Why has the free boson a charge $c=1$ in 2D CFT?

In the free scalar field theory in 2D conformal field theory, we consider the correlation functions of the derivatives of the fields, i.e.

$$\langle \partial \phi(z) \partial \phi(w) \rangle, \tag{1}$$

where we use complex coordinates. Then, since

$$:T(w): = - \frac{1}{2} \lim_{z\to w} \left(\partial\phi(z)\partial\phi(w) + \frac{1}{(z-w)^2} \right), \tag{2}$$

we can use the OPEs $$T(z)T(w)$$ and $$\partial\phi(z)\partial\phi(w)$$ in order to find the central charge, which turns out to be

$$c=1. \tag{3}$$

Now, I am wondering why we look at the derivatives of $$\phi$$ in the first place. It seems to me that the reason why we look at them is that, without them, the correlation function looks like

$$\langle \phi(z) \phi(w) \rangle = - \ln(z-w) - \ln(\bar{z}-\bar{w}), \tag{4}$$

and that such an expression diverges at $$z\to\infty$$. On the other hand, the correlator $$\langle \partial\phi(z)\partial\phi(w)\rangle$$ behaves nicely with a $$(z-w)^{-2}$$ dependency, which makes it vanish (or just converge maybe?) at $$\infty$$.

Now the question is: why is it legitimate to say that the central charge of the free boson is $$c=1$$? It seems to me that I could take the correlation function of other fields nicely behaving, and I would get a different central charge. Or what am I missing that makes this the definitive choice?

Thank you in advance, and sorry if this is a dumb question.

• The central charge is a property of the system, not of any field in particular. – AccidentalFourierTransform Jul 10 at 21:37
• @AccidentalFourierTransform I see. So it is a mistake to say something like the free boson has central charge $c=1$ in the first place? (your nickname is epic by the way) – Jxx Jul 10 at 21:40
• "free boson" does not just mean $\phi$; it means "the system described by $\phi$ (and with Lagrangian $L=\frac12(d\phi)^2$)". – AccidentalFourierTransform Jul 10 at 21:47
• @AccidentalFourierTransform Mm okay, but then if we see particles as excitations of fields, what can I say about the excitations of $\phi$ and $\partial\phi$? They are both bosonic, but they are not the same I imagine? – Jxx Jul 10 at 21:50

Indeed as said in the comments, when people say "the free boson has central charge $$1$$" they mean, "the theory with just one free boson has central charge $$1$$." It is an example of "synecdoche" I believe...

Anyway, the reason why we consider $$\partial\phi$$ as a field rather than $$\phi$$ is because, as you noticed, $$\phi$$ has conformal dimension $$h=0$$, so its two point function diverges. You can also think in terms of the mode expansion (e.g. by quantizing the theory on the cylinder). In that case it would be equivalent to disregard the "zero mode," which indeed makes the path integral ill defined.$${}^1$$

By the way, it's not like we completely throw $$\phi$$ in the trash. It is used to create vertex operators $$\mathcal{V}_\alpha(z,\bar{z}) = \,\,:\!e^{i\alpha\,\phi(z,\bar{z})}\!:\,,$$ which are conformal primaries with $$h =\bar{h} = \alpha^2/(8\pi)$$.

But I guess there is some confusion in your statement. We don't choose to look at $$\partial\phi$$ to compute the central charge. The stress tensor $$T$$ is defined regardless of which operators we decide to consider. It can be derived with Noether's theorem or by coupling to gravity in the usual way. Then we take its OPE with itself $$T(z)T(0) \sim \frac{c}{z^4} + \frac{2T}{z^2} + \frac{\partial T}{z} + (\mathrm{regular}) \,,$$ and extract $$c$$ from that. It just so happens that $$T(z) \propto \,:\!\partial\phi\,\partial\phi(z)\!:$$ in this particular model.

$$\quad{}^1$$The mode expansion is something like (modulo factors) $$\phi(z,\bar{z}) \sim \phi_0 + \pi_0 \log(z\bar{z}) + \sum_{n\neq 0}a_n z^{-n} + \bar{a}_n \bar{z}^{-n}$$ where $$\phi_0$$ is a constant field and it's called the "zero mode." This is troublesome in a path integral because since the mode has no energy (i can safely shift $$\phi \to \phi + c$$ since the action depends on its derivatives only), all configurations with different $$\phi_0$$ have the same weight in the path integral, so it's like integrating a constant function in an infinite line.

• Thank you for this very nice answer. Indeed in my course it looked like the motivation was as I stated in the OP, but maybe it is because of the order in which the things were presented. The justification with the momentum-energy tensor makes sense. – Jxx Jul 11 at 7:15
• As a follow-up question, in the fermion field the energy tensor looks like $T \propto : \psi \partial \psi :$. Can I relate this relation to something in the same way as in your answer? – Jxx Jul 11 at 7:16
• Just a comment that could clarify why we use $\partial\phi$ instead of $\phi$: if you define the free boson theory by the action $S = \int d^2x (\partial \phi)^2$, then there is a shift symmetry $\phi \to \phi + c$. Since this symmetry is exact even at the quantum level, it is often assumed that we consider invariant operators only, for instance $\partial \phi$ but not $\phi$. Note that this is different in dimensions > 2, where the shift symmetry of the free boson is broken when you couple the theory to a background metric. – M.Jo Jul 11 at 9:12
• I was also thinking about shit symmetry. But I couldn't figure out why we disregard operators that are not invariant. It's not a gauge symmetry after all. – MannyC Jul 11 at 15:01
• @AccidentalFourierTransform What makes it not a conformal field then? (as opposed to $\partial\phi$) – Jxx Jul 11 at 18:07

It is a common misconception that the free boson has a central charge $$c=1$$ in 2d CFT. Actually the free boson can have an arbitrary central charge. You can define $$T(w)$$ with an extra term $$\partial^2\phi(w)$$ with an arbitrary coefficient, and the central charge is a function of that coefficient. See e.g. the Wikipedia article on Liouville theory, where you just have to omit the interaction term in order to get a free boson instead of Liouville theory.

• But then it's a free boson only in flat space, right? – MannyC Jul 11 at 19:17
• Not just in flat space. If the space is curved you have to add a term in $R(g)\phi$ to the Lagrangian, where $R(g)$ is the curvature of the metric. The theory is still free after that i.e. it has momentum conservation. – Sylvain Ribault Jul 11 at 19:26
• Wouldn't that term give you tadpole diagrams proportional to $R$? – MannyC Jul 11 at 19:32
• No idea. I never use Feynman diagrams in CFT. – Sylvain Ribault Jul 11 at 19:46