# Is the free Bose field a conformal field?

Let's consider the free Bose field $$S=\frac{1}{2}g\int d^2x\ \partial_\mu\varphi\partial^\mu\varphi.$$ The action apparently shows that the system has conformal invariance (at least at the classical level). And we can read off that the Bose field has conformal dimension $0$. But now, I read in a book (An introduction to Conformal Field Theory: With Application to String Theory, R. Blumenhagen & E. Plauschinn, pp 50, below Eq.(2.92)) that:

As we have mentioned earlier, the free boson $\varphi(z, \bar{z})$ is not a conformal field since its conformal dimensions vanish $(h, \bar{h}) = (0, 0)$.

My question is, does the vanishing conformal dimension indicate that the field itself is not a conformal field? How exactly do we define a conformal field?

Furthermore, the two point correlator of a (quasi-)primary field with vanishing conformal dimension should have the form $\langle\varphi(x)\varphi(y)\rangle= \frac{C_{12}}{|x_1-x_2|}$, but for the free Bose field, on the other hand, we have $\langle\varphi(x)\varphi(y)\rangle= -\frac{1}{4\pi g}\ln (x-y)^2$. Why are they different? Maybe you will say, it is the difference for the correlator forms that indicates the free Bose field is not a conformal field. But I would like to know is there any deeper reason to explain why its correlator is different from that of a primary field and hence itself is not a conformal field?

To be precise, the two-point function of (0,0) fields is really

$\langle \varphi(x) \varphi(y) \rangle \sim \log ( \mu^2 |x-y|^2 )$

where $\mu$ is the IR cutoff of the theory. Note that this is the only dimensionally consistent equation one can write down. This immediately indicates why $\varphi(x)$ is not a well-defined operator in the quantum theory - its correlators are regulator dependent! It therefore cannot be an operator in the quantum theory since all physical operators are required to be regulator independent.

That being said, it is convenient to continue to discuss the object $\varphi(x)$ as one may construct actual operators in the Hilbert space out of it, for instance $\partial_\mu \varphi(x)$ or $: \exp [ i k \varphi(x)] :$, etc.

• Then, is this a kind of anomaly in some sense since $\varphi$ possess the conformal invariance classically? Sep 5 '16 at 12:39
• No. I don't think it is an anomaly. $\varphi$ is conformally invariant. You can use this to show that its 2-point function must be $\sim \log |x-y|$. (you can see this by taking the $h \to 0$ limit of $\frac{1}{|x-y|^{2h}}$. Sep 6 '16 at 0:45
• I like to think of this as analogous to confinement: We have a classical observable which isn't present in the quantum theory. Sep 7 '16 at 23:17

Indeed, the most obvious reason a 2D free boson field is not a proper field of a conformal field theory is that its two-point function is "wrong". This, however, should not itself be surprising - if we just start from a field invariant under conformal symmetry (which $\phi$ is), then we would from the "pure" CFT standpoint expect the theory to be trivial too. However, if we switch to a better description of the theory it becomes apparent how to read it as a proper CFT: The idea is that of usual bosonization, and we declare the "true" conformal field to be $V_\alpha = \, :\mathrm{e}^{\mathrm{i}\alpha\varphi}:$.

How do we know there's a "true" CFT hidden here? Well, we know that the space of states is not trivial! We have the currents $j(z) = \mathrm{i}\partial\varphi$ and its antiholomorphic counterpart. Expanding these in modes (since they fulfill $\partial j = \bar{\partial}\bar{j} = 0$) and integrating yields $$\varphi(z,\bar{z}) = \phi_0 - \mathrm{i}(a_0z + \bar{a}_0z) - \mathrm{i}\sum_{n\neq 0}\frac{1}{n}\left(\mathrm{e}^{-nz}a_n + \mathrm{e}^{-n\bar{z}}\bar{a}_n\right)$$ Examining the commutation relations leads us to intepret $a_n$ as creation and $a_{-n}$ as annihilation operator for $n > 0$. The Hilbert space then must carry a representation of this operator algebra, and in particular it must decompose as the direct sum of Fock spaces built from an eigenvector of $a_0$.

Now, the exponentials in the mode expansion look rather "un-conformal". We usually want mode expansions like $a_nz^{-n-h}$. So we apply the cylinder-to-plane map $z\mapsto \mathrm{e}^{-z}$. This leads us to consider the theory of the conformal fields $V_\alpha$, and one could find that these obey the proper n-point functions for conformal fields. That is, the theory of 2D bosons on a cylinder is itself not naturally interpreted as a pure CFT, but its equivalent theory on a plane (with the usual conceit of CFT where we remove the origin and think of it as the infinite past) does possess such an interpretation.

• Isn't the cylinder-to-plane map $z\rightarrow e^{-z}$ simply a kind of conformal transformation? How can this (coordinates) transformation lead a change of the field operators $\varphi\rightarrow :e^{i\alpha\varphi}:$? Sep 5 '16 at 12:37
• Then, is this a kind of "anomaly" in some sense since $\varphi$ possess the conformal invariance classically? Sep 5 '16 at 12:39
• @WeinEld Maybe I formulated that badly. It doesn't lead to it in the sense that the coordinate change induces that, but we need to change to the plane to have the usual interpretation of modes. And yes, this is an anomalous theory since the central charge of the theory turns out the be 1, but this has nothing to do with the boson itself - most of the CFTs are anomalous; the anomaly term is exactly given by the central charge of the Virasoro algebra. Sep 5 '16 at 12:52

This might be an old question but it appeared as a recommendation in a similar question. As my answer there states, it is possible to keep the bosonic field $$\varphi$$ around if you really want to but you must then resort to the language of logarithmic CFT.