What is the string equivalent of the Feynman Propagator?

The Feynman propagator for a massless point particle is proportional to:

$$\Delta(x-y;t_1-t_2)=\frac{1}{|x-y|^2-(t_1-t_2)^2}$$

which is, formerly, the Fourier transform of $$\dfrac{1}{|k|^2-E^2}$$.

For a bosonic string given by the coordinates $$X^\mu(\sigma)$$ and $$Y^\mu(\sigma)$$ with $$\sigma \in [-\pi , \pi]$$, what is the equivalent expression? I can only find formulas which give operator expressions but don't give it in terms of a functional $$\Delta[X,Y]$$. I feel like this functional should be able to be written out in a series of Fourier modes where the first term is the above expression.

My first thought is that it should be a simple product of propagators over all modes of the string corresponding to different energies.

• Just from guessing I would be looking for something like $e^{- \int \int \ln |X(\sigma)-Y(\sigma')| \partial X. \partial Y d\sigma d\sigma' }$ But it is probably much more complicated than that. Aug 7 '20 at 9:54

I will discuss the closed string propagator because this case is pictorially closer to the scalar propagator in quantum field theory case.

The closed string analog of the (two-leg amputated) line of propagation of a scalar field in a Feynman diagram is a cylinder of finite height $$s$$ and twist angle $$\theta$$. At this point you must notice the analogue with the scalar field propagator picture, just send the radii of the cylinder (immaterial parameter in a conformal field theory) close to zero.

Now perform a conformal transformation from the cylinder to the punctured at the origin unit disk and consider the operator $$e^{-sL_{o}^{+}}e^{i\theta L_{o}^{-}},$$ where $$L_{0}^{\pm} = L_{0} \pm \bar{L}_{0}.$$ That operator implements the conformal transformation $$z \rightarrow z$$ that shrinks and rotates by $$\theta$$ the unit disk. That effect precisely corresponds to the free propagation (with a twist) of a closed string at finite distances in target space. With this in mind the closed string propagator reads: $$b_{0}^{+}b_{0}^{-} \int_{0}^{\infty}ds \int_{0}^{2\pi} e^{-sL_{o}^{+}}e^{i\theta L_{o}^{-}}$$ Where the $$b$$-ghosts were inserted to ensure BRST invariance and integration limits were chosen to cover the entire moduli space.

To answer your second question. Where are the worldsheet embeeding functions $$X(\sigma)$$ in this expresion? The answer is that they are implicit in the propagator formula. Recall that the tree level dynamics of a closed string is specified by the only two moduli ($$s$$ and $$\theta$$) present in the propagator formula. Also the computation is performed from the worldsheet perspective were the actual formula must be reparametrization (indeed BRST) invariant. That's somewhat different from the field theoretical computation that only require Lorentz invariance.

References:

• My comments were extracted with minor comments from Four Lectures on Closed String Field Theory . Here you can check (page 17) a detailed comparison between the closed string propagator with ordinary field theory.
• Two-Point String Amplitudes offers a beautiful two-point tree level amplitude comparison between string theory (both open and closed) and field theory.
• Yes, I'm sure you are right. That is my problem, because I can see the expression which is implicit in X but from that I find it difficult to write an expressions explicit in X. To me, that is the difficult part. Aug 6 '20 at 21:12
• Well it would depend on the two end states of the string $X$ and $Y$. It is interesting. Maybe it can't exist. Aug 7 '20 at 0:01
• @zooby Explicit expressions can be obtained on e.g. flat backgrounds: there $L_0^\pm$ is quadratic in mode operators $\alpha_n^\mu$ (spacetime index $\mu$, oscillator number $n$, and each $\alpha_n^\mu$ is given in terms of the tachyon vertex operator and $\partial X$ (for $\alpha_n^\mu$) or $(\bar\partial X)$ (for the antiholomorphic oscillators $\bar \alpha_n^\mu$). I think it's clear that the zero modes $\alpha_0$ will give you a contribution that looks like the particle one (in a Schwinger integral writing: $(p^2+m^2)^{-1}=\int d\tau \exp(-\tau (p^2+m^2))$); best of luck with the rest.
– user21299
Aug 7 '20 at 2:04
• ... from the Siegel gauge propagator appearing in @RamiroHum-Sah's answer.
– user21299
Aug 7 '20 at 2:06
• I would have suggested discussing the open string case, because the twist parameter is purely stringy and has no counterpart in particle QFT. So the open string propagator is more directly related to what we see in usual QFT, at least in the operator language: in the momentum representation, the contribution from the twist effectively disappears in the mixing matrix from the numerator. I would suggest you to look at the draft of my book (in particular, sec. 14.2.2): lpthe.jussieu.fr/~erbin/files/reviews/book_string_theory.pdf Aug 7 '20 at 21:43