Do primary fields (in a CFT) satisfy the wave equation?

In a 2-dimensional CFT, do primary fields generally satisfy the wave equation? I know that if they have either purely holomorphic or purely anti-holomorphic, they do. But what about a general primary field of weights $(h,{\tilde h})$?

• To precisely which wave equation are you referring? – joshphysics Nov 25 '13 at 20:54
• @joshphysics - $\partial_z \partial_{\bar z} \phi = 0$ – Prahar Nov 25 '13 at 22:43
• Ok so to be clear, you are restricting the question to the theory consisting of a free massless boson? – joshphysics Nov 26 '13 at 18:35
• Well, since we are talking about CFTs, I presume all fields are massless. I do restrict to free fields, but not to bosons. For example, even fermions satisfy the wave equation (in addition to the Dirac equation). My question is simply whether primary fields satisfy the wave equation. It may or may not in addition satisfy other equations that determine its spin. – Prahar Nov 27 '13 at 2:13

I don't think so. Take a theory of a free boson $X$, and consider the operator $\exp(ikX)$. This is a primary operator, but it satisfies $\partial \overline{\partial}\exp(ikX)=ik \partial (\overline{\partial} X\exp(ikX) )=-k^2\overline{\partial}X\partial X\exp(ikX)$, where I used the EOM. If $k^2\not=0$ this doesn't vanish.
No, and this is very general. The two-point function of (the left-moving part of) such an operator satisfies $$G(z) = \frac{1}{z^{2h}}$$ which only satisfies a local wave equation if $h=0$. Note: for a primary of weights $(h,\tilde{h})$ the correlator factorises, so it's sufficient to look at just one part. If you're not convinced, calculate the Fourier transform $\hat{G}(p)$ of $G$ and try to build it as a solution of $$(p^2 + \ldots)\hat{G}(p) = 1$$. This generalises to $d$ dimensions.
• You didn't understand my equation, so please don't downvote because of ignorance. The full propagator is obviously $G(z,\bar{z}) = z^{-2h} \bar{z}^{-2\bar{h}}.$ This is really elementary CFT... – Vibert Jan 30 '14 at 2:16
• I was the original downvoter, because I thought you were saying that an operator with weight $(h,0)$ doesn't satisfy the wave equation (your answer suggests that $(0,0)$ is the only possibility). I removed my downvote because I understand that wasn't your intention. – Matthew Jan 30 '14 at 2:20