I'm confused about one aspect of Polchinski's discussion of the state-operator map. He starts with an operator ${\mathscr{A}}(0)$ at the origin and then defines $$\Psi[\phi_b]=\int[d\phi_i]_{\phi_b}\exp(-S[\phi_i]){\scr A}(0)\tag{2.8.17}.$$ Here the path integral is over field configurations $\phi_i$ on the unit disk $|z|<1$ with the property that on the unit circle $|z|=1$ they obey $\phi_i|_{|z|=1}=\phi_b$.

Clearly this is a functional of the values of field configurations at some circle centered on the origin and therefore defines a state in wavefunctional representation. That seems clear.

But on the other direction, he starts with $\Psi[\phi_b]$ at $|z|=1$. Then he writes $\Psi[\phi_b]$ as the evolution of another state $\Psi'[\phi_b']$ at $|z|=r<1$. Indeed $\Psi'=r^{-L_0-\bar L_0}\Psi$ and therefore $$\int[d\phi'_b][d\phi_i]_{\phi_b,\phi'_b}\exp(-S[\phi_i])r^{-L_0-\bar L_0}\Psi[\phi_b']\tag{2.8.18}$$ just gives $\Psi[\phi_b]$. The path integral is now over the annulus $r<|z|<1$ with boundary values $\phi_b'$ at $|z|=r$ and $\phi_b$ at $|z|=1$.

Then he says: "Now take the limit as $r\to 0$. The annulus becomes a disk, and the limit of the path integral over the inner circle can be thought of as defining some local operator at the origin. By construction, the path integral on the disk with this operator reproduces $\Psi[\phi_b]$ on the boundary".

I'm confused about this. Why this defines an operator? In operator formalism, operators act on states, I don't see how this defines any kind of action on states. In path integral formalism, operators arise as insertions in correlation functions $\langle {\cal O}_1(x_1)\cdots{\cal O}_n(x_n)\rangle$. Again I don't see how would his construction define some $\mathscr{A}(0)$ that we can insert in $\langle {\cal O}_1(x_1)\cdots{\cal O}_n(x_n)\rangle$. So it seems a little bit hand-wavying as it is.

My question is: how does one recognize that this construction gives rise to an operator in a more precise way? How does such a construction determines an action on states? And what appears more relevant, how does it give rise to something that can be inserted in a correlation function $\langle {\cal O}_1(x_1)\cdots{\cal O}_n(x_n)\rangle$?

  • $\begingroup$ related/possible duplicate: physics.stackexchange.com/q/146594/50583 $\endgroup$
    – ACuriousMind
    Sep 7, 2021 at 15:18
  • $\begingroup$ note that we usually don't explicitly construct "the operator" in the operator formalism of the correspondence either - we just show that operators are defined by their action on the vacuum and then declare the corresponding operator to be "the operator" that produces a specific state when acting on the vacuum - ,"morally", this is exactly as non-constructive as the path integral version where you don't get an explicit $O(x)$ expression as you would like. $\endgroup$
    – ACuriousMind
    Sep 7, 2021 at 15:22
  • $\begingroup$ @ACuriousMind now that you mention the operator formalism version of the correspondence, could you point me a reference discussing it? I've been able only to find the path integral version in Polchinski's book and Tong's lecture notes. In operator formalism I've only seen the "operator -> state" direction of the correspondence where we define $|{\cal O}\rangle = \lim_{x\to 0}{\cal O}(x)|0\rangle$", but didn't find the discussion of the other direction. I could be helpful to see it. $\endgroup$
    – Gold
    Sep 7, 2021 at 15:35
  • $\begingroup$ See e.g. Schottenloher, remark 9.11 (that one is actually a bit more constructive than I claim, but it assumes you now how to write any given state as a descendant). $\endgroup$
    – ACuriousMind
    Sep 7, 2021 at 15:59
  • $\begingroup$ Thanks @ACuriousMind, I saw his remark and it seems clearer ! Please tell me if I got it right. We have commuting hermitian operators $L_0$ and $\bar L_0$ which give rise to a basis of eigenstates $|h,\bar h\rangle$. Primary operators acting on the vacuum give some of those eigenstates, which are highest weights. Acting with lowering operators we get the others and they can also be constructed acting on the vacuum with the descendant fields of the primaries. So expanding on this basis any state will be a linear combination of primaries and descendants acting on the vacuum. Is that it? $\endgroup$
    – Gold
    Sep 7, 2021 at 21:00

1 Answer 1


Disclaimer: The fact that no other sources seem to do this makes me a little uneasy.

As you point out, we need to be able to think of this "operator" as an insertion into arbitrary correlation functions. To make sense of Polchinski's derivation with the unit disk, I like to consider a correlator of operators where the largest radial co-ordinate involved is $1$. By radial ordering, this will be the last operator, allowing us to write $\left < 0 | \dots \mathcal{O}_0(1) | 0 \right >$.

If we now change the state on the right to $\left | \Psi \right >$, we can write this as \begin{align} \langle 0 | \dots \mathcal{O}_0(1) | \Psi \rangle = \int [\phi^\prime_b] \langle 0 | \dots \mathcal{O}_0(r) | \phi^\prime_b \rangle r^{-L_0-\bar{L}_0} \Psi[\phi^\prime_b]. \end{align} The evolution operator $r^{-L_0-\bar{L}_0}$ arises because $\left | \phi^\prime_b \right >$ is a state in the theory's radius $r$ Hilbert space so it cannot be acted on directly by $\mathcal{O}_0(1)$. Importantly, if we remove insertions above and rewrite the zero point function as $\int [\phi_i] \exp \left ( -S[\phi_i] \right )$, we get back to (2.8.18).

At this point, we can use the fact that specifying operators is the same as specifying their matrix elements. This lets us define an operator $\mathcal{O}$ such that $\left < \phi^\prime_b | \mathcal{O} | 0 \right >$ is equal to the appropriate value of the wavefunctional. Moreover, since we took $r \to 0$, this will be a local operator. The end result is \begin{align} \langle 0 | \dots \mathcal{O}_0(1) | \Psi \rangle = \int [\phi^\prime_b] \langle 0 | \dots \mathcal{O}_0(1) | \phi^\prime_b \rangle \langle \phi^\prime_b | \mathcal{O}(0) | 0 \rangle \end{align} which looks just like $\left < 0 | \dots \mathcal{O}_0(1) \mathcal{O}(0) | 0 \right >$ where a complete set of states has been inserted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.