# Renormalization in string theory

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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?

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retagged Mar 7, 2014

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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.

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answered Nov 2, 2011 by (2,405 points)

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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.

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answered Nov 1, 2011 by (1,120 points)
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.

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@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.

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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.

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It is fun to look at those "explanations" from my point of view. Nobody understands that space-time we operate in mathematics does not exist in Physics without interactions. It is determined with interactions.

Nobody mentions either that we quantize ill-defined Classical Theories. Especially funny is "shitching on" the interaction (actually already present in the "non-perturbed" theory) as a "gauge" (or "compensating") fied. As soon as you postulate it (rather than thinkig in more physical directions), you will never get rid of renormalizations. The problem here is that we have not found a good mathematical description of interactions yet, we still stick to an awkward way, which is wrong as it needs subtractions of wrong parts, etc.

Renormalizations of "bare" constants is only one of many ways of obtaining good results from bad ones - by discarding the unnecessary terms to the initially good constants. There may be a more "generalized" way:  if one knows the good result from something else, one can subtract the wrong part from the calculation results: $$f_{\text{good}}=F_{\text{bad}}-(F_{\text{bad}}-f_{\text{good}}).$$But "doctoring numbers" or functions is not a calculation in a healthy theory, let's not fool ourselves.

answered Jul 7 by (132 points)
edited Jul 7

What is a healthy theory? ( sorry to emphase that ) "Nature speaks maths" is just a theory. Idem for "Nature is built with an unique scheme starting from ( what was ) one fixed point". We can just say, that starting from some {initial conditions}S, maths that are used in physics become providential. With only this minimalist assumption, one has no more reason to find astonishing that analytic fitting expressions have a limited domain of application. Linking (*) them by a semi group of transformations was a feat of the same order than the introduction of the Noether theorems in physics. Next steps will follow from some new Einstein/Feynman/etc/T'Hoft which will join ill linked domains to output a precursor to new theories valid for a century. Abusive current claims of completeness and fundamentality don't matter, it's just noise.

* it may happen that 2 errors lead to a consistent result. It is also a path to explore in these difficult papers...

@igael: Wrong, Nature does not speak at all. Physics is a domain of human activity; thus very subjective and never "complete".

It is indeed happening in renormalization that two errors lead to a consistent result. I wrote several popular papers to demonstrate this point, so error compensation cannot be excluded from the true reason of renormalization "success". Read, for example this one and you will see what I mean by a "healthy theory".

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