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  Explicit evaluation of a radially ordered product

+ 1 like - 0 dislike

I am trying to understand the application of the operator product expansion to calculate the radially ordered product in the complex plain of $T_{zz}(z)\partial_w X^{\rho}(w)$ which should result in

$$ \langle R(T_{zz}(z)\partial_w X^{\rho}(w))\rangle = -l_s^2\frac{1}{(z-w)^2}\partial_w X^{\rho}(w) - l_s^2\frac{1}{(z-w)}\partial_z^2 X^{\rho}(z) + \cdots $$

but embarassingly I encounter a stumbling block right at the beginning. After inserting $T_{zz}(z) \doteq \, :\eta_{\mu\nu}\partial_z X^{\mu}\partial_zX^{\nu}:$ one has

$$ \langle R(T_{zz}(z)\partial_w X^{\rho}(w))\rangle = R(:\eta_{\mu\nu}\partial_z X^{\mu}(z)\partial_zX^{\nu}(z):\partial_w X^{\rho}(w)) $$

which can obviously be further expanded to

$$ ... = \eta_{\mu\nu}\langle \partial_z X^{\mu}(z)\partial_w X^{\rho}(w)\rangle \partial_z X^{\nu}(z) + \eta_{\mu\nu}\langle \partial_z X^{\nu}(z)\partial_w X^{\rho}(w)\rangle \partial_z X^{\mu}(z) $$

It is exactly this last step I dont understand. If this initial stumbling block is removed, I can understand the rest of the derivation, so can somebody help me remove it?

To generalize a bit, it seems I do not yet properly understand how such expressions involving normal and radial (time) ordered products are generally evaluated. So if somebody could give me a more general hint about this, I would probably be able to see how the last expression in my particular example is obtained.

asked Mar 24, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
edited May 1, 2014 by Dilaton

Related: physics.stackexchange.com/q/22784/2451 and links therein.

This post imported from StackExchange Physics at 2014-03-12 15:19 (UCT), posted by SE-user Qmechanic

Darn, I forgot the check to edit silently, sorry :-/

@Dilaton Oh yes, editing silently when doing a mass edit-out of attributions of one's own posts, to not flood the main page seems like a good idea.   

I think I should also practise that in my mass-retag of all posts.    

But I think it is fine to not check the box for unanswered questions.  

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