# Why is the limiting operator in the CFT state-operator correspondence well-defined, and why is conformal symmetry necessary?

Consider a Euclidean CFT in radial quantisation, and let $$S$$ be the unit sphere centred on the origin. The state-operator correspondence says that any state $$\Psi_S$$ living on $$S$$ can be prepared by a path integral with an insertion only at the origin. It is proved as follows (see Sec 4.6 in Tong):

Let $$S_r$$ be the sphere of radius $$r$$ centred on the origin. Then we can evolve the state $$\Psi_S$$ radially to get some state $$\Psi_r$$ living on $$S_r$$. This evolution can be written as a path integral on the annulus $$r\leq |\mathbf{x}|\leq 1$$: \begin{align} \tag{1}\Psi_S[\phi_S] &= \int D\phi_{r} \space\Psi_r[\phi_r] \int_{\phi|_{S_r}=\phi_r}^{\phi|_S=\phi_S}D\phi e^{-S} \\ \tag{2}&=\int_{\phi|_S = \phi_S} D\phi \space\space\Psi_{S_r}\big[\phi|_{S_r}\big]\space e^{-S}. \end{align}

(I'm working in the wavefunctional picture, where $$\Psi_S$$ is a functional of the field configuration $$\phi_S$$ on $$S$$, and likewise for $$\Psi_r$$. The second line follows since integrating over $$\phi_r$$ removes the inner boundary condition.)

Eq (2) tells us that any state $$\Psi_S$$ can be prepared by a path integral on the annulus $$r\leq|\mathbf{x}|\leq 1$$ with some appropriate insertion $$\Psi_r[\phi|_{S_r}]$$ on the inner boundary.

Now take $$r\to 0$$, so the inner boundary shrinks to the origin. Hence we conclude that $$\Psi_S$$ can be prepared by an insertion at the origin.

Question 1: How exactly is the $$r\to 0$$ limit defined? Each $$\Psi_r$$ is a functional depending on $$\phi|_{S_r}$$, whereas in the $$r\to 0$$ limit we generally expect to obtain an insertion depending on not just $$\phi(0)$$ but also its derivatives. I think the existence of an appropriate limit is the exact content of what people call the "state-operator correspondence", but I can't find any reference addressing this.

Question 2: Why does the QFT need to be conformal for the above to work? Even without conformal symmetry, the path integral still defines some map from states on $$S$$ to states on $$S_r$$. Then we can take $$r\to 0$$ as before and the rest of the argument seems to work fine, showing that all states can be prepared by an insertion at the origin. In other words, I'm claiming that radial quantisation is perfectly well-defined in an arbitrary QFT. Unlike in CFT, in general the 1-parameter family of evolution maps from $$S$$ to $$S_r$$ will not be the exponential of some conserved charge, but I don't see why this would pose a problem to the above argument.

I'm asking these two questions together because I suspect their answers might be related. For example, perhaps the limiting procedure in the state-operator correspondence is well-defined precisely when the theory is conformal.

EDIT: another confusing point is that usually in QFT operators have to be smeared: they're not well-defined "at a point". So I don't know whether the state-operator correspondence is even well-defined, at least the way it's normally written.

• Honestly I think the operator version of the derivation is clearer than the path integral one. I recommend seeing chapter 6 of the yellow book (Di Francesco), may help you out.
– Gold
Jan 30 at 23:28
• I just think about how a theory on a cylinder can easily have a state which oscillates as $t \to -\infty$ without reaching a limit. This is what the ability to sort states into dilation irreps buys you. Jan 31 at 1:56
• About question (2) here's one intuitive argument. In a standard QFT we also have in/out states defined at $t\to \pm \infty$. These states are created by creation/annihilation operators which can be written as integrals of quantum fields over the Cauchly slices at $t\to \pm \infty$. The point to observe then is that the state is really associated to the quantum field at infinitely many points. In a CFT you can use a conformal map to switch to radial quantization. This shrinks the surfaces $t\to \pm \infty$ to points and allows for the states to be associated to quantum fields at single points.
– Gold
Jan 31 at 3:36
• We ought to be able to do it for the subset of states $\Psi_r$ such that the limit as $r \to 0$ exists right? Jan 31 at 19:22
• Something which many sources gloss over is that this limit fails to exist for many states in a CFT as well. But it does exist for dilation eigenstates so we can decompose more general states in terms of those. It's about the set of states created by a local operator spanning the Hilbert space rather than being equal to it. Feb 1 at 4:27

