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 $$ Z_{P}[J] = Z_{NG}[J] \; ? $$


The path integral involving the Nambu-Goto square root in the exponent is a very complex animal. Especially in the Minkowski signature, there is no totally universal method to define or calculate the path integrals with such general exponents.

So if you want to make sense out of such path integrals at all, you need to manipulate it in ways that are analogous to the transition from Nambu-Goto to Polyakov. The fact that these transitions are justified classically or algebraically is a reason to say that you are giving a reasonable definition to the Nambu-Goto path integral.

If you hypothetically had different values of the Nambu-Goto path integrals (and Green's functions), you could still try to perform the steps, the introduction of the additional $h_{ab}$ auxiliary metric, and the transformations to obtain the Polyakov form. So if there were some other value of the Nambu-Goto path integral, there would have to be a way to see it in the Polyakov variables, too.

But the Polyakov path integral is much more well-behaved (also renormalizable, anomaly-free in $D=26$ etc.), especially when you fix the world sheet metric $h_{ab}$ to a flat or similarly simple Ansatz. The Polyakov path integral is pretty much unambiguous and well-behaved which is why there can't be any other reasonable result coming out of it, and because of the relationship with the Nambu-Goto action, there can't be any other meaningful enough meaning of the Nambu-Goto path integral, either.

I think that instead of asking whether two well-defined objects are the same, the right attitude to this question is to admit that the Nambu-Goto path integral (or quantum theory based on it) is a priori ill-defined, a heuristic inspiration, and we're trying to construct a meaningful well-defined quantum theory out of this heuristic inspiration. And the transition to the Polyakov-like calculus isn't just an option, it's pretty much an unavoidable step in the construction of a quantum theory based on the Nambu-Goto heuristics.

  • $\begingroup$ I think you mean that Nambu-Goto action is the starting point to inspire us the Polyakov action, which is what we shall work on. But, path-integral by Nambu-Go action is still an integral, which can at least work out by numerical computation. $\endgroup$ – shuiruge Jul 23 '16 at 1:12
  • $\begingroup$ (continue) And, Nambu-Goto action is the most natural one that satisfies the symmetry of re-parameterization of world-sheet coordinates (and the most general? I cannot find another one), like Polyakov action the almost most general one that satisfies both symmetries of re-parameterzation and of Weyl-transform. Since they are both have such property, are they related with each other? This is the motivation of my question. $\endgroup$ – shuiruge Jul 23 '16 at 1:13
  • $\begingroup$ Dear Shuiruge, feel free to find another way to calculate the Nambu-Goto path integral so that the result would make sense and could be interpreted physically in any way. You will either figure out how to do it - which is basically equivalent to converting it to the Polyakov-style path - and get the same stringy/Polyakov answer or you will fail. ... The second part of your comment suggests that you haven't understood the equivalence of NG and P at the algebraic/classical level. They surely are formally equivalent. $\endgroup$ – Luboš Motl Jul 26 '16 at 5:29
  • $\begingroup$ To get from NG to P, one introduces a new auxiliary degree of freedom, the world sheet metric, but makes the action so that one may verify that the equations of motion determine it to be proportional to the induced metric up to a Weyl-scaled factor. So the Weyl-scaling scalar in the world sheet metric $h$ is arbitrary and there's a gauge Weyl symmetry; while the remaining two degrees of freedom in $h$ may be integrated out. When integrated out, one gets back to Nambu-Goto. That completes the proof of their equivalence at the classical level. $\endgroup$ – Luboš Motl Jul 26 '16 at 5:31
  • $\begingroup$ It is extremely problematic to "just calculate numerically" the NG path integral. First, in the Minkowski signature, such integrals can't be calculated by the numerical limit at all - one may prove rigorously that path integrals don't exist in the sense of measure. Second, in any signature, one needs to have some boundary conditions on the world sheet. The boundary conditions in "time equal to plus minus infinity" coming from the natural conditions in the Polyakov formalism seem to be the only sensible option. One simply needs to triviliaze the behavior of the path integral - Polyakov is how. $\endgroup$ – Luboš Motl Jul 26 '16 at 5:33

I) Recall that the path integral formulation comes in (at least) two versions: Lagrangian & Hamiltonian. It is often argued that the Hamiltonian version is more fundamental, cf. e.g. this Phys.SE post.

II) On one hand, the Dirac-Bergmann analysis/singular Legendre transformation yields that the Nambu-Goto (NG) Hamiltonian Lagrangian density reads

$${\cal L}_{NG,H} ~:=~ P\cdot \dot{X}-{\cal H}, \qquad {\cal H}~=~\lambda^{\alpha} \chi_{\alpha}, \qquad \alpha~\in~\{0,1\},\tag{1}$$

with two first class constraints

$$ \chi_0~:=~P\cdot X^{\prime}~\approx~0, \qquad \chi_1~:=~\frac{P^2}{2T_0}+\frac{T_0}{2}(X^{\prime})^2~\approx~0,\tag{2} $$

see e.g. this Phys.SE post. Here $\lambda^0$ and $\lambda^1$ are Lagrange multipliers for the first class constraints (2).

