Renormalization Group - Scaling fields and physical critical exponents (1D Ising model)

Question

This is related to this question: Critical exponents and scaling dimensions from RG theory.

TLDR: How to compute physical critical exponents $\alpha, \beta, \gamma, etc$ from the RG exponents when the scaling fields are not the reduced temperature and field?

Consider the 1D Ising model with interactions between first neighbours and a magnetic field, as studied here (page 101). As parameters, we take the "reduced" coupling constant $K=J/(k_BT)$ and field $h=H/(k_BT)$.

$\pmb{\big[}$We then proceed to use RG. We start by performing decimation, which leads us to recursion relations that, after the change of variables $K\rightarrow x=e^{-4K}, \;h\rightarrow y=e^{-2h}\quad \text{(4.72)}$ , look like $4.79$. Linearizing these relations near the critical point, we find the eigenvalues $\lambda_i$ of the linearization matrix ($4.100$ and $4.101$). Since $\lambda_i(b)=b^{x_i}$, where $x_i$ are the critical exponents and $b=2$ (scaling factor), we find the $x_i$ (in this case, $x_1=2,x_2=1)$.$\pmb{\big]}$

Now I want to get the physical critical exponents $\alpha, \beta, \gamma, etc$ from the $x_i$, which is essentially explained in pages 111-112 and 116. When the scaling fields $h_i$ are $t$ and $h$ (reduced temperature and field), as in page 116, they are given by \begin{equation} \alpha=2-\frac{d}{x_t}\quad \beta=\frac{d-x_h}{x_t} \end{equation} and so on. The problem is now my scaling fields are $x$ and $y$ given by the change of variables above. How can I relate the $x_i$ with $x_t,x_h$?

EDIT: For instance, here they say "considering $K$ a coupling constant is equivalent to considering $T$ as such", but there's no explaining.

Answer 1

@xihiro: you have almost answered all the way to the end above, and are just a tiny step away. You have found $x_1$ and $x_2$, which are eigenvalues $b^x_i$ of matrix R that you need for your RG step:

$ \left[ {\begin{array}{cc} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \\ \end{array} } \right] $

But what we really need is the matrix: $ R(g)= \left[ {\begin{array}{cc} \frac{\partial K'}{\partial K} & \frac{\partial K'}{\partial h} \\ \frac{\partial h'}{\partial K} & \frac{\partial h'}{\partial h} \\ \end{array} } \right] $

and eigenvalues of R(g) computed specifically around this point. It is easy to see that they are simply related as, for example:

$\frac{\partial x'}{\partial y}=\frac {\frac{\partial x}{\partial K}}{\frac{\partial y}{\partial h}} \frac{\partial K'}{\partial h}$

So, if you now write matrix R(g) at the critical point, you will find that the eigenvalues are again $b^y_t$ and $b^y_h$, where $y_t=2$ and $y_h=1$. Now you can use usual approach to compute critical coefficients, i.e. for example: $\nu=1/y_t=0.5$, and $\alpha=\frac{d}{y_t}-2$ etc.