Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 69 / 2 / b

Let $b,c>1$, rescale $q_{\parallel }^{\prime }=b{q}_{\parallel }$, ${\mathbf{q}}_{\perp }^{\prime }=c{\mathbf{q}}_{\perp }$, and set $B=b{c}^{d-1}$. Use the Fourier-amplitude conventionThe momentum measure contributes $B^{-1}$, and substitution into the quadratic statistical Hamiltonian givesIn the uniform source the momentum delta function cancels the Jacobian determinant of the measure, explaining why $h^{\prime }$ does not have an additional $B^{-1}$. In a full momentum-shell renormalization group transformation integrate out the fast modes first. Their uncoupled Gaussian integrals contribute an additive free-energy density; the displayed transformations describe the slow-mode statistical Hamiltonian after coordinate and field rescaling.

The field convention must be stated. In real space, $x_{\parallel }^{\prime }=x_{\parallel }/b$ and ${\mathbf{x}}_{\perp }^{\prime }={\mathbf{x}}_{\perp }/c$, and Fourier inversion gives $m^{\prime }(x^{\prime })=Bm(x)/z$. Thus if instead the symbol $m^{\prime }=m/z_x$ refers to real-space amplitudes, $z=B{z}_x$. Then the same transformation reads $K^{\prime }=B{z}_x^{2}K/b^2$, $L^{\prime }=B{z}_x^{2}L/c^4$, $t^{\prime }=B{z}_x^{2}t$, $h^{\prime }=B{z}_{x}h$. These are equivalent normalizations, not different critical exponents.