It is not true that any state on the sphere can be prepared by a local operator at the origin. For example, take the state $$|\Psi\rangle=\phi(x)\phi(-x)|0\rangle$$, defined on the unit sphere, where $$0<|x|<1$$ and $$\phi$$ is some local scalar operator. If it were true that $$|\Psi\rangle=\mathcal{O}(0)|0\rangle$$ for some local operator $$\mathcal{O}$$, then the correlation function $$\langle 0|\mathcal{O}'(y)|\Psi\rangle$$, where $$\mathcal{O}'$$ is a local operator, would be regular for all $$|y|>0$$ (the defintion $$\langle 0|\mathcal{O}'(y)|\Psi\rangle$$ makes sense only for $$|y|>1$$ but since correlation functions are analytic, this is enough to ask questions also at |y|<1). But of course we know that $$\langle 0|\mathcal{O}'(y)|\Psi\rangle=\langle 0|\mathcal{O}'(y)\phi(x)\phi(-x)|0\rangle$$ has singularities at $$y=\pm x$$ and no singularity at $$y=0$$.

As Connor Behan points out in the comments only the dilatation eigenstates can be represented by local operators at a point. For dilatation eigenstates, there is no limit to be taken as they evolve trivially. To make the argument precise, one needs to define what you mean by a local operator.

The objection raised by the OP to Connor Behan's comment is addressed by the following observation. The dilatation eigenstates do not span the Hilbert space, they only span a dense subset. In other words, to reproduce a completely general state, one needs to take infinite sums of dilatation eigenstates. These sums will converge in the Hilbert space, but not in the space of local operators (by the argument in the beginning of this answer).

• The OPE is usually written in terms of operators (it is an "operator" expansion, after all). E.g. we'd usually write $\phi(x)\phi(-x)=\sum_k c_k(x,\partial) \hat{\mathcal{O}}_k(y)\big|_{y=0}$. Are you saying that this is wrong, since the RHS doesn't converge, and that it only becomes correct when we have both sides acting on the vacuum? Mar 13 at 21:32
• @nodumbquestions To start, you would need to define what you mean by "operator" -- this is a massively over-used term -- and what topology you have in mind on the space of "opeators". But yes, it is a well-known fact that the OPE is only guaranteed to converge when acting on the vacuum state, in which case it is a convergence in the Hilbert space. It is important to note, however, that there is a freedom in the choice of quantization, and what doesn't act on the vacuum state in one quantization may act on the vacuum state in another quantization. Mar 13 at 22:46
• A simple example is arranging 4 operators on a line in the order 1,2,3,4 and trying to take the OPE between 1 and 3 in the four-point function. It will not converge. Mar 13 at 22:47
• @nodumbquestions a standard reference is "Convergence of operator product expansions on the vacuum in conformal invariant quantum field theory" link.springer.com/article/10.1007/BF01609130 Mar 14 at 11:27
• @nodumbquestions This reference is in working in Lorentzian signature, but the difference is superficial. The avatar of this in Euclidean signature is that the OPE between two operators converges if you can separate them from the other operators in the correlator by a sphere. This just means that there is a radial quantization in which both operators act on the vacuum. Mar 14 at 11:31

TL;DR: OP's wave functional eqs. (1/2) relies on a sewing axiom/completeness relation, not a limit per se.

1. Review. In a (not necessarily conformal) QFT the wave functional is $$\Psi(\phi,t)~=~{}_H\langle \phi,t|\Psi\rangle_H ~=~{}_H\langle \phi,0|\Psi(t)\rangle_S\tag{A}$$ where the instantaneous Heisenberg eigenstates $$\hat{\Phi}({\bf r},t)|\phi,t \rangle_H ~=~\phi({\bf r})|\phi,t \rangle_H \tag{B}$$ satisfy a completeness relation $$\int\!{\cal D}\phi~|\phi,t \rangle_H {}_H\langle \phi,t|~=~{\bf 1} \tag{C}$$ and orthonormality relation $${}_H\langle \phi_1,t|\phi_2,t \rangle_H~=~\prod_{\bf r}\delta(\phi_1({\bf r})\!-\!\phi_2({\bf r})). \tag{D}$$ The overlap/kernel is given by the path integral \begin{align}K(\phi_f,t_f;\phi_i,t_i) ~=~&{}_H\langle \phi_f,t_f|\phi_i,t_i \rangle_H \cr ~=~&\int_{\Phi|_{\partial B(0,t_i)}=\phi_i}^{\Phi|_{\partial B(0,t_f)}=\phi_f}\! \left[ \prod_{x\in B(0,t_f) \backslash B(0,t_i)} d\Phi(x)\right]~e^{-\frac{1}{\hbar}S[\Phi]} \cr ~=~&\int_{\Phi(\cdot,t_i)=\phi_i}^{\Phi(\cdot,t_f)=\phi_f}\! \left[\prod_{{\bf r},~t_i\leq t\leq t_f}d\Phi({\bf r},t)\right]~e^{-\frac{1}{\hbar}S[\Phi]} \cr ~=~&\int_{\Phi(\cdot,t_i)=\phi_i}^{\Phi(\cdot,t_f)=\phi_f}\! {\cal D}\Phi~e^{-\frac{1}{\hbar}S[\Phi]}.\end{align}\tag{E} OP's eq. (1) follows from eqs. (A), (C) & (E) without using conformal symmetry. \begin{align} \Psi(\phi_f,t_f)~\stackrel{(A)+(C)}{=}&\int\!{\cal D}\phi_i~ {}_H\langle \phi_f,t_f|\phi_i,t_i\rangle_H \Psi(\phi_i,t_i) \cr ~\stackrel{(E)}{=}~&\int\!{\cal D}\phi_i~\Psi(\phi_i,t_i) \int_{\Phi(\cdot,t_i)=\phi_i}^{\Phi(\cdot,t_f)=\phi_f}\! {\cal D}\Phi~e^{-\frac{1}{\hbar}S[\Phi]}. \end{align}\tag{1}

