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, 2015 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, 2015 at 11:18

1 Answer 1


Questions 1 and 2 (i.e. "whence (3.26)?")

$Q$ is conserved in the sense that it's a "surface" integral of a conserved current. This is analogous to Gauss's law, where the enclosed conserved charge inside a co-dimension $1$ surface is the surface integral of the electric current (up to some constants). It is similarly possible to interpret those integrals in eq. (3.26) as integrals over a one-dimensional "surface". To see this, think of $\epsilon(z)$ as the z-component of a vector in complex coordinates, i.e. $\epsilon(z)\equiv\epsilon^z(z)$. Similarly, think of $\tilde{\epsilon}(\bar{z})$ as the $\bar{z}$ component, i.e. $\epsilon^{\bar{z}}(\bar{z})$. Using the fact that a CFT has $T_{z\bar{z}}=0$, we can re-write $Q$ as \begin{equation} Q=\frac{1}{2\pi}\oint (-idz)\left(T_{zz}\epsilon^z+T_{z\bar{z}}\epsilon^{\bar{z}}\right)+\frac{1}{2\pi}\oint (-id\bar{z})\left(T_{\bar{z}z}\epsilon^z+T_{\bar{z}\bar{z}}\epsilon^{\bar{z}}\right) \equiv \frac{1}{2\pi}\left[ \int |dz|n^z J_z + \int |d\bar{z}|n^{\bar{z}} J_{\bar{z}}\right]\end{equation} The factor of $-i$ is placed next to the differential element to explicitly draw parallels with surface integrals in higher dimensions which are written is $\int dS n^\alpha J_\alpha$. When $dz$ is directed along the circular contour, $-idz$ is directed radially, i.e. along the normal $n^\alpha$.

Thus, following our initial discussion, this charge is conserved if $\partial^zJ_z+\partial^{\bar{z}}J_{\bar{z}}=0$. Using the fact that metric tensor in the complex space has only a non-zero off-diagonal component $g_{z\bar{z}}=1/2=g_{\bar{z}z}$, we can write $\partial^z=2\partial_{\bar{z}}\equiv 2\bar{\partial}$ and similarly $\partial^{\bar{z}}=2 \partial$. Thus we need to check if $\bar{\partial}J_z+\partial J_\bar{z}=0$. By substituting $J_z=T_{zz}(z)+\epsilon(z)$ and $J_\bar{z}=T_{\bar{z}\bar{z}}(\bar{z})+\tilde{\epsilon}(\bar{z}) $, we can easily check that this holds.

You can see a similar story written out explicitly for higher dimensions in David Simmons-Duffin's TASI lectures on bootstrap (sections 2.4 and 3).

Question 3

The commutator can be interpreted quite naturally in radial quantization. This interpretation is nicely spelled out in David Tong's string theory notes (section 4.5.2, for example). An operator with weight $(h,\tilde{h})$ has a specific transformation rule, as stated in Tong's eq. (4.16). This can be related to an OPE with the stress tensor, as discussed in section 4.2.3 of Tong's notes, \begin{align} T(z)\mathcal{O}(w,\bar{w}) &= \cdots + h \frac{\mathcal{O}(w,\bar{w})}{(z-w)^2}+\frac{\partial\mathcal{O}(w,\bar{w})}{z-w}+\cdots\\ \tilde{T}(\bar{z})\mathcal{O}(w,\bar{w}) &= \cdots + \tilde{h} \frac{\mathcal{O}(w,\bar{w})}{(\bar{z}-\bar{w})^2}+\frac{\partial\mathcal{O}(w,\bar{w})}{\bar{z}-\bar{w}}+\cdots \end{align}

These along with the above-mentioned interpretation of the commutators should be enough to derive eq. (3.27).

Your interpretation in terms of canonical commutation relations is actually equivalent to this discussion because radial quantization is just one way to quantize the theory by choosing the radial direction as the "time" direction. If you translate your interpretation into the path integral formalism, you should be able to see this more clearly.

A general discussion of quantization is given in Simmons-Duffin's section 2.2.


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