Conformal weights in Laurent expansions

Question

Currently I am taking an introductory class in conformal field theory and ran into some confusion. Everywhere I run into Laurent expansions of several important quantities, and (mostly) all of them boil down to the following:

Suppose $\phi (z)$ is a primary field with conformal dimension $h$, then it may be Laurent expanded as:

$$\phi(z) = \sum_{n \in \mathbb{Z}} c_n z^{-n-h}$$

With $c_n$ the Laurent modes (constants). For a particular reference of this, see for example di Francesco's book eqn. (6.7) or Blumenhagen eqn. (2.42) in which it is claimed (without motivation beforehand):

$$T(z) = \sum_{n \in \mathbb{Z}} z^{-n-2} L_n$$

Is there any mathematical or physical reason why we include these conformal weights of the objects in the Laurent expansion (why not simply $\sum_{n \in \mathbb{Z}} z^{n} c_n$) ? Or is it simply convenient in calculations?

Bonus (related) question: in Blumenhagen's book (page 13) the following expression is stated:

$$z'=z + \epsilon(z) = z + \sum_{n \in \mathbb{Z}}\epsilon_n (-z^{n+1})$$

Where $\epsilon(z)$ is assumed to be meromorphic and $\epsilon_n$ constant. Is this particular expansion (so with $z^{n+1}$ instead of $z^n$) similarly done simply for convenience?

Answer 1

The conformal weights are included because they come from the conformal transformation which maps the cylinder to the radial plane. Start with the cylinder where $x \sim x + 2\pi$. Define complex coordinates $w=t+ix$ so $w \sim w + 2\pi i$. Then, all operators admit the expansion $$ \Phi(w,{\bar w}) = \sum_{m,n} \phi_{m,n} e^{-m w} e^{-n {\bar w} } , \qquad m,n \in {\mathbb Z} $$

Now, we perform the conformal transformation which moves us to the radial plane. In particular, we define $$ z = e^{ w} = e^{ i x} e^{t} , \qquad {\bar z} = e^{ {\bar w}}. $$ Then, note that $t \to -\infty$ maps to the origin and $t \to \infty$ maps to infinity. Now, under conformal transformations a primary field transforms as \begin{align} \Phi(z,{\bar z}) &= (\partial_z w)^h(\partial_{\bar z} {\bar w})^{{\bar h}}\Phi (w,{\bar w}) = \sum_{m,n} \phi_{m,n} z^{-m-h} {\bar z}^{-n-{\bar h}} \end{align}

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ASIDE

In Euclidean signature, the adjoint property is $$ \Phi(t,x)^\dagger = \Phi(-t,x) ~~\implies~~ \Phi(w,{\bar w})^\dagger = \Phi(-{\bar w},-w) ~~\implies~~ \phi_{m,n}^\dagger = \phi_{-m,-n} . $$ This is because time evolution is defined by $\Phi(t,x) = e^{H t} \Phi(0,x) e^{-H t}$. We assume here that $\Phi(0,x)$ is Hermitian. On the radial plane, the adjoint property then reads \begin{align} \Phi(z,{\bar z})^\dagger &= \sum_{m,n} \phi^\dagger_{m,n}{\bar z}^{-m-h} z^{-n-{\bar h}}\\ &= {\bar z}^{-2h} z^{ - 2 {\bar h} }\sum_{m,n} \phi_{m,n} (1/{\bar z})^{-m-h} (1/z)^{-n-{\bar h}} \\ &= {\bar z}^{-2h} z^{ - 2 {\bar h} } \Phi(1/{\bar z},1/z). \end{align}

Answer 2

Let me use the book by di Francesco et al. Including the conformal weight in the power of $z$ in the Laurent expansion, as in equation (6.8), allows you to have the nice property (6.10), namely $\phi_{-m} = \phi^{\dagger}_m$. For the energy momentum tensor, it gives $L_{-m} = L^{\dagger}_m$.

It is just a nice convention.

Answer 3

This is simply a convenience, and in the end you should be able to work with any convention. In fact, in the mathematical literature on vertex operator algebras you will often encounter,

$$ Y(\omega,z) = \sum \omega_n z^{-n-1} = \sum L_n z^{-n-2}$$

where $Y$ maps a state to a field, i.e. the stress-energy tensor has modes expanded as if $h=1$ in your notation. This might seem odd but it is actually useful to have a consistent convention to use for any state, regardless of conformal dimension, when you have operators in your algebra of varying conformal dimension.