$bc$ CFT Energy-momentum tensor from Noether's theorem

You got just the wrong coefficient in front of $\partial_z(bc)$, you expected twice as high coefficient due to $\lambda=2$, right? But that's because since the very beginning, your Noether derivation was completely insensitive to the value of $\lambda$, so you got a random value of $\lambda$ by this derivation.

Your derivation is not sensitive to the right value of $\lambda$ because the initial action may be changed by any multiple of $\partial_z(bc)$ which is a "total derivative", and similarly for the antiholomorphic terms. Such total derivatives additions don't change the equations of motion but they change the coefficientin front of the $\partial(bc)$ term in your Noether-derived stress-energy tensor.

It's not hard to see that already eqn 2.5.4 is sloppy about this issue: it only differentiates $c$ but not $b$, by choice, although these two fields apparently play the same role and get "exactly" interchanged when $\lambda\leftrightarrow 1-\lambda$.

There exists the "right" coefficient in front of $\partial(bc)$ that one should add in the action that yields the correct Noether stress-energy tensor, and this choice could be perhaps "justified" as being special. But this is a waste of time. Finally, you want to derive a quantum stress-energy tensor (there is normal ordering etc. in it) and the Noether procedure is only OK to guess the right "classical limit" of the tensor without all the quantum subtleties.

The natural coefficient in front of $\partial(bc)$ in the stress-energy tensor cannot be canonically guessed by any of the classical methods because this term is automatically conserved, anyway: note that $\partial_z\partial_{\bar z}$ acting on a holomorphic or antiholomorphic function vanishes. The Noether method is a method to calculate a conserved current, so if there are many, it doesn't necessarily pick the right one. The right addition must be determined by demanding the right OPEs or some other "fully quantum" condition.

It's an old question though, I would write down my calculation for my own reference.

According to the textbook, $b$ and $c$ fields transform as tensors with weights $(\lambda,0)$ and $(1-\lambda,0)$, so we can get the following infinitesimal variations of conformal transformation. $$ z'=z+\epsilon v(z)\\ \delta b\equiv b'(z)-b(z)=-\epsilon(v\partial b+\lambda\partial v b)\\ \delta c\equiv c'(z)-c(z)=-\epsilon(v\partial c+(1-\lambda)\partial v c) $$ Multiply $\delta b$ and $\delta c$ by $\rho(\sigma)$ and plug into the action, we get $$ -\delta S=\frac{\epsilon}{2\pi}\int d^2z\left(\lambda (\bar\partial\rho) v\partial (bc)-(\bar\partial\rho)v(\partial b)c-\lambda (\partial\rho) v\bar\partial (bc)+(\partial\rho)v(\bar\partial b)c\right). $$ Compare this with Eq. 2.3.4 in the textbook, and note that $d^2z=2d^2\sigma$, $j^z=2j_{\bar z}$, $j^{\bar z}=2j_z$, we can get $$ T(z)=\frac{j_z}{iv(z)}=(\partial b)c-\lambda\partial(bc)~~(\text{before normal ordering})\\ \tilde T(\bar z)=\frac{j_{\bar z}}{iv^*(\bar z)}=-(\bar\partial b)c+\lambda\bar\partial(bc)~~(\text{before normal ordering}) $$ Now put on normal ordering and note that $:bc:$ is purely holomorphic, we get Eq. 2.5.11.

There is still a tricky problem here. The above calculation only considers the variation of the action and forgets about the integral measure. Since $b$ and $c$ fields have a nontrivial scaling here, I would expect there be a nontrivial Jacobian, but I don't know how to compute it.

I am adding my own computation to the mix since it took me a while to follow the Noether method of deriving eq. (13.17) in Blumenhagen, Lüst, Theisen. The setup is the same as in Polchinski.

The fields $b$ and $c$ have conformal weight $\lambda$ and $1-\lambda$ and their action is given by

$$S = \frac{1}{2\pi} \int d^2z \, b \bar \partial c$$

Recall that under conformal transformations $z \rightarrow z+\epsilon(z)$

$$b(z) \rightarrow \left(1 + \frac{\partial \epsilon}{\partial z} \right)^\lambda b(z+\epsilon(z)) = b(z) + \lambda \partial\epsilon(z)b(z) + \epsilon(z)\partial b(z) + \mathcal{O}(\epsilon^2)$$

and similarly for $c(z)$. The way to derive the Noether currents is to pretend that the parameter $\epsilon(z)$ depends on the coordinates $z$ and $\bar z$. I believe this point to be a bit subtle since I was assuming $\bar \partial \epsilon = 0$ using the fact that it is a conformal transformation which should obviously not depend on $\bar z$. Anyway, allowing such behavior, the variation of $S$ then is

$$\begin{align} \delta S &= \frac{1}{2\pi} \int d^2z \,[\lambda \partial \epsilon b + \epsilon \partial b]\bar \partial c + b\bar\partial [(1-\lambda)\partial \epsilon c + \epsilon \partial c] \\ &= \frac{1}{2\pi} \int d^2z \, \left(\lambda \epsilon \partial(b\bar\partial c) + \epsilon \partial b \bar \partial c + (1-\lambda)(b\bar \partial \partial \epsilon c + b\partial \epsilon \bar \partial c) \\+ b \bar \partial \epsilon \partial c + \epsilon b \bar \partial \partial c \right) ~. \end{align}$$

It is not hard to check that all terms cancel after integrating by parts as usual except for the two terms containing barred derivatives of $\epsilon$. One is left with the following expression

$$\begin{align} \delta S &= \frac{1}{2\pi} \int d^2z \, (1-\lambda)b \bar \partial \partial \epsilon c + b \bar \partial \epsilon \partial c \\ &= \frac{1}{2\pi} \int d^2z \, \epsilon \bar \partial[(1-\lambda) \partial (bc) - (b\partial c)] \\ &= \frac{1}{2\pi} \int d^2z \, \epsilon \bar \partial[-\lambda b \partial c + (1-\lambda) \partial b c] \\ &\cong \frac{1}{2\pi} \int d^2z \, \epsilon(z)\partial_{\bar z} T(z)~, \end{align}$$

where $T(z)$ now has the desired form after decorating it with normal-ordering symbols:

$$ T(z) = - \lambda :b \partial c: + (1-\lambda) :(\partial b) c: $$