I have a question about Eq. (4.3.3) in Polchinski's string theory book volume I, p. 131. It is said

Replacing the $X^{\mu}$ with a general matter CFT, the BRST transformation of the matter fields is a conformal transformation with $v(z)=c(z)$, while $T^m$ replaces $T^X$ in the transformation of $b$. Noether's theorem gives the BRST current $$ j_B = c T^m + \frac{1}{2} : cT^g : + \frac{3}{2} \partial^2 c, $$ $$ = c T^m + : bc \partial c : + \frac{3}{2} \partial^2 c, \tag{4.3.3}$$

My question is, what is the explicit expression of $T^m$?

According to this thesis, p 29, $$-\frac{1}{\alpha'}: c \partial X \cdot \partial X = :c T_X:$$

Suppose this expression is correct, I cannot use it to vertify Eq. (4.3.11) $$T(z) j_B(0) \sim \frac{ c^m - 26}{2z^4} c(0) + \frac{1}{z^2} j_B(0) + \frac{1}{z} \partial j_B(0) \tag{4.3.11}$$

if in (4.3.11), $T(z)= -\frac{1}{\alpha'} : \partial X^{\mu} \partial X_{\mu} : \tag{2.4.4}$ and I applied contraction Eq. (2.2.11).

  • 1
    $\begingroup$ Comment to the question (v2): $T^{(\rm m)}$ is the (bosonic) matter part, quadratic in $X$, as written in eq. (2.4.4), with Fourier coefficients called $L^{(\rm m)}_n$, cf. this Phys.SE answer. $\endgroup$
    – Qmechanic
    Commented Aug 21, 2013 at 21:06

1 Answer 1


I did not furnish all the details because it would be too long, but I give some hints at the end of the answer.

I have used the formulae $:T^g: ~= ~:2(\partial c) b + c(\partial b):$ and $:\frac{1}{2}cT^g: ~= ~:bc \partial c:$, when there is an ambiguity in the calculus.

We begin by : $$j_B = cT^m+:\frac{1}{2}:cT^g:+\frac{3}{2}\partial^2c=cT^m+:bc\partial c:+\frac{3}{2}\partial^2c \tag{4.3.3}$$

We have $T(z) = (T^m+ T^g)(z)$, we want to compute the OPE $T(z)j_B(0)$.

Note that $T^m$ has zero OPE with the ghost fields $c,b$ or $T^g$. Note that $c$ has holomorphic weight $-1$ and $\partial^2c$ has holomorphic weight $+1$

We have :

$$T(z)j_B(0) = T^m(z)c(0)T^m(0)+T^g(z)c(0)T^m(0) + T^g(z)c(0)T^g(0) \\+ T^g(z)\frac{3}{2}\partial^2c(0) \tag{1}$$ The first term is : $$T^m(z)c(0)T^m(0)\sim [\frac{c^m}{2z^4} + \frac{2}{z^2}T^m(0)+\frac{1}{z}\partial T^m(0)]~c(0) \tag{2}$$ The second term term is : $$T^g(z)c(0)T^m(0) \sim [\frac{-1}{z^2}c(0)+\frac{1}{z}\partial c(0)]~T^m(0) \tag{3}$$ The third term term is : $T^g(z)c(0)T^g(0) =:2(\partial c(z)) b(z) + c(z)(\partial b(z)): :b(0)c(0) \partial c(0):\tag{4}$

The part concerning one contraction is :

$$\frac{1}{z^2}:b(0)c(0)\partial c(0) :+ \frac{1}{z}:\partial(b(0)c(0)\partial c(0)):\tag{4a}$$

The part concerning 2 contractions is :

$$-\frac{4c(0)}{z^4}+\frac{3\partial c(0)}{z^3} \tag{4b}$$

The fourth term term is : $:T^g(z):\frac{3}{2}:\partial^2c(0)): = :2(\partial c) b + c(\partial b):\frac{3}{2}:\partial^2c(0):$, and this gives :

$$\frac{3}{2}[-\frac{6c(0)}{z^4}-\frac{2\partial c(0)}{z^3}+\frac{\partial^2c(0)}{z^2}+\frac{\partial^3c(0)}{z}]\tag{5}$$

Summing all the terms $(2), (3),(4a), (4b), (5)$, we get the desired result :

$$T(z) j_B(0) \sim \frac{ c^m - 26}{2z^4} c(0) + \frac{1}{z^2} j_B(0) + \frac{1}{z} \partial j_B(0) \tag{4.3.11}$$

Some hints :

The result $(5)$ is obtained by starting from :

$$:T^g(z)::c(w): = - \frac{1}{(z-w)^2} c(z) + \frac{2}{z-w} \partial c(z) \tag{6}$$ then deriving $2$ times relatively to $w$, and finally doing a Taylor expansion of $c(z), \partial c(z)$ around $w$, and finally putting $w=0$.

The results $4a$ and $4b$ are quite long and fastidious, you have to remember that, before doing one contraction or 2 contractions, you have to re-order the terms, and this may give a minus sign because of the anticommutation in the ordered product. For instance, if you have $:ab:~:cde:$, and you have a contraction $ac$ with a contraction $be$, you reorder by $acbed$, you have 2 transpositions, this will get a sign $(-1)^2 = 1$


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