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.


4 Answers 4


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.

  • $\begingroup$ 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. $\endgroup$
    – Prahar
    Nov 15, 2014 at 15:19

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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Prahar
    Nov 15, 2014 at 13:50
  • $\begingroup$ @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. $\endgroup$
    – user10001
    Nov 15, 2014 at 14:29
  • $\begingroup$ Wait, I don't understand that. The wave-functional of a system at time $r$ is $\psi[ \phi_i(\sigma)]$ where $\sigma$ is the coordinate on a circle at fixed radius $r$. Now, if we take $r \to 0$, the field is now only valued at a point $\phi_i(0)$ and the wave function is $\psi [ \phi_i(0) ]$. Now, since in a CFT, there is a one-to-one correspondence between states and local operators, the map $T_D$ is definitely invertible. So my question is valid. $\endgroup$
    – Prahar
    Nov 15, 2014 at 15:06
  • $\begingroup$ 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. $\endgroup$
    – Prahar
    Nov 15, 2014 at 15:17
  • 1
    $\begingroup$ I'm pretty sure the map is surjective as well. The state-operator correspondence forms the basis of several techniques that are commonly employed in CFTs. So, simply by its usage and proliferation in CFTs, I'm sure that the map is invertible. I just want to be able to prove it to myself. $\endgroup$
    – Prahar
    Nov 15, 2014 at 15:51

There is a dictionary that translate the path integral formalism to the operator formalism. A matrix element of a string of local operators in time ordering are translate to a path integral with fields insertions, e.g.

$$\langle \psi_2|\mathcal{T}\mathcal{O}_1[\phi(x_1),x_1]\mathcal{O}_2[\phi(x_2),x_2]|\psi_1\rangle\rightarrow \int_{BC} \mathcal{D}\phi...\mathcal{O}_1[\phi(x_1),x_1]\mathcal{O}_2[\phi(x_2),x_2]e^{iS}$$

with $BC$ standing as a suitable boundary condition, and possibly weights at different boundary conditions too. $BC$ are related to the $|\psi_i\rangle$'s states. This functions $\mathcal{O}_i$ should depend explicity over the fields, otherwise they are simple c-numbers (can be put outside the path integral).

Now, when you have:

$$ \psi[ \phi_f(x),t_f] = \int d\phi_i \int\limits_{\phi(0) = \phi_i}^{\phi(x,t_f) = \phi_f(x) }\mathcal{D}\phi\, e^S\, \psi(\phi_i , 0)\rightarrow \int\limits_{no\,BC\,here}^{\phi(x,t_f) = \phi_f(x) }\mathcal{D}\phi\, e^S\, \Psi[\phi(0),0] $$

i.e. the integral in $d\phi_i$ can be absorbed in the measure, wicth means that there is no boundary condition at $x=0$ anymore, because you are summing over all possible values there. The weight $\psi(\phi_i , 0)$ can be view as a function of the field evaluated at $x=0$, a local operator evaluated at $x=0$.

Now, you can define a state setting $\Psi[\phi(0),0]=1$ $$ |0\rangle \leftrightarrow \psi_0[ \phi(x),t] =\int^{\phi(x,t) = \phi(x) }\mathcal{D}\phi\, e^S\times 1 $$

now, for an arbitrary state you have:

$$ |\psi\rangle \leftrightarrow \psi[ \phi(x),t]=\int^{\phi(x,t) = \phi(x) }\mathcal{D}\phi\, e^S\, \Psi[\phi(0),0] \leftrightarrow \Psi[\phi(0),0]|0\rangle $$

Every state $|\psi\rangle$ can be created out of a local operator $\Psi[\phi(0),0]$ acting on the $|0\rangle$ state:

$$ |\psi\rangle=\Psi[\phi(0),0]|0\rangle $$


Let's consider the vacuum state: $$\Psi_0[\phi_f]=\int d\phi_i \int_{\phi(0)=\phi_i}^{\phi(x,t_f)=\phi_f} [d\phi(x,t)]\exp\left[\frac{i}{\hbar}\int_0^{t_f} dt' L\right]\,.$$ Now let us consider acting on the vacuum with a local operator $O(0)$ at the origin. A local operator in this context is a local function of the fields $\phi$, so I will write it as $O[\phi(0)]$. The resulting state is $$\Psi_O[\phi_f]=\int d\phi_i \int_{\phi(0)=\phi_i}^{\phi(x,t_f)=\phi_f} [d\phi(x,t)]\exp\left[\frac{i}{\hbar}\int_0^{t_f} dt' L\right]O[\phi_i]\,.$$ So now we can attach to the wave function you wrote an operator $O$ via $O[\phi_i]=\psi(\phi_i,0)$.


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