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Conformal weights in Laurent expansions

Physics Asked by Wwhite on January 25, 2021

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?

3 Answers

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.

Answered by Antoine on January 25, 2021

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 $x sim x + 2pi$. Define complex coordinates $w=t+ix$ so $w sim w + 2pi i$. 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}


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}

Answered by Prahar on January 25, 2021

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.

Answered by JamalS on January 25, 2021

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