I am confused about an aspect of coupling constant rescaling in the Wilsonian renormalization group procedure. (I'm following Kardar's "Statistical Physics of Fields, Ch5). I think I understand the basic idea of the renormalization group, but I'm in undergrad and haven't taken field theory or an advanced stat mech course so if I have a conceptual error somewhere I'd really appreciate any corrections.

The partition function for the Landau Ginzburg hamiltonian is written as ($\tilde{\vec{m}}(\mathbf{q}) \ \text{and }\sigma(\mathbf{q})$ are the splitting of the original field into slow and fast components)

$$ \begin{align} Z &= \int D\tilde{\vec{m}}(\mathbf{q})D\sigma(\mathbf{q}) \exp{\bigg\{- \int_{0}^{\Lambda} \frac{d^d \mathbf{q}}{(2\pi)^d} \bigg( \frac{t + K q^2}{2} \bigg) (|\tilde{m}(\mathbf{q})|^2} + |\sigma(\mathbf{q})|^2)-U[\tilde{\vec{m}}(\mathbf{q}),\sigma(\mathbf{q})] \bigg\}\\ &= \int D\tilde{\vec{m}}(\mathbf{q}) \exp{\bigg\{- \int_{0}^{\Lambda} \frac{d^d \mathbf{q}}{(2\pi)^d} \bigg( \frac{t + K q^2}{2} \bigg) (|\tilde{m}(\mathbf{q})|^2}\bigg\} \exp{\bigg\{-\frac{nV}{2} \int_{\Lambda/b}^{\Lambda} \frac{d^d \mathbf{q}}{(2\pi)^d} \log(t + K q^2) \bigg\}} \bigg\langle e^{-U[\tilde{\vec{m}},\vec{\sigma}]}\bigg\rangle_{\sigma} \end{align} $$ I think I understand the overall procedure: integrate out the momenta above the cutoff; rescale the momenta $\mathbf{q} = b^{-1} \mathbf{q}'$ and the field $\tilde{\vec{m}} = z {\vec{m}\,}'$. Then you get the new hamiltonian:

$$ (\beta H)'[m'] = V(\delta f_b^0 + u \delta f_b^1) + \int_{0}^{\Lambda} \frac{d^d \mathbf{q'}}{(2\pi)^d} b^{-d}z^2\bigg( \frac{\tilde{t} + K b^{-2} q'^2}{2} \bigg) |m'(\mathbf{q'})|^2 +u b^{-3d} z^4 \int_{0}^{\Lambda} \frac{d^d \mathbf{q}'_1 d^d \mathbf{q}'_2 d^d \mathbf{q}'_3 d^d \mathbf{q}'_4}{(2\pi)^d} \vec{m}(\mathbf{q}'_1)\cdot \vec{m}(\mathbf{q}'_2)\vec{m}(\mathbf{q}'_3)\cdot\vec{m}(\mathbf{q}'_4) \ \delta^d(\mathbf{q}'_1+\mathbf{q}'_2+\mathbf{q}'_3+\mathbf{q}'_4) $$

where the $t$ is $$\tilde{t} = t+4u(n-2) \int_{\Lambda/b}^{\Lambda} \frac{d^d \vec{k}}{(2\pi)^d} \frac{1}{t+K\ k^2}$$

Then you choose $z=b^{1+\frac{d}{2}}$ so that $K$ stays the same: $K'=K, \ u' = b^{-3d} \ z^4 \ u, \ \text{and} \ t'= b^{-d} \ z^2 \ \tilde{t}$.

My question is: why doesn't the $u$ inside $\tilde{t}$ become a $u'$ ? As I understand it, the couplings change with the cutoff, so shouldn't the $u$ be replaced with $u'$ wherever it appears? If not, why not, and what is the physical meaning of this? (Originally asked here but I decided to split into separate questions.)


