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I'm studying how the Renormalization Group treatment of the simple Gaussian model, $$\beta H = \int d^d r \left[ \frac{t}{2} m^2(r) + \frac{K}{2}|\nabla m|^2 - hm(r)\right]$$

In momentum space, the Hamiltonian reads $$\beta H = \frac{1}{(2\pi)^d} \int d^d q \left[\frac{t + q^2 K}{2} |m(q)|^2\right] - hm(0)$$ and the $q$-integral runs from $0$ to some long-wavelength cut-off $\Lambda$.

The coarsening is done by splitting this integral into one from $0$ to $\Lambda/b$ and one from $\Lambda/b$ to $\Lambda$, and because the Gaussian model is so simple, the two integrals don't mix and decouple nicely. The high-momentum integral contributes just a constant additional term to the free energy, so we ignore it, and then we are left with $$\beta H = \frac{1}{(2\pi)^d} \int_0^{\Lambda/b} d^d q \left[\frac{t + q^2 K}{2} |m(q)|^2 \right] - hm(0).$$

The rescaling is done by introducing a new momentum $q' = bq$. For the order parameter $m(q)$, one makes the scaling assumption $m'(q') = m(q)/z$. Then I can rewrite $\beta H$ in terms of the new momentum variable $q'$, and then I demand that the rescaled Hamiltonian has the same functional form as the old Hamiltonian, which allows me to read off

$$t' = b^{-d} z^2 t$$ $$K' = b^{-d-2} z^2 K$$ $$h' = zh$$

Now we don't know $z$, and in the literature I found that one somehow demands that $K' = K$, so that $z = b^{d/2 + 1}$, and I don't really understand why we can make that demand, and if there are other possibilities. Could we also demand that $t' = t$ and read off a different $z$?

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3 Answers 3

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OK, so I usually think of the RG analysis as (i) as splitting the field into 'fast' and 'slow' modes:

$$m(q) = m_<(q) + m_>(q)$$

where $m_<(q) = 0$ for $q> \Lambda$ and $m_>(q) = 0$ for $q<\Lambda$.

Then (ii) you integrate out the fast modes $m_>(q)$, for which you conveniently split the action in the fast, slow and mixed terms. And then (iii) scale the momentum, scale the fields and absorb any changes into the coupling constants.

It's the scaling of the fields which you are concerned with. The motivation comes from the fact that you want the massless Gaussian model to be a fixed point. In other words, the model for $t=h=0$ is the kinetic term,

$$\beta H = \int d^dr \frac{K}{2}|\nabla m|^2 $$

of which we now it is invariant under an RG transformation (= integration of fast modes + rescaling).

Now, I'm not sure what happens if you choose a different scaling parameter -- you'll most definitely get a different set of RG equations, and a coupling constant space with a different set of fixed points. The original massless Gaussian model is no longer critical though.

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Yes. It is a merely a convenient choice to fix the term in front of the gradient. You may make some other choice and get some other equivalent RG flow. An interesting toy to play with is a field with kinetic term $K_1 q^2 + K_2 |q|^\alpha$ where $\alpha >0$ is some parameter. You can write an RG flow choosing to fix either $K_2$ or$K_1$. Look at the resulting fixed points and their stability in either flow.

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  • $\begingroup$ Thank you. Related to this, I've started another question for a slightly more complicated model. $\endgroup$
    – Lagerbaer
    Nov 25, 2011 at 19:37
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You have to remember what $t$ means in the Hamiltonian. Usually it stands for $t\equiv(T - T_c)/T_c$, i.e., $t$ is the reduced temperature. Now, since we are interested in the critical behavior, the physical relevant fixed points have to have $t = 0$.

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    $\begingroup$ It's unclear exactly what your answer is trying to address. $\endgroup$ Jan 3, 2019 at 8:25

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