I've been working through a textbook and course on conformal field theory recently.

However in a section illustrating how to calculate correlators for secondary fields (using the free boson as an example CFT), there is one line of working I don't understand.

Given a secondary field $ \vert \varphi \rangle $, and a not necessarily primary field $ \vert \phi \rangle$ related by:

$$ \vert \varphi \rangle = a_{-r} \vert \phi \rangle : r > 0 $$

apparently we can reverse the state-field correspondence to write:

$$ \varphi \left( w \right) = \frac{1}{2 \pi i} \oint_{w} \frac{ \partial \Psi \left( z \right) \phi \left( w \right) } { \left( z - w \right)^r } \mathrm{d}z $$

I have no idea how they managed to get this from the above field. Note that here $a_{-r}$ is a creation operator, and $\Psi$ is some arbitrary field (I assume? it wasn't actually specified where $\Psi$ came from).

This step seemed like bit of a jump, and had no other explanation. However my attempt at a solution was:

Given that, for the free boson: $$\partial \Psi \left( z \right) = \sum_{r \in \mathbb{Z}} a_{r} z^{-r-1}$$

We can invert this to get:

$$ a_{-r} = \frac{1}{2 \pi i} \oint_0 \frac{\partial \Psi \left(z \right)} {z^{-r+1} } \mathrm{d} z $$


$$\varphi \left( w \right) = \left( \frac{1}{2 \pi i} \oint_0 \frac{\partial \Psi \left(z \right)} {z^{-r+1} } \mathrm{d} z \right) \ \ \phi \left( w \right) $$

$$ = \frac{1}{2 \pi i} \oint_0 \frac{\partial \Psi \left(z \right) \phi \left( w \right)} {z^{-r+1} } \mathrm{d} z $$

Now clearly this isn't quite the same as what I need to get, as I showed above. The integral is contoured around $0$ not $w$, and has the wrong denominator.

It seems like this could be fixed by using a Laurent series for $\partial \Psi \left( z \right)$ centred around w. However surely this would mean the coefficients would no longer be $a_{-r}$. Further, this would fix the contour so it's now around $w$, and turn the $z$ into a $\left( z-w \right)$ in the denominator, but wouldn't fix the incorrect power. This makes me think that perhaps this isn't the correct approach then.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.