# Can you naïvely reduce the dimensionality of a QFT?

I want to study a QFT, given by an action $$S$$ which is defined in $$(2+1)$$ dimensions, i.e. $$S=\int d^3x \mathcal{L}[\phi,\,\partial\phi].$$

This QFT is invariant under rotations, i.e. in radial coordinates $$r=\sqrt{x^2+y^2}$$ and $$\theta=\tan^{-1}(y/x)$$, the action is invariant under a transformation $$\theta\rightarrow\theta+\Delta\theta$$. My question (quite naïve) is: can you forget about the angular coordinate and study a sort of "effective" $$(1+1)-$$dimensional QFT that is defined only on that radial direction? Something like $$S=\int dtdr\mathcal L[\phi,\partial\phi].$$I think there are some subtelties that prevents me to do this directly, e.g. the dimensions of fields and parameters change, and some properties (e.g. the equivalence between fermions/bosons, which is customary in $$(1+1)-$$dimensions, would be recovered although it is absent in the original $$(2+1)-$$dimensional QFT).

Yes, you could consider such a dimensional reduction. The idea is to decompose $$\phi(t, \mathbf{x})$$ as $$\phi(t, r, \theta) = \sum_{k \in \mathbb{Z}} \phi_k(t, r) e^{i k \theta}.$$ You plug this Ansatz into your action, which then becomes $$S = \int\!d^3x \, L[\phi] = \int\!dt r dr d\theta\, L'[\{\phi_k\}]$$ where $$L'[\{\phi_k\}]$$ is a 3d Lagrangian that depends on the infinite set of modes $$\{\phi_k\}$$ with $$k \in \mathbb{Z}$$.

Finally, you will find that it's easy to perform the $$\theta$$ integral. After all $$\int\!d\theta\, e^{ik\theta} e^{in\theta} = 2\pi \delta_{k+n,0}.$$ You will see that the kinetic term becomes something like $$S = \sum_k \int\!dt r dr\, \phi_{-k} (\Box_\text{2d} + k^2) \phi_k + \text{interactions}.$$ So modes with large $$k$$ contribute a lot to the action and are exponentially suppressed at the level of correlation functions. Interactions will mix modes with different $$k$$, in a way that's ultimately consistent with the $$SO(2)$$ invariance of the $$3d$$ theory.

What makes this dimensional reduction a bit funny is that you give up translation invariance in space. The only manifest spacetime symmetries of the final theory in the $$(t, r)$$ domain are time translations.

Finally, your picture of naive dimensional reduction amounts to keeping only the zero mode $$\phi_0$$. The full theory is described by the infinite set of modes $$\{\phi_k\}$$. If you want to understand this reduction quantitatively, you could truncate the theory by keeping only modes with $$|k| \leq K$$ for some fixed integer $$K$$. Using that cutoff, you would compute things like correlation functions of $$\phi_0$$ (which will depend on $$K$$). Finally, you could change $$K$$, which would allow you to derive a set of renormalization group equations for the couplings in your $$2d$$ theory. By following the renormalization group flow down to $$K = 0$$, you would be able to systematically compute the couplings of the naive theory in terms of those in the full theory with a cutoff. However, that computation would be the length of a small research paper, so it's not in the scope of a StackExchange answer.

• +1, I’ll add that this dimensional reduction is a standard method used to study field theory in a finite volume. See for example the chapter “Finite-size scaling” in Zinn-Justin’s textbook, or this post on the volume dependence of barrier penetration in symmetry-broken states: physics.stackexchange.com/questions/723235/… Apr 25 at 0:51

I think the answer by Hans Moleman is pretty good, so I'll only add a few comments.

Naively changing the dimension of a theory can lead to disastrous results. A good example is General Relativity. In four spacetime dimensions, we have regular GR and everything works as usual. In three spacetime dimensions, though, it happens that the so-called Weyl tensor (which is a part of the Riemann curvature tensor complementary to the Ricci tensor) always vanishes identically. Physically, this means that there is only curvature where the Ricci tensor doesn't vanish, which due to the Einstein equations means that there is only curvature in places where matter is present. Hence, a consequence is that three-dimensional gravity doesn't include propagation: if spacetime were three-dimensional, the Earth wouldn't orbit the Sun. The situation is even worse in two dimensions, where the Einstein tensor vanishes identically, and hence the Einstein equations imply the stress tensor must vanish identically too.

Despite this, it is common in Relativity to consider two-dimensional analogues of four-dimensional spacetimes. This, however, is done by considering an analogue metric, not by solving the Einstein equations in two dimensions. Thus, it is possible to perform a dimensional reduction, but it should be made with care.

The lesson is: sometimes it is possible to perform dimensional reductions in field theory, but you should avoid doing so naively, since the number of dimensions can have quite an impact on the physics.

• Exactly - the naive $\phi_0$ theory is an effective $2d$ field theory that is valid at low energies $E \to 0$. But since you've integrated out all the other modes, you have generated an infinite tower of complicated interactions. These interactions are not renormalizable so they become more and more important as you try to extend the theory to higher energies. Apr 24 at 19:40