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We are told that in a 2D CFT, all primary operators can be written in the form of \begin{equation} \tag{1} \mathcal{O}(z, \bar{z} ) = \sum_{m,n \in \mathbb{Z} } \frac{\mathcal{O}_{m,n} }{z^{m + h} \bar{z}^{n + \bar h}} \end{equation} Recently I have been wondering about what happens when $h$ and $\bar{h}$ are not integers. (For instance, I wonder how $\mathcal{O}(0,0)$ when acting on the ground state $|0\rangle$ is not trivially $0$.) In order to look at a specific example, I tried to see what the mode expansion of the operator \begin{equation} \mathcal{O}(z, \bar{z}) = \; :e^{ i k_\mu X^\mu (z, \bar z)} : \end{equation} is in the free boson CFT. When this operator acts on the ground state $|0\rangle$ when $z = \bar{z} = 0$, it creates a state of weight \begin{equation} (h, \bar{h} ) = \left( \frac{\alpha' k^2}{4}, \frac{\alpha' k^2}{4} \right) \end{equation} from equation 2.4.17 in Polchinski. As $h$ and $\bar{h}$ aren't integers, I therefore tried to see what the mode expansion of this operator is. However, it doesn't quite seem to take the form of $(1)$, which makes me wonder if all operators really can be expressed as $(1)$. I present my work below.

Equation 2.7.4 in Polchinski reads \begin{equation} X^\mu(z, \bar{z}) = x^\mu - i \frac{\alpha'}{2} p^\mu \ln(|z|^2)+ i \left( \frac{\alpha'}{2} \right)^{1/2} \sum_{m \in \mathbb{Z} - \{0\} } \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha_m^\mu} }{\bar{z}^m} \right). \end{equation} Note that $\alpha^\mu_m$ is an annihilation operator for $m > 0$ and a creation operator for $m < 0$. Normal ordering pulls all the annihilation operators to the right and all the creation operators to the left. In this procedure $p^\mu$ is regarded as an annihilation for this procedure and $x^\mu$ is regarded as a creation operator. (The normal ordering proscriptions are related by equation 2.7.12 in Polchinski.)

We now find \begin{align} &:e^{ i k_\mu X^\mu (z, \bar z)} : \\ &= e^{i k_\mu x^\mu} \exp(\sum_{m<0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) e^{ \alpha' k_\mu p^\mu \ln|z|} \exp(\sum_{m>0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) \end{align} As brief check, note that \begin{align} p^\mu |0\rangle &= 0 \\ \alpha^\mu |0\rangle &= 0 \hspace{0.5 cm} m > 0 \end{align} so \begin{equation} : e^{i k_\mu X^\mu(0,0)} : |0\rangle = e^{i k_\mu x^\mu} |0\rangle = |k; 0\rangle \end{equation} as expected.

We now use the commutation relation (2.7.5b in Polchinski) \begin{equation} [x^\mu, p^\nu] = i \eta^{\mu \nu} \end{equation} and Baker-Campbell-Hausdorff to write \begin{align} e^{i k_\mu x^\mu} e^{\alpha' k_\nu p^\nu \ln|z|} &= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| + \tfrac{1}{2} [i k_\mu x^\mu , \alpha' k_\nu p^\nu \ln|z| ] ) \\ &= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| - \tfrac{1}{2} \alpha' k^2 \ln|z| ) \\ &= \exp( i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z| ) |z|^{ - \tfrac{1}{2} \alpha' k^2 } \end{align} making \begin{align} &:e^{ i k_\mu X^\mu (z, \bar z)} : \\ &= e^{i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z|} \\ &\exp(\sum_{m<0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) \exp(\sum_{m>0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) \frac{1}{z^{\alpha' k^2/4 }}\frac{1}{{\bar z}^{\alpha' k^2/4 }} \end{align} This is a somewhat confusing answer. It almost looks to be of the form \begin{equation} :e^{ i k_\mu X^\mu (z, \bar z)} : \; \stackrel{?}{=} \sum_{m \in \mathbb{Z} } \frac{\mathcal{O}_{m,n} }{z^{m + h} \bar{z}^{n + \bar h}} \end{equation} but not quite, due to the $e^{i k_\mu x^\mu + \alpha' k_\nu p^\nu \ln|z|} $ out front. Does anyone have any ideas?

