I am trying to derive the following result (formula 3.7a from the book "Basic Concepts of String Theory") $$\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle = \frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{z} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{z}-\bar{w}) \quad(1) $$ where $\bar{z} = e^{2\pi i(\tau +\sigma)/l}$ and $\bar{w} = e^{2\pi i(\tau' +\sigma')/l}$ with $(\tau,\sigma)\sim(\tau,\sigma + l)$ parametrizing the closed string world sheet.

The propagator is defined as: $$\langle X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma') \rangle = \mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')] - :X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma'):\quad (2)$$ where $\mathcal{T}[\dots]$ denotes time-ordering and $:\dots:$ denotes normal-ordering.

I'm trying to derive (1) from the following mode expansion $$X^\mu_L(\tau+\sigma)=\frac{1}{2}x^\mu + \frac{\pi\alpha'}{l}p^\mu(\tau+\sigma)+i\sqrt{\frac{\alpha'}{2}}\sum_{n>0} \left(\frac{1}{n}\bar{\alpha}^\mu_n e^{-\frac{2\pi}{l} in(\tau+\sigma)} -\frac{1}{n}(\bar{\alpha}^\mu_n)^\dagger e^{\frac{2\pi}{l} in(\tau+\sigma)} \right) \quad(3) $$ with the commutation relation \begin{align} [\bar{\alpha}^\mu_m, (\bar{\alpha}^\nu_n)^\dagger] &= m\delta_{m,n}\eta^{\mu\nu}, \quad (m,n > 0)\\ [x^\mu, p^\nu] &= i\eta^{\mu\nu} \end{align} and we define $:p^\nu x^\mu:=x^\mu p^\nu$.

From the definition (2) and using (3) with the commutation relations, I was able to show that for $\tau > \tau'$, \begin{align} \langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle &= -i\frac{\pi\alpha'}{2l}\eta^{\mu\nu}(\tau+\sigma) + \frac{\alpha'}{2}\eta^{\mu\nu}\sum_{n>0}\frac{1}{n}\frac{\bar{w}^n}{\bar{z}^n} \\ &=\frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{z} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{z}-\bar{w}) \end{align} which agrees with (1). But the same derivation would lead to that for $\tau <\tau'$, $$\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle = \frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{w} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{w}-\bar{z})$$

I wonder if there is any subtlety in my derivation which I have neglected or what would be the correct way to derive (1) from the expansion (3).


1 Answer 1


Maybe I'm missing something but you can get as far as \begin{align} \langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle &= -i\frac{\pi\alpha'}{2l}\eta^{\mu\nu}(\tau+\sigma) + \frac{\alpha'}{2}\eta^{\mu\nu}\sum_{n>0}\frac{1}{n}\frac{\bar{w}^n}{\bar{z}^n} \end{align} without assuming anything about the ordering yet. Then there are two cases.

  1. If $\tau^\prime - \tau$ has a positive imaginary part, the propagator above is the expression you're trying to derive.

  2. If $\tau^\prime - \tau$ has a negative imaginary part, the sum in the propagator above doesn't even converge.

This looks like a version of the statement that only time ordered correlators make sense in Euclidean CFT. This also seems to be the conclusion from $$\langle X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma') \rangle = \mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')] - :X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma'):$$ if we take the vacuum expectation value of both sides.


I would say the plane co-ordinates $(z, \bar{z})$ are not more or less meaningful than the cylinder co-ordinates $(\sigma, \tau)$. There is some semantics going on about what we call Euclidean but I think I see what we're doing differently now.

  1. You're actually computing $\mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')]$ which means you're treating equation (2) as an equation which should be used.

  2. I'm just shoving $X^\mu_L(\bar{z}) X^\nu_L(\bar{w})$ between two vacuum states, as CFT people always do, which means I'm treating equation (2) as just something that we can keep in mind for later if the two-point function needs to be interpreted.

The second approach might be more subtle than I realized due to the $:p^\nu x^\mu: = x^\mu p^\nu$ convention for zero modes. But I agree that, when taking the first approach, I get your two expressions with $\ln(\bar{z} - \bar{w})$ and $\ln(\bar{w} - \bar{z})$. So that supports the expression in the previous question with the two Heaviside functions.

If this happened for vertex operators, I would be more worried. But $X_L$ and $X_R$ are not genuine CFT operators so maybe their correlators are more sensitive to how you quantize the theory.

  • $\begingroup$ I started with the R.H.S expression in eq(2), which is the definition of this correlation function (which is in fact the time-ordered two point function). This definition involves the time-ordering product of the two X_L. That's why my result depends on the ordering. $\endgroup$ Dec 23, 2021 at 4:24
  • $\begingroup$ Also how is the imaginary part related to the time ordering? tau here is the Lorentzian time, not the Euclidean time. $\endgroup$ Dec 23, 2021 at 4:27
  • $\begingroup$ No, the $\tau$ you wrote is Euclidean. You can tell because $\bar{z} = e^{2\pi i (\sigma + \tau) / l}$ is periodic in both $\sigma$ and $\tau$, not just $\sigma$. $\endgroup$ Dec 23, 2021 at 12:27
  • $\begingroup$ And I agree that this two point function is necessarily time ordered. But this means its value should not depend on the ordering of arguments because $\mathcal{T}$ puts them in order anyway... $X_L(\tau) X_L(\tau^\prime) \neq X_L(\tau^\prime) X_L(\tau)$ but $\mathcal{T}[X_L(\tau) X_L(\tau^\prime)] = \mathcal{T}[X_L(\tau^\prime) X_L(\tau)]$. $\endgroup$ Dec 23, 2021 at 12:31
  • $\begingroup$ I think $\tau$ is Lorentzian because you can see $e^{2\pi i (\sigma + \tau)/l}$ satisfy the Lorentzian wave equation $(-\partial^2_\tau + \partial^2_\sigma) f(\tau, \sigma) =0$. I think in the later chapter of the book, you rotate $\tau$ to Euclidean time so that the modes become $e^{2\pi\tau_E +2\pi i \sigma}$ where $\tau_E$ becomes the radial direction. $\endgroup$ Dec 23, 2021 at 16:37

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