This question relates to this link. But I still don't understand it >_<

In Polchinski's string theory vol I, p. 22, there is a divergence term (when $\epsilon \rightarrow 0$) in the zero point energy of open-string (1.3.34), $$ \frac{D-2}{2} \frac{ 2l p^+ \alpha'}{ \epsilon^2 \pi} .\tag{1.3.34}$$

It is said

The cutoff-dependent first term is propotional to the length $l$ of the string and can be canceled by a counterterm in the action proportional to $\int d^2 \sigma (-\gamma)^{1/2}$. In fact, Weyl invariance requires that it be cancelled.

I have a couple of questions related to this statement and I don't know whether it is good to split into several threads.

  1. About the underlying logic of regularization and renomalization. In quantum field theory, we meet a couple of divergences. One is in quantizing the scalar field. There is an infinity, we argue that is the zero-point energy. We throw it away since energy is a relative quantity. And we left the cosmological constant problem.

If the divergence in (1.3.34) is in this sense, string-theory is a quantum gravity theory. We cannot simply throw it away like QFT.

Later in QFT, we meet other divergences in the loop calculations. The reason is, the presented QFT is a low-energy theory. It has certain scale that the theory is not applicaple. Therefore we met the divergence. We regularalize it and renormalize it, where we met the terminology "counterterm".

But, string theory is regarded as a final theory(?!), if I follow the logic of QFT, why there is still divergence in (1.3.34)? What is the scale string theory is not applicable?

  1. How the counterterm works in $\int d^2 \sigma (-\gamma)^{1/2}$?

  2. Why "Weyl invariance requires that it be cancelled"?

  • 2
    $\begingroup$ For 2), all terms in the action have this form (see $1.2.13$ or $1.2.31$ for instance) For 3) the theory is conformal invariant, so scale invariant, you cannot accept that the theory depends on a scale $\frac{1}{\epsilon}$ $\endgroup$
    – Trimok
    Aug 17, 2013 at 17:15
  • $\begingroup$ Hi user26143, +1, and I'm posting this comment here just in case you didn't notice this comment by Dilaton yet, as you are not very active on TeX.SE (nor am I). $\endgroup$ Nov 11, 2013 at 15:47
  • 1
    $\begingroup$ @DIMension10, thank you very much! I have noticed and replied the comments from Dilaton! $\endgroup$
    – user26143
    Nov 11, 2013 at 15:52

1 Answer 1

  1. It is always legitimate to demand a symmetry, especially a gauge symmetry, because it may be a physical constraint defining/restricting a theory. In this case, the scale invariance or Weyl invariance or conformal invariance is enough to see that there only exists one value of the coefficient of the counterterm for which the symmetry is kept. It's the same value for which the counterterm cancels the divergent term in the ground state energy. In general 4D QFTs coupled to gravity, the scale invariance is clearly impossible due to the fields' nonzero masses etc., so we can't demand it, and there's no symmetry reason to determine that the cosmological constant is one thing or another.

  2. You wrote the right counterterm. It adds a multiple of the metric tensor (with the same coefficient) to the stress-energy tensor, like the extra C.C. term in Einstein's equations, and only the $T_{00}$ component matters here and its integral over $\sigma$ is what modifies the energies including the zero-point energy.

  3. Weyl invariance is needed because it's the part of the local symmetries that remove unphysical polarizations of the graviton on the world sheet etc. and this symmetry is needed for consistency. If the ground state energy depended on $\epsilon$, then it would fail to be Weyl-invariant because $\epsilon$ or $\ell$ scale under the Weyl scaling. Because the ground state energy or the dimension of the ground state operator is physically observable, it just cannot depend on dimensionful quantities because that would prove that the theory is not Weyl-invariant. When the counterterm is properly fine-tuned, the divergent term dependent on the dimensionful parameters is cancelled and the Weyl symmetry is restored - that's the point that is relevant for string theory.

  • $\begingroup$ 1. Does the scale invariance imply the beta function is zero. Therfore, the theory is capable in arbitary energy scale? Not like QED, there is Landau pole. $\endgroup$
    – user26143
    Feb 1, 2014 at 12:00
  • 1
    $\begingroup$ Yes, the scale invariance implies that the beta-function and all conceivable versions or generalizations of it are equal to zero. Yes, scale invariance means that the theory is equally well-defined at all scales, so in particular, it can't have the Landau pole or any other breakdown point. That's also why scale-invariant theories (RG fixed points, in another terminology) are so important to understand both the short-distance and long-distance limits of other theories, and as other limits, they may be extrapolated arbitrarily far. $\endgroup$ Feb 1, 2014 at 13:29
  • 1
    $\begingroup$ congrats on your gold badge for answering over 200 questions with the string-theory tag, with over 1k of points. $\endgroup$ Feb 1, 2014 at 16:32

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