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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.

Then the author says that for the Ising model the relevant couplings are the temperature and the magnetic field, and here I'm already confused. First, because $A$ depends on what renormalization scheme we choose, so how can we predict what the eigenvectors are without saying more about $R$? Second, the Ising Hamiltonian (if we absorb temperature in it) is

$$ H_I=-\beta J \sum_{\langle i,j\rangle}s_is_j-\beta h\sum_i s_i$$

I don't see how $\beta$ and $h$ could ever appear as in $(\star)$ as two linearly superposed terms. We would need something like

$$ H_I=H^*+\beta\Phi_\beta+h\Phi_h$$

but I don't see how this can be true because $\beta h$ appears in the Hamiltonian, it seems to me that $\beta h$ must be treated as a single coupling, and you can't take $\beta$ and $h$ separately. In short, my first question is

How can one treat temperature as a coupling constant if it appears in all the coupling constants?

There were originally two parts of this question, following advice of Adam I posted the second part in a separate question

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  • $\begingroup$ When you perform renormalization, you've to first coarse grain and then do it more & more. So, you'd start with the Landau Free energy or what is known as the Wilsonian effective action in field theory. In that form, the temperature also appears in the coupling constants (which is what leads to the phase transition), apart from its usual role as a multiplicative constant to the overall free energy in the argument of the exponential of $Z$. $\endgroup$ – Vivek Jun 22 '20 at 12:02
  • $\begingroup$ I answered the first part of the question. Since on SE we usually prefer one question per post, maybe the second one could be posted as a separate question? (That I would happily answer.) $\endgroup$ – Adam Jun 23 '20 at 9:35
  • $\begingroup$ @Adam thanks again for the answer, I posted the second part as a separate question here $\endgroup$ – user2723984 Jun 23 '20 at 9:56
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While saying that the temperature and the magnetic field are relevant parameters is standard, it is somewhat sloppy, and confusing at first.

What is really meant is that both the temperature (more precisely $\beta J$ for the Ising model) and the magnetic field (or $\beta h$ for the Ising model) have non-trivial projections onto the two distinct relevant directions of the fixed point. Calling $t$ the coefficient of the operator $\Phi_t$ associated with the exponent $\nu$, and $H$ the coefficient of the symmetry breaking operator $\Phi_H$ associated to the exponent $\delta$, what is meant is that $t$ depends on $\beta J$ (but not $h$), and $H$ depends on $\beta J$ and $\beta h$ (since $H$ does not vanish at the critical temperature if $h$ is finite, its temperature dependence will give rise to corrections to scaling).

But in fact, all the coefficients $\mu_k$ do depend on $\beta J$ (and $\beta h$ if they correspond to symmetry breaking operators), but since most are irrelevant, we don't really care (unless we want to describe correction to scaling, see also below).

Furthermore, if the model is more complicated (with for instance more interactions $J'$, $J''$ between spins), then generically any of the couplings will be relevant (that is, one can induce a transition by changing $J'$, keeping $J$ constant), though the phenomenology might be complicated.

Finally, why do we usually assume that $t$ is linear in the temperature? For this, we use our physical intuition. First, the easiest thing we can change experimentally is most of the time the temperature, so it is the relevant physical quantity. Since changing the temperature does induce a transition (otherwise, we would use another physical quantity), the relevant parameter $t$ must depend on temperature. Furthermore, since to get close to the fixed point, only a finite number of RG transformation is necessary, and since each transformation is analytic in all quantities, $t$ is an analytic function of the temperature. By definition, it vanishes at the critical temperature. By Taylor expanding around $0$, it will thus be a linear function of temperature close enough to the critical temperature. In principle, the linear coefficient could vanish, but this is not the case generically (it can happen if there are some additional constraints that are robust, due to some additional symmetry, for instance). The fact that $t$ is in fact a complicated function of the temperature will induce correction to scaling is we are not close enough to the transition.

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  • $\begingroup$ Thank you very much for taking the time to answer! You clarified lots of things, I want to read your book on the RG instead of the one I have. Just a clarification: how do we know that $\Phi_t$ does not break the symmetry, and hence does not depend on $h$? Why couldn't $\Phi_t$ and $\Phi_h$ both be non trivial combination of symmetry breaking and non breaking operators? $\endgroup$ – user2723984 Jun 23 '20 at 9:48
  • $\begingroup$ Also if $h$ (the scaling) depends on both $\beta h$ and $\beta J$, why do we only care about its $h$ (the field) dependence? $\endgroup$ – user2723984 Jun 23 '20 at 9:51
  • $\begingroup$ @user2723984 concerning your second question, see my edit. Concerning the first, it is less trivial, and probably the symmetry argument is not enough. For a linear symmetry breaking term (such as a magnetic field), I know how to show that all other parameters are independent of $h$ in a more advanced RG setting (it amounts to some kind of non-renormalizability of the $h$ term). It might be doable in a Kadanoff like setting, but I've never thought about it. $\endgroup$ – Adam Jun 23 '20 at 10:18
  • $\begingroup$ thank you for the answer, this was very helpful. You said that you would also try to answer what used to be the second part of the question and is now here, it would be very helpful if you could look at it if you have time. Thanks! $\endgroup$ – user2723984 Jun 26 '20 at 7:40
  • $\begingroup$ @user2723984 : no problem. I've answered the other question. $\endgroup$ – Adam Jun 26 '20 at 13:46

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