It's a well known elementary fact that the Nambu-Goto action $$S_{NG} = T \int d \tau d \sigma \sqrt{ (\partial_{\tau} X^{\mu})^2 (\partial_{\sigma} X^{\mu})^2 - (\partial_{\sigma} X^{\mu} \partial_{\tau} X_{\mu})^2}$$ and Polyakov action $$S_{P} = - \frac{T}{2} \int d \tau d \sigma \sqrt{h} h^{a b} \eta_{\mu \nu} \partial_{a} X^{\mu} \partial_{b} X^{\nu}$$ are equivalent at the classical level. More precisely, by solving $\delta S_{P} / \delta h_{a b} = 0$ for $h_{ab}$ and plugging it back to $S_{P}$, we get $S_{NG}$.
However, my question is whether they are equivalent at the quantum level or not. That is, if let $$ Z_{P}[J] := \frac{\int D[h_{a b}] D[X^{\mu}] \exp(-i S_{P} [h_{a b}, X^{\mu}] + i \int d\tau d\sigma J_{\mu} X^{\mu})}{\int D[h_{a b}] D[X^{\mu}] \exp(-i S_{P} [h_{a b}, X^{\mu}])} $$ and $$ Z_{NG}[J] := \frac{\int D[X^{\mu}] \exp(-i S_{NG} [X^{\mu}] + i \int d\tau d\sigma J_{\mu} X^{\mu})}{\int D[X^{\mu}] \exp(-i S_{NG} [X^{\mu}])} \;, $$ do we also have $$ Z_{P}[J] = Z_{NG}[J] \; ? $$
No comments:
Post a Comment