Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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. – Qmechanic Aug 21 '13 at 21:06
@user26143 : I deleted the answer, there were several errors. I am trying to elaborate a correct answer... – Trimok Aug 23 '13 at 9:01
Thank you very much for letting me known! Anyway I was quite delighted by your answer (if not mistaken)! – user26143 Aug 23 '13 at 12:05
@user26143 : I gave a new answer. – Trimok Aug 23 '13 at 15:13
up vote 3 down vote accepted

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 $(-1)^2 = 1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.