I've been looking at the literature on quantizing the bosonic string, and I noticed that there was a change made in the definition of the BRST current around 1992. However, I haven't found any illuminating discussion about why the change was made.

In Green-Schwarz-Witten (and other mathematics and physics literature from the 1980s, presumably originating from Kato-Ogawa), we have $j^{BRST} = cT^{(x)} + \frac{1}{2}:cT^{(bc)}:$, where $T^{(x)}$ and $T^{(bc)}$ are the stress-energy tensors of the matter representation and the bc-system, respectively, and $c$ is a ghost field. In more recent sources like Polchinski, one adds $\frac{3}{2}\partial^2 c$ to this. There was a brief remark in d'Hoker's IAS string theory notes (Lecture 7 page 14) that the extra term "ensures that the current is a $(1,0)$-form as a quantum operator".

In addition, there is a shift in the ghost number. In the older literature, the space of physical states has ghost number $-1/2$, i.e., it is given by the degree $-1/2$ BRST cohomology group. In the newer literature, the physical states have ghost number 1.

Now, the questions:

  1. What is the significance of the extra $\frac{3}{2}\partial^2 c$? If I'm not mistaken, one can add arbitrary total derivatives to the current, and still get a well-behaved BRST operator, although I must confess that I never actually worked out the symmetries of the matter-ghost action by myself. At any rate, there must be a reason to choose this particular total derivative. If it amounts to just d'Hoker's remark about the $(1,0)$-form, I'd be interested in some enlightenment about what it means for a quantum operator to be a $(1,0)$-form, and why that is important enough for everyone to switch conventions.

  2. How/why does the ghost number change?

Any pointers to papers/books with clear explanations or computations would be really appreciated.

  • $\begingroup$ Comment to the question(v1): Presumably much of the early literature did not mention the BRST current, but focused on the BRST charge, where the aforementioned term is invisible. As far as I can tell Green-Schwarz-Witten only briefly mention the BRST current in eq. (3.2.14) while ignoring normal ordering. $\endgroup$
    – Qmechanic
    Nov 6, 2012 at 22:48

1 Answer 1


As you already wrote, the $(3/2)\partial^2 c$ term is needed for the current to be a one-form i.e. $(1,0)$ tensor field; see also page 131 of Polchinski's String Theory, volume 1.

This means that if you compute the OPE $$ T(z) j^{BRST}(0)\sim \dots, $$ you want to get $$\dots \sim \frac{1}{z^2} j^{BRST}(0)+\frac{1}{z}\partial j^{BRST}(0), $$ see e.g. equation (4.3.11) in Polchinski's book. The cancellation of $c(0)/z^4$ singular term in the OPE requires the central charge of the total OPE to vanish (i.e. 26 dimensions for the bosonic string). However, the right coefficient of $\partial^2 c$ is also needed for this tensor relationship to hold. Why?

Just compute the OPE $$ T(z) \partial^2 c (0) $$ to see that $\partial^2 c(z)$ is actually not a $(1,0)$ tensor field: the OPE doesn't have the fully determined form of the $1/z^2$ and $1/z$ singular terms with nothing else. You will get some wrong term, a $\partial c(0)/z^3$ one with some numerical prefactor, I guess.

The reason why this $\partial^2 c$ is not a tensor field is somewhat analogous (although with the CFT flavor) to the fact that in GR, while $\partial_\mu S$ for a scalar $S$ is a tensor because the partial derivative is the same thing as the covariant one in this case, $\partial_\mu\partial_\nu S$ is no longer a tensor because now the covariant derivatives matter.

Because $\partial^2 c$ itself isn't a tensor field, it's clear that $j^{BRST}+K \partial^2 c$ may only be a tensor field at most for one correct value of the coefficient $K$ and a more complete calculation you are invited to do shows that the right constant $K$ is one given by the modern formulae.

At any rate, concerning your question 2., the total derivative doesn't change the total BRST charge, by Gauss' law. That's why the obsolete form of the current wasn't such a big mistake. But in modern times, we prefer to work with mass eigenstates or, in the case of the corresponding operators, with operators of well-defined dimensions, i.e. tensor fields, and the right term $(3/2)\partial^2 c$ is needed for that.

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    $\begingroup$ Thank you! Your answer is very helpful. Do you happen to know where the correction was first worked out? $\endgroup$ Jun 8, 2012 at 9:04
  • $\begingroup$ Thanks for your interest, Scott. No, unfortunately I don't know where it appeared first but it is really a straightforward 5-minute calculation. Let me also mention that the very appearance of some corrections of the form $\partial^2 c$ would be understood to be possible for decades - this expression may come from normal ordering of $\partial (bc)c$ terms because $bc$ may "cancel" (via their anticommutator which is a c-number) to another $\partial$. So this is just another case of a normal ordering ambiguity that may be chosen according to some more physical criteria - e.g. tensor character. $\endgroup$ Jun 8, 2012 at 9:18

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