Compact or non-compact boson from bosonization?

Question

In some discussions of bosonization, it is stressed that the duality between free bosons and free fermions requires the use of a compact boson. For example, in a review article by Senechal, the following statement is made:

In order for bosonization to be rigorously defined, the field $\phi$ must have an angular character. In other words, the target space of the field must be a circle of radius $R$ (to be kept general for the moment), so that $\phi$ and $\phi + 2\pi R$ are identified. We say that the boson is compactified on a circle.

In other sources, however, no mention of the compactness of $\phi$ is made. For example, in the review article by von-Delft and Schoeller, there is no mention of the word "compact". Moreover, books like Fradkin's and Shankar's respective condensed matter field theory books seem to make no mention of this issue, to the best of my knowledge.

As far as I understand, theories of compact and non-compact bosons are quite nontrivially different. At first glance, both theories appear to have the same action: $$ S[\phi] = \frac{K}{2} \int d^2x (\partial_{\mu} \phi)^2. $$ However, a massless non-compact boson is a trivial free theory, and the coefficient $K$ can always be rescaled away by changing our normalization of $\phi$. On the other hand, rescaling a compact boson nontrivially changes the theory, and the theory can undergo the famous KT transition at a critical value of $K$.

Other purely circumstantial evidence that the boson obtained by bosonizing fermions is compact includes:

• In the central bosonization formula, $\psi \sim e^{-i \phi}$, $\phi$ appears inside an exponential. As a result, all the observables one considers in the bosonized side of the theory seem to respect the periodicity of $\phi$. This is transparent from the form of the bosonized action as well: it's easy to find terms in the fermionic theory which translate to objects like $\cos \phi$ in the action of the bosonized theory, but one can never generate a mass term like $m \phi^2$ to my knowledge.

• In books on the Luttinger liquid, the parameter $K$ above is said to be related to the compactification radius of the boson. In Fradkin's book, for example, $K$ is referred to as the compactification radius, but then the compactness of $\phi$ is never used thereafter.

• There is a self-duality which trades $\phi$ for its dual field $\theta$, and correspondingly $K$ to $1/K$. As far as I know, this is again related to the fact that $\phi$ is compact; if $\phi$ were non-compact, we could rescale $K$ to whatever we want without changing the physics. It's only when $\phi$ is compact that this duality becomes nontrivial.

However, the above are all only circumstantial evidence, and don't explain why a compactified boson is necessary. Moreover, books like Fradkin's and review articles like von-Delft's, which put quite a bit of work into proving the bosonization formulae, never seem to run into trouble by not using a compact boson.

So to summarize, here is the question: in abelian bosonization, where a Dirac fermion is equivalent to a massless boson, is the boson compact or noncompact? Is there ever freedom to use one or the other, or must it be compact (or noncompact)? And most importantly, how can I see why the boson is compact or not?

Answer 1

It is compact.

The simplest approach is to look at symmetries. The most obvious symmetry has algebra $\mathfrak{u}_1$, given by rephasing the fermion or shifting the boson. What is the global form of this symmetry? (i.e., what is the group for this algebra?)

It is clear that the fermion has group $U(1)$, as the group element is a phase.

For the boson, the group is $\mathbb R$ for the non-compact case and $U(1)$ for the compact case. The reason is obvious: the set of transformations $\phi\mapsto\phi+\alpha$ has the structure of $\mathbb R$ if $\alpha$ is arbitrary, but colapses to $U(1)$ if $\alpha\cong\alpha+2\pi$.

Hence, if these symmetries are to be identical (which is a necessary condition for duality), the boson must be compact.

(You can also convince yourself that these theories are CFTs and their chiral algebra is precisely the affine extension of $\mathfrak u_1$; both are rational with respect to this algebra and in fact holomorphic. For the fermions this is obvious but for the bosons this is only true at $R=1$ (in some normalization).)

Answer 2

When in doubt, you can go back to the torus partition function. For the free Dirac fermion, it is \begin{equation} Z = \frac{1}{2} \left ( \left | \frac{\vartheta_2(\tau)}{\eta(\tau)} \right |^2 + \left | \frac{\vartheta_3(\tau)}{\eta(\tau)} \right |^2 + \left | \frac{\vartheta_4(\tau)}{\eta(\tau)} \right |^2 \right ). \end{equation} This agrees with the partition function of the free boson at $\sqrt{2}$ times the self dual radius. Some techniques for evaluating related partition functions (along with this specific example) are given in Chapter 8 of Ginsparg.

This can be used as a starting point to find the fermionic description of compact free bosons at other radii. The point is that the radius can be tuned by adding the current-curent deformation $\partial \phi \bar{\partial} \phi$. After applying the duality, this becomes a four-fermi interaction which defines the massless Thirring model. You can then find another pair of theories related by abelian bosonization if you add a fermion mass. Since this breaks conformal symmetry, it leads to a non-trivial S-matrix. This massive Thirring model S-matrix is the same as the one for the sine-Gordon model as shown by Coleman.