1
$\begingroup$

In the standard treatment of bosonic string theory the “heuristic” argument for the critical dimension goes as follows (see Ref. 1-4).

Upon quantization the mass-squared operator becomes normal ordered and an a priori unknown constant is added, just in case the normal ordered expression is not the true expression,

$$ \tag{1} M^2 = \sum_{n=1}^\infty \alpha_{-n} \cdot \alpha_n - a $$

(up to multiplicative constants).

(Note: $M^2$ has to differ only by a finite value from the normal ordered expression in order to be well defined on the Fock space vaccuum, so (1) makes sense.)

The next step is (GSW): “Let us try to calculate the normal-ordering constant $a$ directly. This normal-ordering constant arises from the formula”

$$ \tag{2.3.15} \frac{1}{2} \sum_{n=-\infty}^\infty \alpha_{-n} \cdot \alpha_n = \frac{1}{2} \sum_{n=-\infty}^\infty :\alpha_{-n} \cdot \alpha_n : + \frac{D-2}{2} \sum_{n=1}^\infty n $$

Then the LHS of (2.3.15) is suggested to be the “true“ $M^2$ so that a comparison yields $a = - \frac{D-2}{2} \sum_{n=1}^\infty n = \frac{D-2}{24}$.

Now my question: Why is the last step (equality of (1) and (2.3.15)) valid? We introduced $a$ because we don't know the true ordering, so why should the one in (2.3.15) be correct?

Even stronger: The note above implies that the LHS of (2.3.15) is no well defined operator – it really should not be the true form of (1).

References:

  1. Green, Schwarz, Witten: Superstring Theory, Vol. 1 (p. 96)
  2. D. Tong: Lectures on String Theory (p. 38 f.)
  3. J. Polchinski: String Theory, Vol. 1 (p. 22)
  4. Blumenhagen, Lüst, Theisen: Basic Concepts of String Theory (p. 44)
$\endgroup$
  • $\begingroup$ Equality (2.3.15) is just a consequence of the commutation relations between the oscillators. Now from Lorentz invariance, the mass of a string with excitation number 1 should be 0 (in light-cone gauge, for instance, you have only the transverse oscillators, and only a massless particle has $D-2$ degrees of freedom), so $a=1$. But if you apply the relation (2.3.15) on such a state, you see that the second sum is precisely $a$, because the first sum is 1 thanks to normal ordering. Hence the second sum equals 1 as well. $\endgroup$ – Antoine Mar 12 '15 at 13:56
  • 1
    $\begingroup$ This question (v1) seems to essentially asking Why is it legitimate to regularize the sum $\sum_{n=1}^\infty n=-\frac{1}{12}$? This particular sum is also discussed here, and on Math.SE here. See also this, this and this Phys.SE posts and links therein. $\endgroup$ – Qmechanic Mar 12 '15 at 14:14
  • $\begingroup$ @user40085: One cannot apply (2.3.15) on any state with finite excitation level because the LHS is not well defined (doesn't give a finite result). $\endgroup$ – Florian Oppermann Mar 12 '15 at 14:53
  • $\begingroup$ @Qmechanic: No, that's not my question. My question is, why it is justifiable to claim that the LHS of (2.3.15) equals $M^2$. Any way I can clarify my OP? $\endgroup$ – Florian Oppermann Mar 12 '15 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.