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 $z$?