I have a (maybe silly) question regarding the expression of the energy momentum tensor of the $bc$-system in equations $(2.5.11a)$ and $(2.5.11b)$ in Polchinski's String Theory, page 50. I know that this has already been discussed here: $bc$ CFT Energy-momentum tensor from Noether's theorem, but I really cannot understand if the result I got is correct or if I am missing something.

Let us start from the transformations of the fields b and c: under a conformal transformation $z'=z'(z)$, a primary field transforms as in equation $(2.4.15)$. Since $z$ and $z'$ are just labels for points, one can switch their names, which leads to \begin{align}\nonumber \mathcal{O}'(z,\bar z)&=\left(\frac{\partial z}{\partial z'}\right)^{-h}\left(\frac{\partial \bar z}{\partial \bar z'}\right)^{-\tilde h}\mathcal{O}(z',\bar z'), \\ \nonumber &=\left(\frac{\partial z'}{\partial z}\right)^{h}\left(\frac{\partial \bar z'}{\partial \bar z}\right)^{\tilde h}\mathcal{O}(z',\bar z'). \end{align} Let us apply it for the $bc$-system case, where $(h_b=\lambda, \tilde h_b=0)$ and $(h_c=1-\lambda, \tilde h_c=0)$. Let us start from the b field and we will just state the result for c. If $z'=z+\epsilon v(z)$ \begin{align}\nonumber b'(z)&=(1+\epsilon v(z))^\lambda b(z')\cong(1+\epsilon\lambda v(z))(b(z)+\partial b(z)\epsilon v(z))\cong b(z)+\epsilon \partial b(z) v(z)+\epsilon\lambda \partial v(z)b(z), \end{align} which implies \begin{align}\nonumber \delta b(z)=\epsilon\partial b(z) v(z)+\epsilon\lambda \partial v(z)b(z). \end{align} For the c field we have \begin{align}\nonumber \delta c(z)=\epsilon\partial c(z) v(z)+\epsilon(1-\lambda) \partial v(z)b(z). \end{align} Multiply now these transformation rules by a function $\rho(z,\bar z)$ and vary the action \begin{align}\nonumber \delta S&=\frac{1}{2\pi}\int d^2z(\epsilon\rho\partial b v +\epsilon\rho\lambda\partial v b)\bar\partial c+b\bar\partial(\epsilon(1-\lambda)\rho\partial v c+\epsilon\rho\partial c v),\\ \nonumber &=\frac{1}{2\pi}\int d^2z(\epsilon\rho\partial b v\bar\partial c+\underline{\epsilon\rho\lambda\partial v b\bar\partial c}+\epsilon(1-\lambda) b\bar\partial\rho\partial v c+\epsilon(1-\underline{\lambda})b\rho\partial v\bar\partial c+\epsilon b\bar\partial \rho\partial c v+\epsilon b\rho\bar\partial\partial c v), \\ \nonumber &=\frac{1}{2\pi}\int d^2z(-\epsilon\partial\rho b v\bar\partial c-\underline{\epsilon\rho b \partial v\bar\partial c}-\underline{\epsilon\rho b v\partial \bar\partial c}+\epsilon(1-\lambda) b\bar\partial\rho\partial v c+\underline{\epsilon b\rho\partial v\bar\partial c}+\epsilon b\bar\partial \rho\partial c v+\underline{\epsilon b\rho\bar\partial\partial c v}), \\ \nonumber &=\frac{1}{2\pi}\int d^2z(-\epsilon\partial\rho b v\bar\partial c-\epsilon(1-\lambda) \partial b\bar\partial\rho v c-\epsilon(1-\lambda) b\partial \bar\partial\rho v c-\epsilon(\underline{1}-\lambda) b\bar\partial\rho v \partial c+\underline{\epsilon b\bar\partial \rho\partial c v}), \\ \nonumber &=\frac{1}{2\pi}\int d^2z \epsilon\bar\partial\rho(-(1-\lambda)\partial bc v+\lambda b\partial cv) -\epsilon\partial \rho b v\bar\partial c-\epsilon b\partial\bar\partial \rho(1-\lambda) vc\\ \nonumber &=\frac{1}{2\pi}\int d^2z \epsilon\bar\partial\rho(-(1-\lambda)\partial bc v+\lambda b\partial cv) +\epsilon\partial \rho(\bar\partial b(1-\lambda)vc-\lambda b v \bar\partial c), \end{align} where underlined terms cancel. Then from equation $(2.3.4)$ on Polchinski's book I would get \begin{align} \label{eq1} T(z)=-(\partial b c-\lambda\partial(bc)),\\ \label{eq2} \tilde T(z)=-(\bar\partial b c-\lambda\bar\partial(bc)). \end{align} The first equation is exactly minus what Polchinski gets. Why is that so?

The second equation should be zero as in $(2.5.11b)$: the only way I can explain this is that $\rho(z,\bar z)$ should instead depend only on $\bar z$ or that I should apply the equations of motion $\bar\partial b=\bar\partial c=0$, but I don't really see why.

  • $\begingroup$ i don’t understand why you think it’s ok to replace z and z’ but not O and O’. z and z’ are not just variables, they label the frames in which the various fields are defined, so O’ is defined in z’ frame and not z as you have written. $\endgroup$ Aug 12, 2019 at 6:40
  • $\begingroup$ For what I understand z and z' are just names for points, right? It seems fair to me to decide to call them both in a different way. $\endgroup$
    – samario28
    Aug 12, 2019 at 6:49
  • $\begingroup$ i’m talking about the fact you wrote O’ instead of O in the transformation law. you can call the variables what you like but you have to be consistent. O’ is shorthand for $O^{(z’)}$, so when you change variables you need to take this into account $\endgroup$ Aug 12, 2019 at 6:54
  • $\begingroup$ I did that because if you don't, when you compute the variations of the fields, you have to Taylor expand b' and c', instead of b and c. Isn't O' just the transformation of O? You could decide for example to evaluate O' in z, being it a function.. Could you then show me how to obtain the variation of the fields starting from (2.4.15) in Polchinski's book? $\endgroup$
    – samario28
    Aug 12, 2019 at 7:00
  • 2
    $\begingroup$ Polchinski says that (2.5.7) satisfies the naive equations of motion, which means that $\bar\partial_1:b(z_1)c(z_2):$=0 for any $z_1$ and $z_2$ as a consequence of (2.5.6b) and (2.5.8). After your suggestions now $\tilde T(\bar z)=:\bar\partial bc(z):-\lambda\bar\partial(:bc(z):)$. The first term vanishes for (2.5.6a) while the second vanishes as a consequence of $\bar\partial_1:b(z_1)c(z_2):$=0. Do you agree? $\endgroup$
    – samario28
    Aug 12, 2019 at 8:59

1 Answer 1


Because the transformation of the field is induced from the coordinate change, the function $\rho(z,\bar{z})$ might have been attached to \begin{eqnarray} z'=z-\rho(z,\bar{z})\epsilon v(z). \end{eqnarray} Then there would be additional term induced for $\delta b$ and $\delta c$, and one can readily prove the undesirable term will be cancelled and $\tilde{T}(z)=0$.

Perhaps I am wrong and it is just a coincidence, so please point out my mistake.

Maybe I misunderstand the comments above, but if we concentrate on the form of the posted $\tilde{T}(z)$ above, it would be zero after normal ordering. Thus it means $\tilde{T}(z)=0$ in the sense of no other operator overlapping with it.


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