2
$\begingroup$

In the context of quantization in string theory, the modern approach is the path integral/modern covariant quantization approach. As known from QFT, we fix our gauge and represent the arising Fadeev-Popov determinant through a Grassmann-integral with ghost variables $c_\beta$ and $b^{\alpha \beta}$ that are new fields of our theory, see for example the usual Tong.

In all cases of quantization, one finds that the modes of the energy-momentum tensor fulfill the Virasoro algebra (we neglect the normal ordering $a=1$ constant for simplicity) \begin{equation} [L^{(X)}_m,L^{(X)}_n] = (m-n) L^{(X)}_{m+n} + \frac{c^{(X)}}{12} m(m^2 - 1) \delta_{m+n,0}\,. \end{equation} The subscript will distinguish these generators from the ones we introduce in a moment. Namely, in modern covariant quantization one also gets an independent (slightly different) Virasoro algebra for generators $L_m^{(g)}$ built from the modes of our ghost fields, with corresponding central charge $c^{(g)}$. One then defines the total generator for both the Polyakov and the ghost action \begin{equation} L_m := L^{(X)}_m + L^{(g)}_m \end{equation} with central charge $c := c^{(X)} + c^{(g)}$. Now, as we have fixed our gauge, we should not have a Weyl anomaly, i.e. we demand that $c=0$ and as we can calculate $c^{(g)} = -26$, one gets the usual $c^{(X)} = 26$. The vanishing central charge then of course means that the total system does only transform under the Witt algebra.

We also demand that the other quantizations "produce" no Weyl anomaly, i.e. that our gauge redundancies are not broken through quantization, i.e. again $c=0$. My, probably rather simple, question now is how this was/is ensured in old covariant or light-cone quantization, without the use of ghosts, i.e. $c=c^{(X)}$. Do we necessarily need ghosts? The literature I have looked through always only demands a vanishing central charge when later discussing modern covariant quantization, but this may be due to a lack of research or understanding on my side.

$\endgroup$
1
  • $\begingroup$ +1. Same question was asked here, but I have never seen a satisfactory resolution in any source. $\endgroup$ Aug 6 at 9:51

1 Answer 1

3
$\begingroup$

In lightcone quantization the mass of a level $N$ state is given by $m^2 \sim ( N - \frac{D-2}{24} )$. The $N=1$ transforms in a rep of $SO(D-2)$. This is because in lightcone quantization there are $D-2$ oscillators $X^i$ ($X^+$ and $X^-$ are effectively gauged away).

Now, representations of the Poincare group are labelled by a mass $m^2$. If $m^2 > 0$, then they are further labelled by representation of the massive little group $SO(D-1)$. If $m^2 = 0$, then they are labelled by representations of $SO(D-2)$. Consequently, since the $N=1$ mode transforms in a rep of $SO(D-2)$ it must be massless. Therefore, $$ 1 - \frac{D-2}{24} = 0 \quad \implies \quad D = 26. $$ Once we deduce this, we must verify for consistency that all the modes with $N>1$ (which are now all massive) actually transform under reps of $SO(25)$ and not under $SO(24)$.

I don't remember the details here but if I recall correctly, in old covariant quantization, there are spurious states which have negative/zero norm. We must require that these states decouple completely from the rest of the Hilbert space so that we have positive norm. This requires that the spurious states have zero overlap with ALL states in the theory, $$ \langle \psi | \chi \rangle = 0 , \qquad |\psi\rangle \in \text{all}, \qquad | \chi \rangle \in \text{spurious}. $$ I think this condition gives $D=26$.

$\endgroup$
1
  • 2
    $\begingroup$ Maybe there is a misunderstanding from my side, you summarized how both quantizations derive that $c=26$ but my question was precisely how we now deal with the fact that $c = 26 \neq 0$, i.e. that we have a Weyl anomaly in our description. Modern Covariant Q. circumvents the fact that the $c=0$ CFTs are trivial by having a ghost theory with $c=-26$. How do old covariant and light-cone quantization deal with the Weyl anomaly? $\endgroup$
    – horropie
    Aug 6 at 13:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.