In 2D a scalar field is dimensionless so terms in the Lagrangian $\phi^n$ of arbitrary power are renormalizable (indeed we can even have two derivatives). This has two consequences that seem to be in tension to me, so maybe I am missing something important.

Let's say we start with a $\phi^4$ theory. What does that mean? On the one hand, in perturbation theory we can calculate any correlation function in terms of a finite set of renormalized parameters, so the theory is well defined. All the dependence on the cutoff is shifted to our bare parameters.

Another way to define the theory is to put it on a lattice and measure the correlation functions. I would imagine the formulas for the bare parameters in the perturbative approach tell you how to adjust your parameters on the lattice so that you get the same low energy theory for different lattice spacings. At every lattice spacing the theory is a $\phi^4$ theory, only the parameters change.

On the other hand, if we consider Wilsonian renormalization, and integrate over a shell of momentum, all sorts of different terms are generated in the Lagrangian. Unlike in higher dimension, these extra terms, e.g. $\phi^6$ stick around as you go to lower energy since they are renormalizable.

From this point of view I would imagine if you simulate the $\phi^4$ theory at one lattice spacing, if you want to change the lattice spacing and get the same theory at low energy you had better include these extra renormalizable terms in the Lagrangian. So it only looks like a pure $\phi^4$ theory at one scale.

How is this resolved?


1 Answer 1


I gave a much more detailed answer at What is the Wilsonian definition of renormalizability? so try to read it.

But a quick explanation is as follows: Let $\Lambda=2^n$ be your UV cutoff. You are dealing with a sequence $n=1,2,\ldots$ of bare pure $\phi^4$ theories which are starting points for RG transformations, say a transformation $T$ corresponding to zooming by a factor of 2. The effective theory at say unit scale is the limit of $T^n$ applied to the starting point corresponding to $\Lambda=2^n$, when $n$ goes to infinity. So the starting points are pure (no $\phi^6$ junk etc.) but the effective theories will usually contain in the Lagrangian all terms allowed by the symmetries.

Edit: The OP asks me to point out where the OP's confusion is, so I'll try. I will use the notations of my long answer linked to above. The UV cut-off in momentum space is $\Lambda=e^{-t}$ where $t$ will eventually be taken to $-\infty$ in order to perform the construction of a continuum QFT. The bare action is $$ V_{t}^{\rm bare}(\phi)=\sum_{i\in I}g_i^{\rm bare}(t)\int \mathcal{O}_{i}(x)\ d^dx $$ for Wick-ordered local operators $$ \mathcal{O}_{i}(x)=:\partial^{\alpha_1}\phi(x)\cdots\partial^{\alpha_k}\phi(x):_t $$ and bare couplings $g_i^{\rm bare}(t)$. Here $I$ is just an index set of labels for operators. Note that Wick-ordering is with respect to the Gaussian theory with cut-off at scale indicated by $t$, hence the $:\cdots:_t$ notation.

Regardless of one's choice of method, modern Wilsonian RG or old BPHZ counterterm renormalization, what needs to be done to construct the continuum QFT is to choose the right ansatz for the bare coupling dependence on the UV cut-off, i.e., the functions $g_i^{\rm bare}(t)$, so that $$ \forall t_2, \ \lim_{t_1\rightarrow -\infty}\ RG[t_2,t_1](V_{t_1}^{\rm bare})=:V_{t_2}^{\rm eff}\ {\rm exists}. $$ Here $RG[t_2,t_1](V_{t_1}^{\rm bare})$ is the (temporary) effective theory at (log, position space) scale $t_2$ for the bare theory with cut-off at scale $t_1$. What matters in the end is $V_{t_2}^{\rm eff}$ which is the effective theory at scale $t_2$ of the wanted continuum QFT.

The OP's confusion is to think that $V_{t}^{\rm bare}$ and $V_{t}^{\rm eff}$ are the same when they are not.

If one is constructing a $\phi^4$ theory, this means that the only nonzero bare couplings $g_i^{\rm bare}(t)$ one is allowed to play with are the couplings for the operators $(\partial\phi)^2$, $\phi^2$ and $\phi^4$. It is fine to transgress this rule, but then one would be constructing some other theory like a $\phi^6$ model, etc.

Now even if one adheres to the rule and succeeds in removing the cut-off, the effective theories $$ V_{t}^{\rm eff}(\phi)=\sum_{i\in I}g_i^{\rm eff}(t)\int \mathcal{O}_{i}(x)\ d^dx $$ will have nonzero couplings for pretty much all operators allowed by symmetry. This is because the RG is a highly nonlinear transformation.

Finally, the long answer I gave is more adapted to the generic situation (like $\phi^4$ in 3D) where the perturbed CFT has finitely many relevant operators. The 2D situation mentioned by the OP is special: all powers $\phi^k$ are relevant at the Gaussian fixed point. So the procedure to find the right bare ansatz $g_{i}^{\rm bare}(t)$ is different. The easiest choice which works is to take $g_{i}^{\rm bare}(t)$ constant with respect to the UV scale $t$. All the counterterm renormalizations (just tadpoles) are hidden in the Wick-ordering $:\cdots:_t$.

  • $\begingroup$ Yes that is the Wilsonian side. I understand. So why in the perturbative side can I choose a regularization, find a correlation function and then re-express that in terms of non-cutoff dependent things like the correlation length, and shift all regularization dependence to the parameters of $\phi^4$ only? That implies we can change the scale of regularization of the $\phi^4$ theory and get the same theory. $\endgroup$
    – octonion
    Oct 10, 2018 at 23:35
  • $\begingroup$ Are you saying what we call the $phi^4$ theory at unit scale is necessarily defining the theory at some kind of limit as $\Lambda$ goes to infinity? So the bare parameters, which have $\Lambda$ dependence have no meaning? This is what I was thinking of as the `old' way of thinking about renormalization. I thought that the bare parameters tell you how to define your theory at a finite $\Lambda$. $\endgroup$
    – octonion
    Oct 10, 2018 at 23:54
  • $\begingroup$ Well, try to read the answer I linked to. That should remove the confusion you seem to be struggling with. $\endgroup$ Oct 11, 2018 at 14:49
  • $\begingroup$ Well I started to read it, and after a couple pages of translating I still didn't see anything I didn't know. Now I'm clearly misunderstanding something, and it's quite possible somewhere buried in a long answer to a completely different question I will discover something. But it would be quite helpful if you could point out where my confusion is, or even give some indication you understood my question. $\endgroup$
    – octonion
    Oct 11, 2018 at 14:58
  • 1
    $\begingroup$ Thank you, that was very clear and I upvoted. Basically I am content to deal with 'temporary' effective theories rather than a true continuous theory. I was indeed assuming it is the case that $RG[t',t_1](V_{t_1}^{bare})=V_{t'}^{bare}$ for $t'$ arbitrarily close to $t_1$, because after all $RG[t_2,t']RG[t',t_1]=RG[t_2,t_1]$ (the path integral is the same after we integrate a shell). And my problem is that already $RG[t',t_1](V_{t_1}^{bare})$ contains those other terms. But you are saying that the the temporary effective theories at $t_2$ are not quite equal for different $t_1$, only the limit? $\endgroup$
    – octonion
    Oct 12, 2018 at 19:40

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