In Polchinski's first volume on String Theory, it is claimed in exercise 2.3 (or analogously in eq.(6.2.17)) that $$ \Bigl\langle \prod_{i=1}^N : e^{i k_i \cdot X_i(z_i,\bar{z}_i)} : \Bigr \rangle = i C^X (2\pi)^D \delta^D \left(\sum_{i = 1}^N k_i\right) \prod_{i,j=1, i<j}^n |z_{i j}|^{\alpha' k_i \cdot k_j} $$ This is considering a free 2-dimensional CFT in flat metric for the free fields $X^\mu_i$.

However, it is also claimed in eq.(2.2.13) that

$$ : e^{i k_1 \cdot X_1(z, \bar{z})} : : e^{i k_2 \cdot X_2(0, 0)}: \ = |z|^{\alpha' k_1 \cdot k_2} : e^{i k_1 \cdot X_1(z, \bar{z})} e^{i k_2 \cdot X_2(0, 0)}: $$ so that taking the correlator on both sides, I would obtain

$$ \langle : e^{i k_1 \cdot X_1(z, \bar{z})} : : e^{i k_2 \cdot X_2(0, 0)}: \rangle \ = |z|^{\alpha' k_1 \cdot k_2} \langle : e^{i k_1 \cdot X_1(z, \bar{z})} e^{i k_2 \cdot X_2(0, 0)}: \rangle = |z|^{\alpha' k_1 \cdot k_2} \langle 1 \rangle $$ where I used in the last step that $X_i$ being free, the exponentials can be expanded in creation and destruction modes and normal ordering just amounts to putting the creation operators on the left, as claimed in eq.(2.7.12), so that $\langle : O : \rangle = 0$ when $O$ is a product of free fields, as in common QFTs. So only the constant part $1$ in the series expansion of the exponential contributes.

This last equation is inconsistent with the first equation, and I don't understand why. There is a missing delta for momentum conservation. For the two equations to be consistent we would need $\langle : e^{i k_1 \cdot X_1(z, \bar{z})} e^{i k_2 \cdot X_2(0, 0)}: \rangle = \delta(k_1 + k_2) \langle 1 \rangle \neq \langle 1 \rangle$. This makes sense from the OPE and state-operator map point of view, but from my calculation I don't understand why this should be true. Can't we trust that $\langle : O : \rangle = 0$ when $O$ is a product of free fields ? Is this because the expansion of $X^\mu(z)$ in eq. (2.7.4) involes the term $x^\mu$ that does not necessarily vanish when acting on the ground state ?

  • $\begingroup$ Don't forget that $X$ also contains zero modes (momentum) in addition to creation and annihilation; you can define states like $|0;k\rangle = e^{ik \cdot P} |0\rangle$ which are orthogonal. This kind of thing is superbly explained in GSW vol 1. $\endgroup$ Jul 13, 2023 at 20:50

1 Answer 1


I think when you expand the exponential, you will get a big sum of terms, and some of these terms will contain only creation operators (importantly no annihilation operators), so the result (whatever it is) is probably not going to equal <1>!

  • $\begingroup$ But creation operators also have vanishing VEV. Schematically, $\langle 0 | (a^\dagger)^n a^m | 0 \rangle = 0$ for any $(n,m) \neq (0,0)$, because creation operators acting on their left kill the vacuum, for example $\langle 0 | a^\dagger |0 \rangle = 0$ because $\langle 0 | a^\dagger = (a | 0 \rangle)^\dagger = 0$ $\endgroup$ Jul 13, 2023 at 19:51
  • $\begingroup$ Oops my bad. I'll dig out my copy of Polchinski and have a re-read $\endgroup$ Jul 13, 2023 at 20:32
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    $\begingroup$ Okay I think the answer to your question is explained in section 6.7 of Polchinski's book. Essentially you have to be careful how you interpret your two equations. The first one you wrote is derived using path integral methods, and the second one you wrote is in terms of operators. The process of relating operator matrix elements to their path integral counterparts is explained in this section. In particular, this explains where the delta function comes from. $\endgroup$ Jul 13, 2023 at 21:07

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