7
$\begingroup$

In introductions to string theory it is common to start with the quantum theory of a single worldline by quantization of the Polyakov action for some $0$-brane.

What does this theory correspond to, exactly? I know it's not the theory of a single point particle, since those are not well defined for a variety of reasons, and from some arguments of ACuriousMind$^{[1]}$ it may not have a position operator defined. I also know that the full theory of worldlines can be considered as equivalent to QFT, with every worldline diagram corresponding to a Feynman diagram.

Would the theory of a single worldline correspond to the theory of a single QFT particle state because of this?

$ [1]$ private correspondance

$\endgroup$
  • 1
    $\begingroup$ I would help if you could shed some lite on that private correspondence of yours. I see no reason why you shouldn't identify 1-dimensional Polyakov action theory with the relativistic particle. Take a look at my path integral derivation of the Klein-Gordon propagator from the 1-dimensional Polyakov action: solenodonus.com/file/path-integral-for-point-particle.html $\endgroup$ – Prof. Legolasov Jun 6 '17 at 15:08
  • $\begingroup$ @SolenodonParadoxus I guess the issue is what one means by "quantum theory of the relativistic particle". I agree it has the KG propagator, but I believe Slereah's notion of a "particle theory" includes some way to localize said particle, i.e. some sort of position operator. I.e. the question here is more whether the quantum theory of the 1d Polyakov action corresponds to the delocalized QFT notion of a particle state or to something closer to a localizable particle state. $\endgroup$ – ACuriousMind Jun 6 '17 at 15:40
  • $\begingroup$ @ACuriousMind please elaborate. What does it mean to have a position operator? Doesn't passing to the Gelfand triple given by the Schwartz space on $R^4$ (together with its algebraic dual and $L_2[R^4]$) make the position operator well-defined? I can also promote the constraint to an operator acting on this algebraic dual space to give the reparametrization invariant dynamics. Or are we talking about something completely different? $\endgroup$ – Prof. Legolasov Jun 7 '17 at 2:01

Your Answer

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

Browse other questions tagged or ask your own question.