Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So I'm following Szabo's book "An Introduction to String Theory and D-brane Dynamics (2nd ed, 2011); still on the canonical treatment in chapter 3. After doing a mode expansion, we get (up to a constant) a set of nice ladder operators \begin{equation}[\alpha_m^\mu, \alpha_n^\nu] = m\delta_{m+n}\eta^{\mu\nu}\end{equation} where $\delta_{m+n}$ is unity if $m=-n$, zero otherwise; $\eta$ is the Minkowski metric in D dimensions, $m$ and $n$ are integer labels and the $\alpha$s are Fourier coefficients. I'm omitting the LM/RM sets and concentrating on the open string.

We use these to define these composite operators: \begin{equation} L_n = \frac{1}{2}\sum\limits_{m=-\infty}^\infty{\alpha_{n-m}\cdot\alpha_m}\end{equation} which we want to use for generate conformal transformations. Cool.

So we need some Witt-like algebra, and because we're quantising we're expecting some weirdness, and it turns out you get the Virasoro algebra. But I just can't get the central term. For a start, it doesn't look right: each pair of $\alpha$s can give me a factor of m, so having \begin{equation}[L_m,L_n]\end{equation}, we have four $\alpha$s and hence we'd expect to be able to extract some central term of order $n^2$. But the Virasoro central term is $n(n^2-1)$ - so where does the extra n come from? There are no conveniently-placed sums that might give rise to this. Every time I attempt to plug through the algebra I end up with something silly, for example $\sum\limits_{p=-\infty}^\infty p$. So no central term at all!

So the root of my question is: where does the extra n come from? If I can understand that, hopefully I'll be able to do the rest of the derivation myself (don't take away my fun:)).

share|improve this question

1 Answer 1

The derivation of the Virasoro algebra is obviously a very important calculation in ST/CFT. For starters, OP (v1) does not mention explicitly the normal ordering ": :" in the definition of the Virasoro generators

$$ L_n ~=~ \frac{1}{2}\sum_{k\in\mathbb{Z}} :\alpha^I_{n-k}\alpha^I_k: $$

Normal ordering is important for $L_0$. Here the index $I$ runs over transversal dimensions. We recommend to divide the calculation into two cases: $n+m\neq 0$ and $n+m=0$. We will not further spoil the fun for OP, but just mention that a pedagogic derivation can e.g. be found in Ref. 1.


  1. B. Zwiebach, A first course in ST, 2nd edition, 2009, p. 256. (Note that the presentation in 2nd edition is improved as compared to 1st edition.)
share|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.