1. In QFT, does the “sum over paths” of the Feynman formulation requires that all the possible world lines share the same proper time for a given process?


  1. Since one acknowledges the fact that this integration is performed in accordance to the underlying principle of extremal action, is there any constraint(s) for the allowed proper time interval(s) spawn by this sum?

  2. How the ensemble of encompassed world lines is related to the uncertainty (position-momentum or time-energy) principles (for instance, considering virtual exchange quanta being on or off mass shell)?

  • $\begingroup$ Do you set up your experiment in such a way that only paths with the same proper time count? In that case, yes. Imagine what happens to a classical wavefront when you add that requirement. Whether quantum or classical domain, it's a different experiment. $\endgroup$
    – CuriousOne
    Aug 14, 2016 at 4:55
  • $\begingroup$ @CuriousOne : Thanks for your comment. As it may appear, I am new to QFT. What do you mean by " Whether quantum or classical domain, it's a different experiment."? I am considering the quantum domain for a given experiment performed in a fixed stationnary lab. $\endgroup$
    – Wilj
    Aug 14, 2016 at 5:16
  • $\begingroup$ You have probably seen these videos: youtube.com/watch?v=1zmRulUaDk8, and then there is the issue that fixing the time will lead to an energy uncertainty. Whether we do this in QFT or classical theory, the effects are similar. $\endgroup$
    – CuriousOne
    Aug 14, 2016 at 5:25

1 Answer 1


In QFT, it isn't the paths of particles that were summed. Rather, it is the field configurations. So proper time is not relevant here. What happened for the usual amplitude computation in QFT is that one specifies field configuration at $t=-\infty$ and $t=\infty$ in a specific coordinate, and 'path integral' considers all possible fields with the specified boundary condition weighted by $e^{iS[\phi,J]}$.

  • $\begingroup$ What happens to energy-time uncertainty in this picture? $\endgroup$
    – CuriousOne
    Aug 14, 2016 at 5:30
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    $\begingroup$ It's not necessarily correct that one must sum over field configurations. Another way to derive Feynman diagrams is to literally consider particles propagating along trajectories. Sums over these trajectories give the propagators of the Feynman diagrams and these particles are allowed to split/merge at the vertices. So your excuse to avoid to consider the proper time isn't really right, and you avoided the propert time altogether with this excuse, so I don't think that it may be considered a right answer. $\endgroup$ Aug 14, 2016 at 8:52
  • $\begingroup$ @LubošMotl: could you perhaps elaborate on situation in QFT where one does not sum over field configurations? To me it seem contrary to the whole idea of a path integral in QFT not to have a sum (or integral) over field configurations. I can understand that there would be situations in a different context where one would have such trajectories, but that is not QFT. Or am I mistaken? $\endgroup$ Aug 14, 2016 at 12:01
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    $\begingroup$ @user110373: about the remark of Lubos, see for instance sns.ias.edu/sites/default/files/files/Witten-FeynmanString.pdf $\endgroup$
    – Trimok
    Aug 16, 2016 at 9:12
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    $\begingroup$ @user110373: No, it is about QFT , and about its wordline formalism (ie first quantized presentation) (linked to the Schwinger representation of the propagator), see chapter 2 of arxiv.org/pdf/hep-ph/9205205v1.pdf. See sns.ias.edu/ckfinder/userfiles/files/2015_Phys_Today.pdf. See ncatlab.org/nlab/show/worldline+formalism. $\endgroup$
    – Trimok
    Aug 18, 2016 at 8:32

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