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  Operator-State Correspondence in CFT

+ 1 like - 0 dislike

I'm trying to understand a proper derivation of the operator-state map in CFT. Here is what I think I've understood so far, which is mostly based on David Tong's string theory lectures, and a little of Polchinski volume 1.

We consider a generic state in the Schrödinger representation, obtained from evolving some initial wavefunction $\Psi_i[\phi_i]$ at time $t_i$ to some time $t_f$ via the path integral

$$ \Psi_f[\phi_f] = \int\mathcal{D}\phi_i\int_{\phi_i}^{\phi_f}\mathcal{D}\phi\, e^{-S[\phi]} \Psi_i[\phi_i]$$

where the inner integral with goes over all configurations with boundary condition $\phi(t_i) \equiv \phi_i$, $\phi(t_f) \equiv \phi_f$, and the outer one integrates over all boundary configurations $\phi_i$.

We can then take the limit $t_i\to-\infty$, which in radial quantization corresponds to sending $r\to0$, which just corresponds to the origin. The wave function then looks like

$$ \Psi_f[\phi_f] = \int^{\phi_f}\mathcal{D}\phi\, e^{-S[\phi]} \mathcal{O}(0). $$

where we have renamed the initial condition $\Psi[\mathrm{const}]$ to $\mathcal{O}(0)$ (which we can do, since it is just a constant). This then looks like a cutout of a correlation function involving a local operator $\mathcal{O}(0)$, which justifies interchanging states and local operators inside correlation functions, where the state corresponds to $\mathcal{O}(0) | 0 \rangle$ (where $|0\rangle$ is the state corresponding to the identity operator).

However there are a few things I don't really understand yet.

- Where does the necessity for conformal invariance come in? Evidently we need radial quantization for this to work, but afaik, I can in principle apply radial quantization to any Euclidean field theory, conformal or not (maybe this is what I'm getting wrong?).

- This gets us a correspondence between states and local operators, but generally in CFT what we really want is a more specific correspondence between dilation eigenstates $|\Delta\rangle$ and primary (or quasi-primary in 2d) operators $\mathcal{O}_\Delta(x)$. How does this follow from the above formulation? Is the idea that I can, if $|\Delta\rangle$ is a dilation eigenstate, always extend $\mathcal{O}(0)$ to a corresponding unique primary operator $\mathcal{O}_\Delta(x)$ through a translation?

Would be very grateful for some explanations! 

P.S. I only really want to apply it to 2d CFT but I've kept the notation more general, in case the higher dimensional case is a better starting point for some people. I hope I haven't made any mistakes in my formulation.

asked Jul 7, 2022 in Theoretical Physics by justsomethingidc (15 points) [ no revision ]

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