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Extending the Gaussian model by introducing a second field and coupling it to the other field, I consider the Hamiltonian

$$\beta H = \frac{1}{(2\pi)^d} \int_0^\Lambda d^d q \frac{t + Kq^2}{2} |m(q)|^2 + \frac{L}{2} q^4 |\phi|^2 + v q^2 m(q) \phi^*(q)$$

Doing a Renormalization Group treatment, I integrate out the high wave-numbers above $\Lambda/b$ and obtain the following recursion relations for the parameters: $$\begin{aligned}t' &= b^{-d} z^2 t & K' &= b^{-d-2}z^2 K & L' &= b^{-d-4}y^2 L \\ v' &= b^{-d-2}yz v & h' &= zh \end{aligned}$$ where $z$ is the scaling of field $m$ and $y$ is the scaling of field $\phi$.

One way to obtain the scaling factors $z$ and $y$ is to demand that $K' = K$ and $L' = L$, i.e., we demand that fluctuations are scale invariant.

But apparently, there is another fixed point if we demand that $t' = t$ and $L' = L$ which gives rise to different scaling behavior, and I wonder

a) why I can apparently choose which parameters should be fixed regardless of their value ($K$ and $L$ in one case, $t$ and $L$ in the other case)

b) what the physical meaning of these two different fixed points is...

(My exposure to field theory/RG is from a statistical physics approach, so if answers could be phrased in that language as opposed to QFT that'd be much appreciated)

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The short answer is that if you have coefficients for all the terms, you have two independent exact fake scale invariance for the field $\phi$ and $m$ which just rescales the fields and the coefficients appropriately to keep the Hamiltonian exactly the same. This is not a real invariance of the action, since it changes the parameters of the action, it is best thought of as choosing the dimensional scale of the two fields. You usually do this by fixing the terms "L" and "K", but you get a different scaling if you fix the "t" and "L" terms, which is physical in different limits.

I should point out that this model is exactly solvable, there are no real interactions in this model, so the Renormalization Group analysis is just dimensional analysis in disguise. There are two rotated q-modes mixing $m(q)$ and $\phi(q)$ which are completely free.

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