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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 $\langle {\cal O}_1(x_1)\cdots{\cal O}_n(x_n)\rangle$?

Gold
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1 Answers1

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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.

Connor Behan
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