# BRST quantization (Green, Schwarz, Witten)

In Green, Schwarz, Witten Volume 1, section 3.2, BRST quantization is introduced in a general way. A Lie algebra $G$ is defined with elements $$[K_i,K_j] = f_{ij}{}^k K_k \tag{3.2.1}$$ where $f_{ij}{}^k$ is the structure constant. Antighosts $b_i$ and ghosts $c^i$ transform in the adjoint and dual adjoint representations respectively. They obey $$\{c^i,b_j\} = \delta^i_j\tag{3.2.2}$$ The nilpotent BRST operator is $$Q = c^i K_i - \frac{1}{2}f_{ij}{}^kc^i c^j b_k \tag{3.2.4}$$ The indices here are all summed.

This is then applied to the Virasoro algebra without central charge $$[L_m,L_n] = (m-n)L_{m+n}$$ where $$L_m = L^{(\alpha)}_m + L^{(c)}_m - a\delta_m \tag{3.1.58}$$ The ghost contribution is $$L^{(c)}_m = \sum_{n=-\infty}^\infty(m-n)b_{m+n}c_{-n} \tag{3.1.49}$$ According to the book, the BRST operator is \begin{align*} Q &= \sum_{-\infty}^\infty L^{(\alpha)}_{-m}c_m - \frac{1}{2}\sum_{-\infty}^\infty (m-n):c_{-m} c_{-n} b_{m+n} : -ac_0 \tag{3.2.11}\\ &= \sum_{-\infty}^\infty :(L^{(\alpha)}_{-m} + \frac{1}{2}L^{(c)}_{-m} - a\delta_m)c_m: \tag{3.2.12} \end{align*}

It seems that the antighosts $b_i$ are now $b_m$, and the ghosts $c^i$ are now $c_{-m}$, so that $$\{c_m,b_n\}=\delta_{m+n}\tag{3.1.44}.$$ The non-zero structure constants are $f_{mn}^{(m+n)} = (m-n)$.

My question is: how is the normal ordered ghost term derived? I think the ghost contribution to the Virasoro operator should be normal ordered, since the ordering ambiguity is adsorbed into $a$. So, following equation $(3.2.4)$, \begin{align} Q &= \sum_{-\infty}^\infty c_{-m}L_m - \frac{1}{2}\sum_{-\infty}^{\infty} (m-n)c_{-m}c_{-n}b_{m+n} \\ &= \sum_{-\infty}^\infty L^{(\alpha)}_{-m} c_{m} + \sum_{-\infty}^\infty c_{-m}L^{(c)}_m - ac_0 - \frac{1}{2}\sum_{-\infty}^{\infty} (m-n)c_{-m}c_{-n}b_{m+n} \end{align}

Inserting $L^{(c)}_m$, it seems that \begin{align} \sum_{-\infty}^\infty (m-n) c_{-m}:b_{m+n}c_{-n}: - \frac{1}{2}\sum_{-\infty}^{\infty}(m-n)c_{-m}c_{-n}b_{m+n} &= -\frac{1}{2}\sum_{-\infty}^\infty (m-n) : c_{-m}c_{-n}b_{m+n}: \end{align}

I am unsure as to how to relate these. It seems that trying to combine the two terms will give a $-3/2$ since the operators are anticommuting, as well as a divergence sum. Perhaps equation $(3.1.49)$ is off by some factor? Above that equation it is written that $L_m = \frac{1}{\pi}\int^\pi_{-\pi} d\sigma e^{im\sigma} T_{++}$, and computing this I get $i$ times equation $(3.1.49)$.

Any help is appreciated, thanks.

1. Well, the string BRST operator (3.2.11-12) with an infinite-dimensional Lie algebra cannot be adapted from the BRST operator (3.2.4) for finite-dimensional Lie algebras as it stands, mostly because the second term in the latter does not use normal ordering. The issue is quite grave since the difference between the second term $$\tag{1} Q_2~=~-\frac{1}{2}\sum_{n,m\in\mathbb{Z}} (m-n) c_{-m}c_{-n}b_{m+n}$$ without normal ordering and the corresponding second term with normal ordering is infinite:

$$\tag{2} Q_2~-~:Q_2:~=~c_0 \sum_{n\in\mathbb{Z}} n c_{-n} b_n ~-~ :c_0 \sum_{n\in\mathbb{Z}} n c_{-n} b_n: ~=~ \infty~c_0.$$

1. The most powerful method is instead to formulate the BRST symmetry with the help of OPEs, radial ordering ${\cal R}$, and Wick's theorem between radial and normal ordering. See e.g. this and this Phys.SE posts and links therein. The BRST charge operator is $$\tag{3} Q~=~\oint \frac{dz}{2\pi i}~ j_{\rm BRST}(z),$$ where the BRST current operator $$\tag{4} j_{\rm BRST}(z)~=~ c(z)T^{(X)}(z) + \frac{1}{2} :b(z)c(z)\partial c(z): + \frac{3}{2} \partial^2 c(z)$$ is a primary field of conformal weight $h=1$, which satisfies the OPE $${\cal R}~ j_{\rm BRST}(z)~j_{\rm BRST}(w) ~=~ -\frac{c^{(X)}-18}{2(z-w)^3}c(w)\partial c(w)- \frac{c^{(X)}-18}{4(z-w)^2}c(w)\partial^2 c(w) - \frac{c^{(X)}-26}{12(z-w)}c(w)\partial^3 c(w) +\text{non-singular terms}.\tag{5}$$ Here $$\tag{6} c(z)~=~ \sum_{n\in\mathbb{Z}} c_n z^{-n+1}, \qquad b(z)~=~ \sum_{n\in\mathbb{Z}} b_n z^{-n-2}.$$ The BRST charge operator $Q$ is famously nilpotent $Q^2=0$ iff the matter central charge $c^{(X)}=26$ is twenty-six.

References:

1. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; Sections 3.1 & 3.2.

2. J. Polchinski, String Theory, Vol. 1, 1998; Section 4.3.

• 3. R. Blumenhagen, D. Lust and S. Theisen, Basic Concepts of String Theory, 2013; Section 5.2. – Qmechanic Mar 25 '16 at 0:01