In the 2007 "String Theory and M-Theory" textbook by Becker, Becker, Schwartz there is the following claim about the canonical first quantization of a bosonic string: the quantization of the mode expansion operators (Virasoro operators) for a bosonic string is $$L_{m}=\frac{1}{2} \sum_{n=-\infty}^{\infty} :\alpha_{m-n}\cdot \alpha_{n}:$$ where $m\in \mathbb{Z}$, $:(\cdots):$ denotes the normal ordering of $(\cdots)$, meaning mode operators $\alpha_{k}$ with positive indices, which are canonical lowering operators for the number of excitations of a given mode, are moved to the right of operators with negative indices, which are canonical rising operators. The book then claims that:

  1. $L_{0}= \frac{1}{2}\alpha_{0}\alpha_{0}+\sum_{n=1}^{\infty} \alpha_{-n}\cdot \alpha_{n}$ is the only operator $L_{k}$ for which the normal ordering matters.

  2. An arbitrary constant could have appeared in this expression.

I don't understand these claims. Why would $L_{0}$ be the only operator for which the normal ordering matters? For example, $:\alpha_{-3+5}\cdot \alpha_{3}:$ is not the same as $\alpha_{-3+5}\cdot \alpha_{3}$. And where does the "arbitrary constant" come from? I am aware that differently ordered creation and annihilation operators differ by a constant, and that in QFT we get rid of this constant using the Wick theorem and declaring the constant difference insignificant; I don't understand why specifically in the case of the $L_{0}$ Virasoro operator we have to write down this constant explicitly, while in $L_{k}, k\ne 0$ the constant is not present.

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/10804/2451 $\endgroup$
    – Qmechanic
    Commented Mar 25 at 19:23
  • $\begingroup$ I was NOT asking about the meaning of the normal ordering, I asked specifically about where does the arbitrary constant come from in this specific case (and also why the ordering does not matter in all operators but one); this constant is not present in QFT (or, rather, we get rid of it but saying that its a constant infinity that does not matter), although in QFT we use the same operation, and I don't understand why here the arbitrary constant is present specifically in $L_{0}$ (and not in other terms). The linked question does not contain an answer to mine, only a reference to this problem. $\endgroup$ Commented Mar 25 at 20:12

1 Answer 1


Starting from the classical Poisson algebra of $L_m$ quantities (which form a Witt algebra), we would like to quantize it, i.e. define operators $\hat{L}_m$. Due to the operator ordering ambiguity, the definition of the operators $\hat{L}_m$ is ambiguous up to a term $a\hbar\delta_m^0\hat{\bf 1}$ proportional to the identity operator $\hat{\bf 1}$, where $a\in\mathbb{R}$ is the so-called intercept parameter $a$. (A monomial in the quadratic expression $L_m$ contains non-Poisson-commuting $\alpha_n$-modes iff $m=0$.) We can shift the meaning of the intercept parameter $a$ (by an infinite amount) by introducing normal ordering, but the ambiguity is still there in principle, to be fixed at a later stage in the quantization process. See also this and this related Phys.SE posts.

  • $\begingroup$ Yes, now, reading your comment with clear head, I can see that the original answer contained all that is needed to understand the answer to the first question, I kept missing that $\alpha_{i}$ commute in all other terms. However, I still don't quite understand why the constant can't be ignored here for the same reason it was in QFT. $\endgroup$ Commented Mar 26 at 17:21

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