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 $z$?
Answer
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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