Relating scaling and critical exponents in the Ising model

Question

I'm reading the chapter about the renormalization group in Yeoman's book "Statistical mechanics of phase transitions" and I'm puzzled about how the author relates the scaling of the RG with the critical exponents. We have some RG map on the Hamiltonian $H\rightarrow R(H)$. We suppose that we are close to the fixed point $H^* $, so

$$H'=R(H^*+\delta H)=H^* + A(H^*)\delta H$$

where $A$ is a matrix and $\delta H$ is seen as a vector with the coupling constants as components. This matrix can be diagonalized and we can write

$$ A(H^*)\delta H= A(H^*)\sum_k\mu_k \Phi_k=\sum_k\lambda_k\mu_k \Phi_k\tag {$\star$}$$

where $\Phi_k$ are functions of the lattice and $\lambda_k$ are the eigenvalues of $A$. It's easy to argue that they must have the form

$$ \lambda_k=b^{y_k}$$

where $b$ is the scaling factor of the map. No problem until here. If $y_k>0$ we call it relevant, otherwise irrelevant. Consider the 2D Ising model $$ H_I=-\beta J \sum_{\langle i,j\rangle}s_is_j-\beta h\sum_i s_i$$ We know that one of the two relevant direction is $t\sim \frac{T-T_c}{T_c}$ as temperature controls the phase transition and $t$ must vanish at $T_c$, and the other can be identified with the magnetic field $h$.

The author then gives the scaling form of the free energy, which I agree with

$$ f(t,h,g_3,g_4,\dots)=b^{-d}f(b^{y_T}t,b^{y_h}h, b^{y_3}g_3,\dots)$$

differentiating twice wrt $t$

$$ f_{tt}(t,0,0,0,\dots)=b^{2y_T-d}f_{tt}(b^{y_T}t,0, 0,\dots)$$ from this the author wants to extract the critical exponents for the specific heat capacity, she writes (page 116)

this can be done because the scaling factor $b$ is arbitrary. Choosing $b^{y_1}|t|=1$ transfers all the temperature dependence to the prefactor and leaves it multiplied by a function of constant argument $$f_{tt}(t,0,0,0,\dots)=|t|^{(d-2y_T)/y_T}f_{tt}(\pm 1,0, 0,\dots) $$

and here I'm completely lost: what happened? The scaling factor is not arbitrary - it depends on my choice of renormalization map (for example, choice of block size). Much less it is dependent on $|t|$! In fact the whole $b^{y_T}$ cannot depend on $|t|$, because it is an eigenvalue of a matrix that's independent of the Ising model or its temperature. How can I make sense of all this?

Answer 1

The argument is easier to follow from the continuous RG point of view (a la Wilson), though it can be adapted in the block spin picture. I'll take the former point of view here.

The RG procedure gives rise to a flow of the Hamiltonian, $H_s$, with $H_{s_0}$ the initial Hamiltonian, and $\partial_s H_s = R(H_s)$ (to connect with the OP notations $b=e^{s}$). Assume that $H_{s_0}$ is fine-tuned such that the flow brings $H_s$ very close to the fixed point Hamiltonian $H^*$, and call $s^*$ a RG time at which the distance between $H_s$ and $H^*$ is small enough so that we can linearize the flow.

We find that for $s$ not too far from $s^*$, $$ H_s=H^*+\sum_k \mu_k(s) \Phi_k, $$ with $\mu_k(s)=e^{(s-s^*)y_k}\mu_k(s^*)$. Assume that there is only one relevant direction, parametrized by the term $k=1$. Physically, we know that if the system was at $T=T_c$, the flow would go to the fixed point, and therefore, $\mu_1(s^*)$ must vanish with $t=(T-T_c)/T_c$. Assume that it does so linearly, then $\mu_1(s)=a e^{(s-s^*)y_k} t$, with $a$ a constant. In principle, $\mu_{k>1}(s^*)$ will also depend on $t$ (i.e. on the initial condition of the flow), but they do not have to vanish to get to the fixed point, so we can forget about this dependence.

Now, for the linearization to be valid, we need to stay close to the fixed point, this means that we need both $|t|\lesssim 1$ (to be close to the fixed point at $s^*$) and $e^{(s-s^*)y_1} t\lesssim1$, assuming that $a$ is of order one, so that the flow is still close to the fixed point.

The free energy scales as $$ f(t) = e^{-s d} f(\mu_k(s)), $$ where we only write its dependence on $t$ in the rhs of the equation. Assuming we can linearize the flow as discussed above, we have $$ f(t) = e^{-sd}f(e^{(s-s^*)y_1} t, \ldots). $$ Now using the fact that $s$ is arbitrary, and choosing $s=(s^*-\log(|t|))/y_1$ which is ok (as we are somewhat still in the regime of linearizability of the flow), we find $$ f(t) \propto |t|^{d/y_1}, $$ where the proportionality constant is $e^{-s^*d/y_1}f(\pm1,\ldots)$, which is finite for all $t$ (since $s^*$ is just a finite non-universal RG time, and $f(\pm1,\ldots)$ is finite and depends on $|t|^{y_k/y_1}\mu_k(s^*)$, $k>1$, which vanishes for $t$ small enough). (However, if one of the irrelevant coupling is dangerously so, one must be more careful.)

Answer 2

The scaling relation must hold for arbitrary scaling factors $b$ and thus it must also hold for the specific choice $$b = |t|^{-\frac{1}{y_{T}}} .$$ Inserting this into the scaling relation you get the above form. I would say this is somewhat similar to parametrising a function of two arguments $f(x,y)$ via setting $y=y(x)$ and effectively getting a function $g(x)=f(x,y(x))$ depending only on $x$.
Since the relation holds for all $b$, we can also choose it to be temperature dependent. I hope this helps.