I'm taking a course in string theory and have encountered renormalization for the first time (and I suspect it isn't the last).

Specifically, while quantizing the bosonic and spinning strings, an ordering ambiguity in the mode expansion gives us a constant (called 'a' in the notes I'm reading) which is a divergent sum. The constant is then fixed to 1 (at least for the bosonic string in light-cone quantization - don't remember the others at the moment) by some very convincing physical arguments reminiscent of renormalization in good ol' QFT.

My question is: why would we expect something like this to be necessary? Isn't the whole philosophy of renormalization that the necessity for adjustment of parameters comes out of the fact that your QFT is an effective low-energy approximation of some complete theory? If this complete theory is a string theory, then why do we still encounter the need for renormalization?

  • $\begingroup$ Related: physics.stackexchange.com/q/9340/2451 $\endgroup$
    – Qmechanic
    Oct 16, 2012 at 10:06
  • $\begingroup$ The normal ordering constant $a$ is equal to 1 due to the Super-Virasoro constraint. And as for the last question, don't worry, there's always Matrix string theory, and BFSS Matrix theory (also, known, as, M(atrix) theory,.,). $\endgroup$ Jun 24, 2013 at 10:26

2 Answers 2


There are two separate issues here:

Worldsheet versus Spacetime:

String theory is a generalization of quantum field theory, but perturbative string theory is formulated in a language pre-dating QFT. In Schwinger‘s proper time method, or “first” quantization, you construct a free field theory by summing over particle trajectories (worldlines) with an appropriate action. This summation is a path integral which formally is a “quantization” of the world-line theory. You then include interactions perturbatively by making the world-lines of those particle form more complicated graphs, which are nothing but the Feynman graphs of the corresponding field theory. This is a particularly cumbersome way to get the usual QFT perturbation theory. In this somewhat confusing language, the order in $\hbar$ is given by the topology of the graph, and has nothing to do with the process of “quantization“ of worldline theory.

(Sadly, physicists tend to call the process of solving any differential equation perturbatively “quantization”, even when it has nothing to do with QM, solving the Schrodinger equation for the time dependence of probability amplitudes, or any of that. Bad terminology is something one needs to get used to in this business.)

When generalizing to string theory, we don’t have the equivalent of the “second” quantized field theory. Instead we generalize worldlines to world-sheets, in other words point particles to strings. The sum over topologically trivial world-sheets gives classical, or free (i.e. leading order in $\hbar$) string theory, and quantum corrections (or interactions) come from including world-sheets of more complicated topology. In studying string perturbation theory it is important not to get spacetime and worksheet confused then.

In order to get free string theory you need to “first” quantize a two-dimensional field theory, which will give you the stringy generalization of a free field theory (with infinitely many fields). Free two dimensional field theory (corresponding to strings in flat spacetime) is complicated enough to require some simple sort of renormalization (e.g. normal ordering, or Polchinski’s “conformal” normal ordering). More complicated worldsheet theories (corresponding to strings in curved spacetime) require full-fledged renormalization, usually done perturbatively in the worldsheet theory as well. Note that in that case the expansion parameter is the string tension, which has nothing to do with $\hbar$, as you are still looking at classical strings which are sufficiently complicated in curved spacetime as to be usually only solvable perturbatively in $\alpha’$.

The ambiguity you noted (e.g. the normal ordering constant) is inherent in renormalization and usually one needs to impose additional physical constraints to fix these constants. In this case the ambiguity is fixed (when this is possible, e.g. with the right critical dimensional) by the requirement of preserving the symmetries of Weyl and diffeomorphism invariance of the worldsheet. Those symmetries are crucial for the spacetime interpretation of the worldsheet theory. It turns out that spacetime consistency conditions (absence of negative norm states) imply that the two dimensional field theory has to be conformal (or Weyl invariant, when formulated on curved worldsheet).

Renormalization in Spacetime

As the first part implies, this is not really relevant for this discussion. Nevertheless I want to comment that renormalization has nothing to do with the short-distance structure of the theory. The modern understanding is that it is efficient to organize QFT calculations by length (or momentum) scale, and that renormalization is the technical way to achieve that. For example, you’d want to renormalize your theory to facilitate calculations even when it is UV finite. Beyond that, probably a second semester‘s worth of QFT should clear this up.

  • $\begingroup$ Nice answer, @Moshe. $\endgroup$ Nov 3, 2011 at 9:57

First, let me say that it has nothing to do with renormalization as in "Renormalization Group". The problem is much simpler and concerns quantization of a classical theory.

Whenever you quantize a theory, the map $quantization:\{\text{classical observables}\}\rightarrow\{\text{quantum observables}\}$ is ill-defined. On the other hand, the map in the other direction is well-defined (if you keep track of $\hbar$), and a quantization consists of choosing some quantum observable (i.e. a preimage of the latter map) for the classical observable of interest.

The problem you mention is in the definition of the quantum $L_0$. Classically, it is $$L_0=\frac{1}{2}\alpha_0^2+\sum_{n=1}^\infty\alpha^i_{-n}\alpha^i_n.$$

Now, let's find one of the possible quantizations. This is (again) $$L_0=\frac{1}{2}\alpha_0^2+\sum_{n=1}^\infty\alpha^i_{-n}\alpha^i_n.$$

Note, that the second sum is secretly finite (that is why we like normal-ordering): all our states have a finite number of particles, so the operator is well-defined. Different preimages correspond to different choices of the operator ordering, so we can shift $L_0\rightarrow L_0-a$ for some (finite) $a$. If you want to keep track of $\hbar$ ($\alpha'$), then the relevant operator is $L_0-\alpha' a$ (possibly with a factor of 2 depending on the normalization).

Finally, you want your quantum theory to be nice (ghost-free), which puts some restrictions on $a$ and the dimension.

Edit: as Moshe points out, I've got my $\hbar$ and $\alpha'$ wrong. The correct quantization parameter here is $\alpha'$. See his answer for the distinction.

  • 1
    $\begingroup$ The dimension of the constant $a$ is given by the string tension, which plays the role analogous to $\hbar$ for the worldsheet theory, so $L_0 \rightarrow L_0 - a\,\alpha’$ really. The Planck constant itself does not make an appearance in tree level string theory. I also think that the OP is thinking correctly about this procedure as some simple form of renormalization. For more complicated string backgrounds one has to calculate the beta functions (and anomalous dimensions for operators) for the worldsheet theory, which is at the heart of renormalization. $\endgroup$
    – user566
    Nov 2, 2011 at 2:59
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    $\begingroup$ @Moshe, thanks for noticing the mistake! About renormalization: I guess I meant that ordering ambiguities are really about properly defining quantum operators. Scale dependence is somewhat secondary. $\endgroup$
    – Pavel Safronov
    Nov 2, 2011 at 3:34
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    $\begingroup$ Yeah, I guess this is right for free field theories, but for more complicated operators (in interacting field theories) the two things are related. Anyhow, probably semantics at this point. $\endgroup$
    – user566
    Nov 2, 2011 at 3:39

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