Critical exponents and scaling dimensions from RG theory

Question

In most books (like Cardy's) relations between critical exponents and scaling dimensions are given, for example $$ \alpha = 2-d/y_t, \;\;\nu = 1/y_t, \;\; \beta = \frac{d-y_h}{y_t}$$ and so on. Here $y_t$ and $y_h$ are scaling dimensions of scaling variables $u_t$ and $u_h$ related to (reduced) temperature $t$ and field $h$. This is always discussed in the context of the Ising model. I am confused about what $y_t$ and $y_h$ are in general? In general you have scaling dimensions $y_1, y_2\dots y_n$ for the scaling fields $u_1, u_2,\dots u_n$, where it's not clear what $y_t$ and $y_h$ are. In cases where the RG is 'diagonal', $t$ and $h$ themselves are scaling variables, the problem does not exist, but that is not the general situation.

For example, the RG equation of the XY model in $d=2+\epsilon$ is $$\frac{dT}{dl} = -\epsilon T+ 4\pi^3 y^2,\;\; \frac{dy}{dl} = \left(2-\frac{\pi}T\right)y,$$ where $T$ is the temperature and $y$ is related to vortex fugacity. There is a finite temperature fixed point for $T^\star=\pi/2$. Imagine we want to calculate $\nu$ and $\alpha$ at this non-trivial fixed-point. By linearizing the above at the fixed point, we get some two dimensions $y_1$ and $y_2$ for two scaling variables $u_1$ and $u_2$. These variables are both linear combinations of $T$ and $y$. How can I know which scaling dimension/eigenvalue corresponds to the thermal eigenvalue $y_t$? What are the values of $\alpha$ and $\nu$?

Answer 1

The OP is right that the couplings $g_1$, $g_2$,... parametrizing the field theory are in general combinations of the scaling field of the RG fixed points $u_1$, $u_2$,...

In the Ising model, the two most relevant scaling field $u_1$ and $u_2$ can be associated with the temperature $t$ and the magnetic field $h$. However the OP is wrong when saying that $u_1$ and $u_2$ are diagonal (meaning that $u_1=t$, $u_2=h$). Indeed, in general, every coupling that respect the Ising symmetry will have a projection onto $u_1$, whereas all couplings that break the symmetry will have a component onto $u_2$. This means for instance that one can in principle drive the transition not by changing $t$ in the bare action, but by changing the interaction $\lambda$. In that case, the correlation length will diverge as $|\lambda-\lambda_c|^{-1/y_t}$. However, this is usually not what happens for microscopic models, and is thus not discussed in most books.

The confusion arises because one usually work in perturbation theory, where very few coupling constants are kept, implying a projection of the whole flow onto a very small subspace of the true space of all coupling constants.

Concerning the flow equation of the XY model close to two dimensions, one should notice that here we only have one relevant field (which one naturally associates with the temperature, as it is the experimentally tunable parameter), corresponding to $u_1$ as it preserves the XY symmetry $y_t$ is thus the value of the positive eigenvalue associated with deviations to the fixed point (one then compute $\nu$ and $alpha$ from the equations given by the OP). There are no symmetry breaking field here, so one does not see the effect of $u_2$, and its eigenvalue $y_h$. Instead, the other direction is an irrelevant one (also associated with a XY symmetric field), the eigenvalue only contributing to correction to scaling.