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In the usual QFT context, the conserved charge is defined to be $$Q=\int \mathrm{d}x^3j^0.$$

Under the radial quantization of the 2d Euclidean CFT, the conserved charge associated with $z\rightarrow z+\varepsilon(z)$ is usually defined as $$Q=\frac{1}{2\pi i}\oint\mathrm{d}z\ T(z)\varepsilon(z).$$

Therefore I would expect the integrand to be the "time" (or radial) component of the conserved current, but how can we think of $T(z)\varepsilon(z)$ as the "time" component of the conserved current?

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$J^0=J^z+J^{\bar{z}}$ (up to some factor of 2 or $\sqrt{2}$ -- edit: not quite correct, see comments for a more complete picture), but for a holomorphic current $J^{\bar{z}}=0$, so these indeed match.

Another way to see that this is a reasonable definition of charge is the fact that it doesn't depend on the contour, so it's constant in time, and generates the symmetry when commuted with the fields in the theory.

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  • $\begingroup$ Thanks! The second point is quite helpful, but what does J^0 mean? Is it the current in the cylindrical coordinates? (If so it's different from my result) $\endgroup$
    – WunderNatur
    Commented Jul 12, 2017 at 1:31
  • $\begingroup$ Hmm. Sorry, I may have missed some transformation factors going between the cylinder and the plane. Either way it should be a function of z so the logic is the same. What did you have? $\endgroup$
    – user121664
    Commented Jul 12, 2017 at 15:30
  • $\begingroup$ I got $J_0=zJ_z+\bar{z}J_\bar{z}$, where $J_z$ is holomorphic, $J_\bar{z}$ is antiholomorphic. Then I thought $Q=\int_0^{2\pi}d\sigma^1J_0$ should be a good starting point. I expressed $Q$ in terms of $z$, and there are correct terms showing up, but still some annoying extra terms. $\endgroup$
    – WunderNatur
    Commented Jul 12, 2017 at 15:43
  • $\begingroup$ The holomorphic and antiholomorphic parts are separately conserved for a cft, so you can turn the antiholomorphic terms off for the holomorphic current. Does that solve it? $\endgroup$
    – user121664
    Commented Jul 12, 2017 at 15:52
  • $\begingroup$ Thanks, but I am still not sure whether my understanding is correct. I think that the holomorphic part is conserved separately is a result of $J_z$ being holomorphic, which means $\partial^zJ_z=0$. Here are more details of my results: $d\sigma^1=\frac{1}{2i}(\frac{\mathrm{d}z}{z}+\frac{\mathrm{d}\bar{z}}{\bar{z}})$. Then $Q=\frac{1}{2i}\oint(\frac{\mathrm{d}z}{z}+\frac{\mathrm{d}\bar{z}}{\bar{z}})(zJ_z+\bar{z}J_\bar{z})$. Do you mean that when I consider the holomorphic part I can set $\mathrm{d}\bar{z}$ and $J_\bar{z}$ to be zero and $Q=\frac{1}{2i}\oint\mathrm{d}zJ_z$? $\endgroup$
    – WunderNatur
    Commented Jul 12, 2017 at 16:09

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