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

\begin{equation} \tag{1} \phi(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\phi_{m,n}}{z^{m + h} \bar{z}^{n + \bar{h} } } \end{equation} 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 \begin{equation} \tag{2} \phi(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\phi_{m,n}}{z^{m} \bar{z}^{n} } \end{equation} 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 $z \to 0$. \begin{equation} \tag{3} \phi_{m,n} |0\rangle = 0 \text{ for } (m,n) > (-h, -\bar{h}) \end{equation} 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? \begin{equation} \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 \end{equation} If $h$ was an integer, then one power would remain that is $z^0$, but I'm assuming $h$ isn't an integer.

  • $\begingroup$ If you don't want an answer to reference the cylinder, then the only correct answer to your question is "it's a convention". $\endgroup$
    – Prahar
    Commented Feb 17, 2021 at 23:45
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    $\begingroup$ 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). $\endgroup$
    – Prahar
    Commented Feb 18, 2021 at 0:05
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    $\begingroup$ @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). $\endgroup$
    – Prahar
    Commented Feb 18, 2021 at 0:30
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    $\begingroup$ @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. $\endgroup$
    – Prahar
    Commented Feb 18, 2021 at 0:58
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    $\begingroup$ 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 $z\rightarrow \lambda z$. $\endgroup$ Commented Feb 18, 2021 at 16:33

1 Answer 1


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

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 $J \in \mathbb{Z}$.


On the Euclidean plane we have the mode expansion \begin{equation} \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}). \end{equation} With the adjoint given by \begin{equation} [\mO^{(p)}(w, \bw) ]^\dagger = \bw^{-2\bh} w^{-2 h} \mO^{(p)}(1/\bw, 1/w) \end{equation} we then have \begin{equation} [\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}) \end{equation} 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}


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 \begin{equation} \mO'(w', \bw') = (\partial_w w')^{-h} (\partial_\bw \bw')^{-\bh} \mO(w, \bw). \end{equation} 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 \begin{equation} \mathcal{O}^{(c)}(\tau, \sigma) = e^{\tau H} \mO(0, \sigma)^{(c)} e^{- \tau H} \end{equation} 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 $\mO^{(c)}$.


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