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When both "space" ($x$) and "time" ($y$) directions are periodic, the Laplacian on torus with coodinate $z=x+iy$ has a normalized zero mode $$ \varphi_0(z) = \frac 1 {\sqrt{{\rm Im}(\tau)}} $$ (Here $\tau$ is the modular parameter defining the torus.) As $$ -\nabla^2 \varphi_0=0. $$ the zero mode means that the Laplace operator is not 1-1 and so prevents ...


4

A descendant is not something that has derivatives in it. It is something that is a total derivative of something else $\mathcal{O}' = \partial_\mu \mathcal{O}$. More precisely, the definition of a primary operator is $$ K_\mu \mathcal{O}(x) = 0\,, $$ where $K_\mu$ is the generator of special conformal transformations. You can prove that you operator is a ...


3

The character of a representation of the Virasoro algebra depends on the whole structure of the representation, not just of the conformal dimension $h$ of the highest-weight state. For a Verma module, the character is simply $$ \chi_{c,h}^\text{Verma}(\tau) = \frac{q^{h+\frac{1-c}{24}}}{\eta(\tau)}$$ as you wrote. However, for generic $c$ there are two ...


2

They have put all the anti-holomorphic dependence into $\bar{v}(\bar{z})$. So for holomorphic modes of stress energy tensor $[L_n, \bar{v}(\bar{z})]=0$. For anti-holomorphic generators $[\bar{L}_n, \bar{v}(\bar{z})]\neq0$. Field $X(z, \bar{z})$ have holomorphic and anti-holomorphic parts in Laurent expansion (2.89). Then from equation (2.40) with $h=\bar{h}=...


2

Your claim that (1.7) only considers infinitesimal conformal transformations with singularity at zero is simply wrong: The most general generator that you get from (1.7) is a linear combination of all the basic generators there, i.e. a Laurent series $\sum a_n z^n$. While the center of a Laurent series may be its only singularity, it need not be - it has an ...


1

First, the one-particle state localized in the point, e.g. $\phi(x)|0\rangle$ is very singular. What behaves more nicely insteas is a smeared state $\int_V d^D x \eta(x) \phi(x)|0\rangle$. However if we look at this "particle at a point" we will see, $$ \phi(0)=\int \frac{d^{D-1}p}{(2\pi)^D\sqrt{2 E_p}} a^\dagger(\vec{p})|0\rangle $$ I.e. to produce ideally ...


1

Indeed there is a version of the state operator correspondence that holds in 1d CFT/Quantum mechanics. Note that operator -> state map is trivially true in any general QFT in arbitrary dimensions. However, state -> operator map is the non-trivial bit that holds only in a CFT, and follows from scale invariance as depicted by the diagram in your question. In a ...


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