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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:)).

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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.

References:

  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.)
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