III) The original Nambu-Goto square root Lagrangian density

$$\begin{align} {\cal L}_{NG}~:=~&-T_0\sqrt{{\cal L}_{(1)}}, \cr {\cal L}_{(1)}~:=~&-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta} ~=~(\dot{X}\cdot X^{\prime})^2-\dot{X}^2(X^{\prime})^2~\geq~ 0, \end{align}\tag{3}$$

is re-obtained by integrating out the variables $P$, $\lambda^0$ and $\lambda^1$ in the Nambu-Goto Hamiltonian Lagrangian density (1) if we restrict the variable

$$\lambda^1~>~0 \tag{4} $$

to be positive. The ineq. (4) is needed in order to exclude an unphysical negative square root branch. Note that the ineq. (4) implies that the corresponding first-class constraint $\chi_1$ is technically speaking not enforced with a Dirac delta distribution in the path integral.

IV) On the other hand, the Polyakov (P) De Donder-Weyl (DDW) Lagrangian density reads

$$\begin{align}{\cal L}_{P,DDW}~=~&P^{\alpha} \cdot \partial_{\alpha}X +\frac{\gamma_{\alpha\beta}P^{\alpha}\cdot P^{\beta}}{2T_0\sqrt{-\gamma}} \cr ~=~& P^{\tau}\cdot \dot{X} +P^{\sigma}\cdot X^{\prime}+ \frac{(P^{\sigma} + \lambda^0 P^{\tau})^2}{2T_0 \lambda^1} - \frac{\lambda^1}{2T_0} (P^{\tau})^2, \end{align}\tag{5}$$

where we have introduced a world-sheet (WS) metric $\gamma_{\alpha\beta}$ and polymomenta $P^{\alpha}=(P^{\tau};P^{\sigma})$. See also e.g. my Phys.SE answer here.

V) In the second equality of eq. (5), we have defined two auxiliary variables, $\lambda^0$ and $\lambda^1$, as follows:

$$\begin{align} \lambda^0~=~&\frac{\gamma_{\tau\sigma}}{\gamma_{\sigma\sigma}} ~=~-\frac{\gamma^{\tau\sigma}}{\gamma^{\tau\tau}}, \cr \lambda^1~=~&\frac{\sqrt{-\gamma}}{\gamma_{\sigma\sigma}} ~=~ \frac{-1}{\sqrt{-\gamma}\gamma^{\tau\tau}}~\geq~0 \quad\Leftrightarrow\quad\gamma~:=~\det\left(\gamma_{\alpha\beta}\right)_{\alpha\beta}~=~-\left(\lambda^1\gamma_{\sigma\sigma}\right)^2~\leq~0 . \end{align} \tag{6}$$

The sign of the variable $\lambda^1$ must be positive in order to make the kinetic term in the Polyakov Lagrangian density positive definite. See also my Phys.SE answer here for more details.

Note the three WS metric components $\gamma_{\tau\tau}$, $\gamma_{\tau\sigma}$, and $\gamma_{\sigma\sigma}$ only enter the Polyakov De Donder-Weyl Lagrangian density (5) via the two combinations $\lambda^0$ and $\lambda^1$! (This neat fact is connected to Weyl symmetry.)

VI) To achieve the Polyakov Hamiltonian Lagrangian density ${\cal L}_{P,H}$ from the Polyakov De Donder-Weyl Lagrangian density (5), we should keep the momentum variable $P^{\tau}\equiv P$, and integrate out the auxiliary variable $P^{\sigma}$. It turns out that the Polyakov Hamiltonian Lagrangian density becomes precisely the Nambu-Goto Hamiltonian Lagrangian density (1)! Note that this off-shell conclusion is stronger than the usual claim that the Polyakov string and the Nambu-Goto string have the same classical EOMs on-shell!

Altogether, the only remaining difference is in the third degree of freedom of the WS metric, which corresponds to the Weyl symmetry, and whose path integral contribution naively factorizes and hence decouples. A more careful analysis reveals regularization issues for both the Polyakov string and the Nambu-Goto string, which potentially leads to a conformal anomaly. In a flat target space (TS), the conformal anomaly famously only vanishes in the critical dimension $D=26$.

TL;DR: In conclusion, since the Hamiltonian Lagrangian density of the Nambu-Goto string and the Polyakov string are identical, then any path integral quantization scheme (that is consistent with the Hamiltonian formulation) must be identical as well.



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.