# Problems with common mode expansion for 2D plane CFT for non integer weight?

Equation (6.7) in the big yellow book (Di Francesco, Mathieu, Senechal) says we can write the field of a 2D CFT with weight $$(h, \bar h) = (\tfrac{1}{2}(\Delta + J), \tfrac{1}{2}(\Delta - J))$$ as

$$$$\tag{1} \phi(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\phi_{m,n}}{z^{m + h} \bar{z}^{n + \bar{h} } }$$$$ I am very confused for this equation due to the appearance of $$(m + h)$$ and $$(n + h)$$ in the denominator instead of $$m$$ and $$n$$ (where $$m$$ and $$n$$ are integers). Note that while $$h - \bar{h} = J$$ is an integer, $$h$$ and $$\bar{h}$$ do not have to be integers individually, which is what confuses me.

I have read this answer which motivates it from a cylinder CFT, but I don't want an answer that references the cylinder, I am interested in a resolution involving only the plane.

I will list my problems with this equation:

1. Wouldn't a more standard expression be $$$$\tag{2} \phi(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\phi_{m,n}}{z^{m} \bar{z}^{n} }$$$$ which is basically a Laurant expansion? For example, if $$\phi(z, \bar{z})$$ were just a function (I know it's really an operator) then a very large class of functions can be written in terms of the equation above, as essentially a Laurent series, but I don't know what class of functions can be written with $$m + h$$ and $$n + \bar{h}$$ instead of $$m$$ and $$n$$? It seems to me that if a function is expressible in terms of equation $$(1)$$, it can't be expressible in terms of equation $$(2)$$, and vice versa. In other words, only one can be correct and the other must be wrong, It can't be a matter of convention. I would like an explanation as to why $$(1)$$ must be correct and $$(2)$$ must be incorrect (without invoking the cylinder), especially given that a Laurent series seems to me to be more natural.

2. If $$m$$ and $$n$$ are too big, the operator $$\phi_{m, n}$$ will annihilate the vacuum state $$|0\rangle$$. This is because the correlation functions must be smoothly varying as $$z \to 0$$. $$$$\tag{3} \phi_{m,n} |0\rangle = 0 \text{ for } (m,n) > (-h, -\bar{h})$$$$ However... doesn't that imply that all states made from acting on the vacuum are $$0$$, just because all the powers of $$z$$ that remain are positive? $$$$\tag{4} \phi(0,0) | 0 \rangle = \lim_{z, \bar{z} \to 0} \sum_{(m,n) > (-h, -\bar{h}) } \frac{\phi_{m,n}}{z^{m + h} \bar{z}^{n + \bar{h} } } |0 \rangle =0$$$$ If $$h$$ was an integer, then one power would remain that is $$z^0$$, but I'm assuming $$h$$ isn't an integer.

• If you don't want an answer to reference the cylinder, then the only correct answer to your question is "it's a convention". Commented Feb 17, 2021 at 23:45
• If $h$ (or ${\bar h}$) is half-integer, then the field $\phi(z,{\bar z})$ is not single-valued on the plane and this is crucial (rotation by $2\pi$ on the cylinder is not the identity for fermionic states). This multi-valuedness is clearly absent in your (2), but is there in (1). Commented Feb 18, 2021 at 0:05
• @user1379857 If you are interested in $h \in {\mathbb R}$ you need to understand what that means. It means that your field has particular properties as you rotate $z \to e^{2\pi i} z$. In particular, for generic $h \in {\mathbb R}$, the field is multi-valued. All of these properties are captured by the expansion (1), but not (2). Commented Feb 18, 2021 at 0:30
• @user1379857 - yes, "it's a convention" was a naive answer (though correct to some extent since the multivaluedness could potentially have been absorbed into a crazy transformation of $\phi_{m,n}$). The better answer is given by the multi-valuedness argument AND my second comment regarding the action of $\phi_{m,n}$ on the vacuum state. Commented Feb 18, 2021 at 0:58
• i think the relevant case might be when $\phi(z,\bar{z})$ are 'components' of a tensor in a given chart, in the sense that the conformally-invariant quantity in that chart is: $\phi(z,\bar{z})(dz)^h(d\bar{z})^{\tilde{h}}$. With the Laurent expansion of $\phi(z,\bar{z})$ as in (1), the modes scale in a convenient way under $z\rightarrow \lambda z$. Commented Feb 18, 2021 at 16:33

$$\newcommand{\bw}{{\bar{w}}} \newcommand{\bh}{{\bar{h}}} \newcommand{\mO}{\mathcal{O}} \newcommand{\vk}{|0\rangle} \newcommand{vb}{\langle 0|}$$

Okay, here is the answer. Here we use \begin{align} h &= \tfrac{1}{2}( \Delta + J) \\ \bar{h} &= \tfrac{1}{2}( \Delta - J) \end{align} where $$\Delta > 0$$ and $$J \in \mathbb{Z}$$.