2. Now let us consider a CFT. A wave functional $$\Psi_{\cal O}(\phi_f,t_f)$$ for a local operator $$\hat{\cal O}$$, or corresponding state $$|{\cal O}\rangle_H~=~\hat{\cal O}(0)| \Omega \rangle_H,\tag{83}$$ is given by \begin{align}\Psi_{\cal O}(\phi_f,t_f) ~=~&{}_H\langle \phi_f,t_f|{\cal O} \rangle_H \cr ~\stackrel{(83)}{=}~&{}_H\langle \phi_f,t_f|\hat{\cal O}(0)|\Omega\rangle_H \cr ~\stackrel{(C)}{=}~&\int\!{\cal D}\phi_i~ {}_H\langle \phi_f,t_f|\phi_i,t_i\rangle_H \Psi_{\cal O}(\phi_i,t_i) \cr ~\stackrel{(E)}{=}~&\int\!{\cal D}\phi_i~\Psi_{\cal O}(\phi_i,t_i) \int_{\Phi(\cdot,t_i)=\phi_i}^{\Phi(\cdot,t_f)=\phi_f}\! {\cal D}\Phi~e^{-\frac{1}{\hbar}S[\Phi]}\cr ~=~&\int^{\Phi(\cdot,t_f)=\phi_f}\! \left[\prod_{{\bf r},~ t\leq t_f}d\Phi({\bf r},t)\right]~e^{-\frac{1}{\hbar}S[\Phi]}{\cal O}(0) \cr ~=~&\int^{\Phi(\cdot,t_f)=\phi_f}\! {\cal D}\Phi~e^{-\frac{1}{\hbar}S[\Phi]}{\cal O}(0),\end{align}\tag{F} due to a sewing axiom/completeness relation. In particular, the middle expressions/right-hand sides of eq. (F) are independent of $$t_i$$, so the limit-process $$t_i\to -\infty$$ is a constant process, and therefore convergent to that constant.

An $$n$$-point correlator function satisfies \begin{align} {}_H\langle\Omega |T\left\{ \hat{\cal O}_1(x_1)\ldots \hat{\cal O}_n(x_n) \right\}|\Omega\rangle_H ~=~&\int\!{\cal D}\Phi~e^{-\frac{1}{\hbar}S[\Phi]} {\cal O}_1(x_1)\ldots {\cal O}_n(x_n)\cr ~=~&\left[ \prod_{b=1}^n \int\!{\cal D}\phi_b~ {}_H\langle \phi_b,R_b |{\cal O}_b\rangle_H \right]\cr &\int_{\Phi|_{\partial B(x_b,R_b)}=\phi_b}\! \left[ \prod_{x\notin\cup_{b=1}^n B(x_b,R_b)} d\Phi(x)\right]~e^{-\frac{1}{\hbar}S[\Phi]}, \end{align}\tag{81} cf. Ref. 2.

• The logic here is unclear to me. I want to show that every state can be prepared by an insertion at the origin (equivalently, acting on the vacuum by some $\hat{O}(0)$). But you seem to be taking this as a starting point in Eq (83), and then using it to prove... what exactly? Note that my original question is basically how you got from the fourth to the fifth line of Eq (F), which requires taking a limit. Feb 12 at 17:28