# Path integral derivation of the state-operator correspondence in a CFT

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Below, I paraphrase the path integral derivation of the state-operator correspondence in David Tong's notes on CFT (see pdf here). This is my interpretation of the text in that pdf, so please correct me if I'm wrong

He starts with the standard formula for time-evolution of the wave-function in the path integral formalism $$\psi[ \phi_f(x),t_f] = \int [d\phi_i(x)] \int\limits_{\phi(x,t_i) = \phi_i(x)}^{\phi(x,t_f) = \phi_f(x) }[d\phi(x,t)] \exp \left[ \frac{i}{\hbar } \int_{t_i}^{t_f} dt' L \right] \psi[ \phi_i(x) , t_i ]$$ Now, we consider a radially quantized CFT, where the time direction is radial. Further, we take $t_i = 0$ in the equation above. Since this corresponds to the origin of the radial plane, the apriori function $\phi_i(x)$ reduces to a number $\phi_i$. The path integral then reduces to $$\psi[ \phi_f(x),t_f] = \int d\phi_i \int\limits_{\phi(0) = \phi_i}^{\phi(x,t_f) = \phi_f(x) }[d\phi(x,t)] \exp \left[ \frac{i}{\hbar } \int_{0}^{t_f} dt' L \right] \psi(\phi_i , 0)$$ Next, he says and I quote

The only effect of the initial state is now to change the weighting of the path integral at the point $z = 0$. But that’s exactly what we mean by a local operator inserted at that point.

Can anyone help me understand why this is what we mean by a local operator inserted at that point? I feel like I understand the statement, in principal, but I would like a more precise description. In other words, what I would really like is an explicit construction of the operator whose insertion in a certain path integral would reproduce the equation above.

PS - A reference to a paper is enough.

This post imported from StackExchange Physics at 2014-11-13 20:51 (UTC), posted by SE-user Prahar

asked Nov 13, 2014
edited Jun 21, 2016

All you need to do is insert the operator at 0 that maps $\phi_i$ to whatever other state you want to put there, though one needs to understand the mapping of field values like $\phi_i$ to states in the Hilbert space. Of course, you probably know all this now (:

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