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  • $\begingroup$ Do the residue theorem and Taylor expansion not help with respect to a direct determination of $\mathcal O_{m,n}$? $\endgroup$
    – user21299
    Feb 22, 2021 at 6:08
  • $\begingroup$ I'm sure there should be a way to use them, but I'm having trouble drawing any conclusions that way $\endgroup$ Feb 22, 2021 at 6:12

2 Answers 2

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Okay, I have the answer. The first place I'd like to direct you to is this question + answer where I explain in a lot of detail how mode expansions work in general, including for non integer weight.

For this problem in particular, I should have just stopped when I obtained the formula. \begin{align} &:e^{ i k_\mu X^\mu (z, \bar z)} : \\ &= e^{i k_\mu x^\mu} \exp(\sum_{m<0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ) e^{ \alpha' k_\mu p^\mu \ln|z|} \exp(\sum_{m>0} \frac{1}{m} \left( \frac{\alpha^\mu_m}{z^m} + \frac{\widetilde{\alpha}^\mu_m}{{\bar z}^m} \right) ). \end{align} This formula is indeed of the form \begin{equation} \mathcal{O}(z, \bar{z})| 0 \rangle = \sum_{m,n = 0}^\infty \frac{\mathcal{O}_{mn}}{z^m \bar{z}^n} |0\rangle \end{equation} because \begin{equation} p^\mu |0\rangle = 0. \end{equation} You see, just because we have \begin{equation} \mathcal{O}(z, \bar{z})| 0 \rangle = \sum_{m,n = 0}^\infty \frac{\mathcal{O}_{mn}}{z^m \bar{z}^n} |0\rangle\hspace{1cm} (\text{true}) \end{equation} does NOT mean we have \begin{equation} \mathcal{O}(z, \bar{z}) = \sum_{m,n = 0}^\infty \frac{\mathcal{O}_{mn}}{z^m \bar{z}^n} \hspace{1cm} (\text{false}). \end{equation}

They are very different expressions. You simply can't do this mode expansion without acting on the vacuum on the right. If you would like to see how to do mode expansions in general, I will once again link you to here where I explain it in detail.

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EDIT: Holomorphicity of the expectation value is not strong enough to argue for holomorphicity of an operator, see user1379857's answer.

I can't comment yet, so I post this as an answer.

Suppose you can write $\mathcal{O}(z,\bar{z}) = O(z) \bar{O}(\bar{z})$, where $O(z)$ is of weight $(h,0)$ and $\bar{O}(\bar{z})$ of weight $(0,\bar{h})$. Then one can argue that $O(z)$ is holomorphic (except at insertion points) by noting that $[\bar{L}_{-1},O(z)]=\bar{\partial}O(z)$ has zero norm. By the residue theroem, it follows that

$O(z) = \sum\limits_{n \in \mathbb{Z}-h} \frac{O_n}{z^{n+h}}$.

Likewise for $\bar{O}(\bar{z})$.

However, $:e^{ikX(z,\bar{z})}:$ cannot be factorized, otherwise the zero mode would be duplicated in the process. Does this answer your question?

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  • $\begingroup$ This is a good answer although I'm still somewhat tentative. Seems weird to me that this simple operator breaks standard CFT expressions. $\endgroup$ Mar 2, 2021 at 21:25
  • $\begingroup$ I'm not sure this non-factorisation argument is the way to go. You could have considered compact scalars, where the zero modes do factorise. $\endgroup$
    – user21299
    Mar 3, 2021 at 17:43
  • $\begingroup$ @alexarvanitakis I agree, I guess zero norm of the derivative is not strong enough to argue for holomorphicity in general. $\endgroup$
    – and008
    Mar 4, 2021 at 7:38

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