.

Question 5: Show that the linearized renormalized group transformation can be put into the differential form
\begin{eqnarray}
\frac{dr}{d\tau}=2r+12C_d(\Lambda^{d-2}u-\Lambda^{d-4}ru)~,~\frac{du}{d\tau}=(4-d)u-36C_d\Lambda^{d-4}u^{2}.
\end{eqnarray}

## The Landau-Wilson model

In the above differential equations we change the variables as

\begin{eqnarray}
x=\frac{r}{\Lambda^2}~,~y=C_d\frac{u}{\Lambda^{4-d}}.
\end{eqnarray}
We obtain then the RG equations
\begin{eqnarray}
\frac{dx}{d\tau}=2x+12y-12xy~,~\frac{dy}{d\tau}=\epsilon y-36y^{2}.
\end{eqnarray}
The fixed point $(x_*,y_*)$ is defined by
\begin{eqnarray}
0=2x_*+12y_*-12x_*y_*~,~0=\epsilon y_*-36y_*^{2}.
\end{eqnarray}
We have the two solutions
\begin{eqnarray}
&&(x_*,y_*)=(0,0)~,~{\rm Gaussian}\nonumber\\
&&(x_*,y_*)=(-\epsilon/6,\epsilon/36)~,~{\rm Wilson-Fisher}
\end{eqnarray}
Linearizing the RG transformation around the fixed point by writing $x=x_*+\delta x$ and $y=y_*+\delta y$ we obtain the differential equations
\begin{eqnarray}
\frac{d\delta x}{d\tau}=(2-12y_*)\delta x+(12-12x_*)\delta y~,~\frac{d\delta y}{d\tau}=(\epsilon -72y_*)\delta y.
\end{eqnarray}
The RG matrix $W$ is given by
\begin{eqnarray}
W= \left( \begin{array}{cc}
2-12y_* & 12-12x_*  \\
0 & \epsilon-72y_*  \end{array} \right)
\end{eqnarray}
with eigenvalues given respectively by $\lambda_1=2-12y_*$ and $\lambda_2=\epsilon-72y_*$. The corresponding left eigenvectors are given  by
\begin{eqnarray}
\phi_1= \left( \begin{array}{c}
a   \\
\frac{12(1-x_*)a}{2-\epsilon+60y_*}   \end{array} \right)~,~\lambda_1=2-12y_*.
\end{eqnarray}
\begin{eqnarray}
\phi_2= \left( \begin{array}{c}
0   \\
1   \end{array} \right)~,~\lambda_1=\epsilon-72y_*.
\end{eqnarray}
The scaling fields are given respectively by
\begin{eqnarray}
v_1=a.\delta x+\frac{12(1-x_*)a}{2-\epsilon+60y_*}.\delta y~,~\lambda_1=2-12y_*.
\end{eqnarray}
\begin{eqnarray}
v_2=0.\delta x+1.\delta y~,~\lambda_2=\epsilon-72y_*.
\end{eqnarray}
These are given explicitly as follows.

Gaussian fixed point $(x_*,y_*)=(0,0)$:
\begin{eqnarray}
&&v_1=\delta x+6\delta y~,~\lambda_1=2\nonumber\\
&&v_2=\delta y~,~\lambda=\epsilon.
\end{eqnarray}

Wilson-Fisher fixed point $(x_*,y_*)=(-\epsilon/6,\epsilon/36)$:
\begin{eqnarray}
&&v_1=\delta x+(6-\epsilon)\delta y~,~\lambda_1=2-\frac{\epsilon}{3}\nonumber\\
&&v_2=\delta y~,~\lambda_2=-\epsilon.
\end{eqnarray}

The scaling fields satisfy
\begin{eqnarray}
\frac{dv_i}{d\tau}=\lambda_iv_i.
\end{eqnarray}
The solution is given by
\begin{eqnarray}
v_i=v_0b^{\lambda_i}.
\end{eqnarray}
Let us assume now that $d<4$ and close to $4$ ($\epsilon>0$ and small). In this case we have $\lambda_1=2-\epsilon/3>1$ and thus $v_1$ is a relevant scaling field whereas $\lambda_2=-\epsilon<1$ and thus $v_2$ is an irrelevant scaling field.

The critical surface corresponding to the Wilson-Fisher fixed point is determined by setting the relevant  scaling field, i.e. $v_1$, to zero. This of equivalent to the line in the $x-y$ plane given by
\begin{eqnarray}
y-y_*=-\frac{1}{6-\epsilon}(x-x_*).
\end{eqnarray}
To the linear order of $\epsilon$ this line passes by the other trivial fixed point at the origin. This line or critical surface represents our first preferred direction in the $x-y$ plane.

The critical surface $v_1=0$ represents the critical temperature $T=T_C$. This can be seen as follows. The other independent preferred direction should be obtained by setting the irrelevant variable $v_2$ to $0$. This gives $\delta y=0$ (the direction is parallel to the $x$ axis) and $v_1=\delta x=\delta r/\Lambda^2$ (this is the direction of increase or decrease of the temperature away from the critical value $T_c$).

The critical surface $v_1=0$ corresponds  therefore precisely to the critical temperature $T=T_C$.  Hence, starting at a point $A$ with $v_1<0$, i.e. with $T<T_C$, the system under RG will approach the low temperature phase where the symmetry is spontaneously broken. Whereas starting at a point $B$ with $v_1>0$, i.e. with $T>T_C$, the system will approach under successive RG transformations the high temperature phase. See figure.

The mass critical exponent is directly deduced to be  ($v_1$ is the relevant scaling field identified as the temperature and it behaves as $v_1\sim b^{\lambda_i}$)
\begin{eqnarray}
D_t=\lambda_1=2-\frac{\epsilon}{3}\Rightarrow \nu=\frac{1}{D_t}=\frac{1}{2}+\frac{\epsilon}{12}.
\end{eqnarray}
The magnetic field scaling field was shown in the exercise to be given by the mean field value:
\begin{eqnarray}
D_h=\frac{d+2}{2}.
\end{eqnarray}
From these two critical exponents we can calculate all other exponents as we have shown in the previous section.

The last remark concerns the dimensions $d>4$. In this case the two scaling fields are both relevant and therefore the Wilson-Fisher fixed point is unstable. The theory is actually dominated by mean field.