## Plane

On the Euclidean plane we have the mode expansion $$$$\mathcal{O}^{(p)}(w, \bw) = \sum_{m,n = 0}^\infty (a^\dagger_{mn} w^{m} \bw^n + a_{mn} w^{-m-2h} \bw^{-n-2\bh}).$$$$ With the adjoint given by $$$$[\mO^{(p)}(w, \bw) ]^\dagger = \bw^{-2\bh} w^{-2 h} \mO^{(p)}(1/\bw, 1/w)$$$$ we then have $$$$[\mO^{(p)}(w, \bw) ]^\dagger = \sum_{m,n = 0}^\infty (a_{mn} \bw^{m}w^n + a_{mn}^\dagger \bw^{-m-2h} w^{-n-2\bh})$$$$ which you can see by inspection is indeed the adjoint of $$\mathcal{O}^{(p)}$$ we have written.

If we also have \begin{align} a_{mn} \vk &= 0 \\ \vb a_{mn}^\dagger &= 0 \end{align} then this implies \begin{align} \mO^{(p)}(0,0) \vk &= a^\dagger_{00} \vk \\ \vb [\mO^{(p)}(0,0)]^\dagger &= \vb a_{00} \\ \vb [\mO^{(p)}(0,0)]^\dagger \mO^{(p)}(0,0) \vb &= \vb a_{00} a_{00}^\dagger \vk. \end{align}

## Cylinder

Let's now transform to the Euclidean cylinder. We have ($$\sigma \sim \sigma + 2 \pi$$) \begin{align} w &= e^{w'} = e^{\tau + i \sigma} \\ w' &= \ln(w) \end{align} and the primary transformation law $$$$\mO'(w', \bw') = (\partial_w w')^{-h} (\partial_\bw \bw')^{-\bh} \mO(w, \bw).$$$$ This means that \begin{align} \mO^{(c)}(\tau, \sigma) &= w^h \bw^\bh\mO^{(p)} (w, \bw) \end{align} Using \begin{align} w^h \bw^\bh &= (e^{\tau + i \sigma})^h (e^{\tau - i \sigma})^\bh \\ &= e^{\tau(h + \bh) + i \sigma ( h - \bh)} \\ &= e^{\tau\Delta + i \sigma J} \end{align} we get \begin{align} \mO^{(c)}(\tau, \sigma) &= \sum_{m,n = 0}^\infty ( a^\dagger_{mn} e^{(m + n + \Delta) \tau + i (m-n + J) \sigma} + a_{mn} e^{-(m+n + \Delta ) \tau - i (m -n + J) \sigma} ). \end{align} Note that, in Euclidean signature, we have $$$$\mathcal{O}^{(c)}(\tau, \sigma) = e^{\tau H} \mO(0, \sigma)^{(c)} e^{- \tau H}$$$$ so \begin{align} [\mathcal{O}^{(c)}(\tau, \sigma)]^\dagger &= e^{-\tau H} [\mO^{(c)}(0, \sigma)]^\dagger e^{ \tau H} \\ &= \mO^{(c)}(- \tau, \sigma). \end{align} Here we used that the operator is self adjoint on the $$t = \tau = 0$$ time slice. The above equation is different from Lorentzian signature, where the adjoint fixes $$t \mapsto t$$, because the lack of an $$i$$ now means that $$\tau \mapsto - \tau$$. In any case, we can see by inspection that our expression for $$\mathcal{O}^{(c)}$$ does indeed satisfy the above relationship.

Now, let's see how to act with this operator at the distant past. We then have \begin{align} \lim_{\tau \to - \infty} \mO^{(p)}(w, \bw) \vk &= \lim_{\tau \to - \infty} w^{-h} \bw^{-\bh} \mO^{(c)}( \tau, \sigma) \vk \\ &=\lim_{\tau \to - \infty} (e^{-\Delta \tau - i J \sigma})e^{\Delta \tau + i J \sigma} a_{00}^\dagger \vk \\ &=a_{00}^\dagger \vk. \end{align} We have to put in the prefactor $$w^{-h} \bw^{-\bh}$$, otherwise the euclidean time evolution will decay away the state to zero because $$\Delta > 0$$. However, this is precisely the factor you would expect to get just from the relationship of $$\mO^{(p)}$$ and $$\mO^{(c)}$$.