I was reading through my notes on the path integral quantization of bosonic string theory when a general question about path integral quantization arised to me.

The widely used intuitive explanation of a path integral is that you sum over all paths from spacetime point $x$ to spacetime point $y$. The classical path has weight one (is this correct?), whereas the quantum paths are weighed by $\exp(iS)$, where $S$ is the action of the theory you are considering. In my current situation we have the Polyakov path integral: $$ Z = \int\mathcal{D}X\mathcal{D}g_{ab}e^{iS_p[X,g]}, $$ where $S_p$ is the Polyakov action. I have seen the derivation of the path integral by Matrix kernels in my introductory QFT course. A problem which occured to me is that if the quantum paths are really "weighted" by the $\exp(iS_p)$, it only makes sense if $\mathrm{Re}(S_p) = 0$ and $\mathrm{Im}(S_p)\neq0$. If this were not the case, the integral seems to be ill-defined (not convergent) and in the case of an oscillating exponential we cannot really talk about a weight factor right? Is this reasoning correct? Is the value of the Polyakov action purely imaginary for every field $X^{\mu}$ and $g_{ab}$?

Secondly, when one pushes through the calculation of the Polyakov path integral one obtains the partition function $$ \hat{Z} = \int \mathcal{D}X\mathcal{D}b\mathcal{D}ce^{i(S_p[X,g] + S_g[b,c])}, $$ where we have a ghost action and the Polyakov action. My professor now states that this a third quantized version of string theory (we have seen covariant and lightcone quantization).


  1. I am wondering where the quantization takes place? Does it happen at the beginning, when one postulates the path integral on the grounds of similarity to QFT? I am looking for a well-defined point, like the promotion to operators with commutation relations in the lightcone quantization.

  2. Finally, in a calculation of the Weyl anomaly and the critical dimension, the professor quantizes the ghost fields. This does not make sense to me. If the path integral is a quantization of string theory, why do we have to quantize the ghost fields afterwards again?

  • 1
    $\begingroup$ The first part of this question seems related to this question: physics.stackexchange.com/q/61139 @LubošMotl gave a nice answer there. Also, it can help to consider a Wick rotation so that the exponent is real and negative, like you naively expect, as mentioned in answer $\endgroup$
    – innisfree
    Commented May 23, 2013 at 18:18

2 Answers 2

  1. In Minkowski spacetime, the action has to be real. Btw, that's necessary for the classical limit to give principle of least action. Yes, such sums are ill-defined, so some might say that the theory is mathematically defined by analytically continuing (Wick rotating) to Euclidean time, where you have nice exponentially decaying weights. You'll get saddle points given by the extrema of the action and you can expand around those solutions and deal with the theory perturbatively.

  2. Think of the path integral in QM. Going from paths of least action (classical mechanics) to a weighted sum over all paths gives us quantum mechanics ("first quantization"). Going from quantum mechanics to QFT involves a sum over all field configurations ("second quantization"). Similarly, the moment you write out the path integral summing over all possible string configurations, you're studying a quantum mechanical system of strings.

  3. Even in QFT when you do the Fadeev-Popov method and introduce ghosts, you have to quantize them so that diagrams with ghosts consistently cancel amplitudes of (some) diagrams with longitudinal gluons. I'm at a loss for a more insightful answer.

  • $\begingroup$ Thanks for the clear answer! Just a small point concerning my third question: indeed it makes sense that if we consider an action of several fields, all of them have to be quantized. My confusion is that after using the path integral to calculate the partition function, only the $X$ and $g$ fields seem to be quantized, whereas the ghost fields have to be quantized separately. Why is this the case? $\endgroup$
    – Funzies
    Commented May 23, 2013 at 17:45
  • $\begingroup$ @Erik: I'm not satisfied with my answer for the third part. I've made an edit. There are examples where we couple a quantized field to classical background sources to get a sensible effective theory, but I guess you can't do anything like that in the QCD example since you want the contributions from the ghosts to cancel contributions from longitudinal gluons. I don't recall the details of string quantization now. I'll respond to your comment if I get time to look through my notes and jog my memory. $\endgroup$
    – Siva
    Commented May 23, 2013 at 21:29
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    $\begingroup$ You could leave the question open for a better answer, or maybe post your exact doubts (as expressed in the comment) as a separate string theory question. My answer was pretty generic, hardly referring to the details of string theory. $\endgroup$
    – Siva
    Commented May 23, 2013 at 21:31

I am afraid, I am just going to provide the standard lore here. I will do so nonetheless as not many have attempted answering this question yet. Allow me look at the questions from simplistic QM(not QFT) picture. I'll take the question in reverse order: When did we quantize? Well, we never have to, that is the beauty(?) of path integrals! You just pretend your p's and q's are real numbers locating the particle in phase space. In absence of "momentum dependent potential", after the path integral program(PIP) all you are left with is sum (superposition) over histories giving you some amplitude. In QFT the extension of this naive picture only works for bosons and you have to introduce complications (grassmann nos for fermions; additional ghosts for gauge fields etc) so that PIP works (i.e. you have some classical field with given properties which is summed over all possible configuration). All this lets you bypass quantisation, deriving appropriate Wick's theorem and Feynman diagrams etc. Now, what if action S is real? Indeed it is real in usual QM, the lore is that only neighborhood of classical path contributes to the total amplitude which becomes superposition of "amplitude" contribution from individual paths. Consider a set of chosen path which is not close to the classical trajectory. Since for them phase factor is not stationary they could in principle have widely different values from one another. One expects that when you sum these phase factors there will be "wild" phase cancellations associated with it.


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