From Becker, Becker and Schwarz String Theory and M-Theory:

For the infinitesimal conformal transformation $$\tag{3.25}\delta z=\varepsilon(z)\quad\text{and}\quad \delta\bar z=\tilde\varepsilon(\bar z),$$ the associated conserved charge that generates this transformation is $$\tag{3.26}Q=Q_\varepsilon+Q_{\tilde\varepsilon}=\frac{1}{2\pi i}\oint \left[T(z)\varepsilon(z)dz+\tilde T(\bar z)\tilde\varepsilon(\bar z)d\bar z\right].$$ The integral is performed over a circle of fixed radius. The variation of a field $\Phi(z,\bar z)$ under a conformal transformation is then given by $$\tag{3.27}\delta_\varepsilon\Phi(z,\bar z)=[Q_\varepsilon,\Phi(z,\bar z)]\quad\text{and}\quad \delta_{\bar\varepsilon}\Phi(z,\bar z)=[Q_{\bar\varepsilon},\Phi(z,\bar z)].$$

I have a few questions about this passage.

  1. In what sense is $Q$ conserved? I can see $$\partial\bar\partial Q=0$$ Is this what they mean? (In QTF, I would call a charge conserved if $\dot Q=0$.)

  2. How does (3.26) even come about? In standard QFT I would write $$Q=\int d^3x\,J^0=\int d^3x\,\frac{\partial\mathcal{L}}{\partial\dot\Psi_a}\delta\Psi_a$$ which is conserved due to the Euler-Lagrange equations and where $J^\mu$ is the Noether current. I don't see how this applies to Wick rotated space.

  3. I image the derivation of (3.27) will clarify itself once 2. has been answered. In QFT I would write $$Q=\int d^3x\,\frac{\partial\mathcal{L}}{\partial\dot\Psi_a}\delta\Psi_a=\int d^3x\,\pi^a\delta\Psi_a$$ By means of the canonical commutation relation we have $$\delta\Psi_a=i[Q,\Psi_a]$$ Is this procedure correct for deriving (3.27)?

Any help greatly appreciated.

  • $\begingroup$ Have a look at Tong's string notes chapter 4. The first few sections answer all your questions and the whole book is very well written. $\endgroup$ – Heterotic Feb 10 '15 at 9:09
  • $\begingroup$ I've been reading those notes too. Is his (4.12), $$\delta\mathcal{O}(\sigma)=-\operatorname{Res}[\varepsilon(z)T(z)\mathcal{O} (\sigma)]$$ what we get when we combine (3.26) and (3.27) in BBS and adjust the contour? I believe it is. $\endgroup$ – Ryan Unger Feb 10 '15 at 11:18

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