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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 2015-03-30 13:51 (UTC), posted by SE-user Prahar
asked Nov 13, 2014 in Theoretical Physics by prahar21 (535 points) [ no revision ]

2 Answers

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Inserting a local operator means multiplying the integrand of the path integral by an operator with fixed position. This way, only the value of the operator at this position contributes to the path integral. If you now assume that the operator is an insertion at the position $z=0$, which in the present context of radial quantization corresponds to the initial point in time, it simply plays the role of a weight factor. The concept is understandable from the formulae you wrote down: in the first one, you have the general form where $t_i$ is left arbitrary, and in the second one you restrict the operator to a certain position $t_i=0$, therefore "localizing" it.

Regarding a reference I can recommend you chapter two of Polchinski: it discusses insertions both in a general context and in their application to radial quantization and the operator/state correspondence.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Frederic Brünner
answered Nov 14, 2014 by Frederic Brünner (1,060 points) [ no revision ]
I have read the chapter that you talk about. Also, you seem to have simply repeated in words that Tong is saying, which I understand. However, what I'm looking for is an explicit construction of the correspondence, for a general class of theories.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prahar
+ 1 like - 0 dislike

The following was meant to be a comment rather than an answer. However, since it was a bit long for a comment so I am writing it in the answer box.

In the case of a field theory, states can be thought of as functions on the space of boundary conditions on a spatial slice. This is so because the space of boundary conditions on a spatial slice is the configuration space and (by the definition of canonical quantization) the quantum states are functions on the configuration space.

Now, in the case of a field theory in the complex plane with the radial direction taken as the time direction, spatial slices are of the form of cirlces. Therefore, the quantum states are now functions on the space of boundary conditions on a circle of fixed radius. We can chose any circle of nonzero radius to define our quantum space.

If, we denote by $H$ our space of states then the path integral on an annulus $A$ with fixed boundary conditions on its inner and outer boundary circles, defines a map

$$T_A : H\to H$$

This is the statement of the first integral in your question. Given a state on the inner boundary circle, we can get a state on the outer boundary circle by doing the path integral.

Now, instead of an annulus, consider a disc. In this case we have only one boundary. If we insert a local functional $\mathcal{O}(\phi(0),\partial_{z}\mathcal{O}(0))$ at the origin and do path integral on the whole disc then we'll of course get a quantum state in $H$ (fixing a boundary condition and then doing the path integral will give us a number. Thus, path integral will give a function on the space of boundary conditions on the boundary of the disc which is, by definition, a quantum state). Thus, path integral on the disc $D$ (of say unit radius) with a local functional inserted at the origin will define a map

$$T_{D}:\{\text{space of local functionals at the origin}\}\to H$$

This holds in any two dimensional field theory. However, in case of a conformal field theory, the above map's dependence on the geometry of the disc is much simpler than in a theory without conformal symmetry.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001
answered Nov 15, 2014 by user10001 (635 points) [ no revision ]
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I agree with everything you said, but its not what I'm looking for. Here's what I want: If I give you the wave-function of the system at the origin, namely $\psi(\phi_i)$, can you construct the corresponding operator?

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prahar
@Prahar At the origin $\psi(\phi_i)$ is not a wave function. Its rather a local functional of the field. Wave functions are assigned to proper boundaries. However, I am not sure if the above map $T_D$ is invertible.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001
I know that the correspondence is 1-1 since for every local operator ${\cal O}(z,{\bar z})$, I can construct a state ${\cal O}(0,0)|0\>$ which defines a state at the infinite past.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prahar
@Prahar the wavefunctions are of the form $\psi(\phi_i(\sigma))$ where $\phi_i(\sigma)$ is field specified on a circle of nonzero radius. On a circle of zero radius (i.e. a point) there are no (nontrivial) boundary conditions to be specified and we can at most associate a local functional depending upon the value of the field and its derivatives at that point. So by taking the r->0 limit of a wave function assigned to the inner boundary of an annulus we may only get a local functional at the origin and not a wave function.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001
Also, the correspondence is 1-1 in one direction i.e. to each local functional we can assign a state on the boundary. However i am not sure if to each state specified on the boundary we can construct a local functional or not.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001
Most recent comments show all comments
@Prahar ya you are right. I read that in a hurry. Perhaps the argument is that to each primary state we can assign a local field (through some algorithm that i don't remember) and then fields corresponding to other states can be generated by applying differential operators (Ln's) to the primary fields. But, I have never encountered any rigorous proof of these statements.

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user user10001
The space of boundary conditions at the origin is anyway much smaller than on a circle of nonzero radius and hence can't be the whole configuration space.

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user user10001

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