1 Answer 1


What you are doing as part of this calculation is deriving the relationship between the coupling constants of the model at the coarse scale and the original scale. The results you get, $$K' = K$$ $$u' = b^{-3d} z^4 u,$$ $$t' = b^{-d} z^2 \left(t + (n-2)\int_{\Lambda/b}^\Lambda \frac{d^d\vec{k}}{(2\pi)^d} \frac{1}{t+K k^2}\right)$$ are the so-called recursion relations between the parameters at the original fine scale ($K$, $u$, and $t$) and the parameters at the coarse scale ($K'$, $u'$, and $t'$). You do not replace $u$ with $u'$ in $\tilde{t}$ for the same reason that you do not replace the $t$ with $t'$. i.e., the right hand sides are the old parameters at the original scale, the left hand sides are the new parameters at the coarsened scale.

The primed quantities are really just a relabeling of terms in your coarsened Hamiltonian so that it matches the original Hamiltonian (up to approximations); they are not a change of variables. The only change of variables you are actually performing is the rescaling of the degrees of freedom.

In case it helps to clarify what you are actually doing during this calculation, I'll elaborate on that last sentence: the renormalization group procedure consists of two distinct steps: 1) averaging over (integrating out) statistical degrees of freedom and 2) making a change of variables on the remaining degrees of freedom to restore the system to its original scale.

I like to think of this in terms of probability distributions: if you have a multivariate distribution $p(m_1,m_2,\dots,\sigma_1,\sigma_2,\dots) \propto \exp(-H(m_1,m_2,\dots,\sigma_1,\sigma_2,\dots))$, then the first step of the renormalization group procedure is to marginalize over (average out) the variables $\sigma_1,\dots$ to obtain the distribution $p(m_1,m_2,\dots)$. Then, you define a new set of rescaled variables $m_1',m_2',\dots$, giving a new distribution $$p'(m_1',m_2',\dots) = \left|\frac{\partial \vec{m}}{\partial \vec{m'}} \right| p(m_1(\vec{m}'),m_2(\vec{m}'),\dots) \propto e^{-H'(m_1',m_2',\dots)},$$ where $\left|\frac{\partial \vec{m}}{\partial \vec{m'}} \right| $ is the Jacobian of the change of variables (I've assumed continuous variables for simplicity here).

Often this change of variables is just a rescaling $m' \sim z m$ (and the Jacobian therefore does not add any important terms to the Hamiltonian). After this, defining the primed coupling constants is in some sense just a matter of notational cleanup. Since we expect that in many cases our coarse-grained Hamiltonian will have more or less the same form as our original Hamiltonian, it makes sense to define new coupling constants so that the form of the two Hamiltonians is superficially similar. e.g., if the pairwise interaction term in the original Hamiltonian were $J m_i m_j$ and the pairwise term in the coarsened Hamiltonian is $f(J,{\rm other~couplings}) m_i' m_j'$, we define $J' = f(J,{\rm other~couplings})$.

The conceptual leap that follows is that, if we are able to perform this procedure in the first place, mashing on the coarsened Hamiltonian until it looks like the original Hamiltonian, then there is nothing* stopping us from doing this again and again, obtaining a new set of coarsened coarsened couplings $J'' = f(J',{\rm other~couplings}')$, which allows us to interpret this relationship between the couplings as a recursion relation at different scales. (*important caveat: if there is only a finite number of degrees of freedom, then we can only perform this procedure a finite number of times).

  • $\begingroup$ Thanks so much for your answer, its extremely helpful. I suspected that something like this was the case and was the reason for my confusion. The suggestion to think about distributions is great! Can you also help with part 2 of my question here? I'd like to write the 'free energy' as $Z' = \int Dm'(\mathbf{q}) e^{(\beta H_{gaussian})'[m'] + U'[m']} \rightarrow F_{gaussian} + F_{corrections}$ but can't see how to expand the log(integral). Thanks again! $\endgroup$
    – user264588
    May 22, 2020 at 14:45
  • $\begingroup$ Sorry, forgot to put the link to my original question: physics.stackexchange.com/questions/552497/… $\endgroup$
    – user264588
    May 22, 2020 at 16:44

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