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As we know, path integral considers the superposition of all the possible paths. That would include paths with velocity more than and equal to that of light.

Momentum of a particle travelling faster than light would not be easily defined because of relativity. And so will be its De Broglie wavelength. So, I am guessing that the path integrals only consider different 'paths', but assume each path to have the same wavelength, corresponding to the energy with which the particles are shot. Am I right?

My question basically is- do we take the change in wavelength into account when the momentum changes for different trajectories? You see, in double slit experiment, the fringe width is hugely determined by the wavelength of the interfering particles. But according to path integral, the momentum of every particle will be different, and so will be its de broglie wavelength. So, the fringewidth will be determined by a complex formula for the wavelength. Does this happens? Or instead, we consider each trajectory to have the same wavelength corresponding to the energy of the input particles?

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Doing path integral in relativistic system is somehow different than non-relativistic system. This is so because in non-relativistic mechanics everyone agree about simultaneous events, and can agree about the time slices on the definition of path integral. This is no longer true in relativistic mechanics. You can try to construct the action with the right symmetries for you particle (this is sufficient to fix an action of a system) and find something as

$$ S[X^{\mu}]=-m \int d{\tau}\sqrt{G^{\mu\nu}\frac{dX_{\mu}}{d\tau}\frac{dX_{\nu}}{d\tau}} $$

Where you need to sum over all maps $\tau\rightarrow X_{\mu}$ modulo diffeomorphisms in your path integral.

You can already see some differences. It is a non bi-linear action on $X$. It is not even polynomial in $X$. In the presence of those terms, the path integral is very hard to make sense. One strategy is define an auxiliary field over the domain of the map, a "metric tensor" $g_{00}$, such that the symmetries are preserved and the action bi-linear in $X$.

$$ S[X^{\mu};g_{00}]=\int d\tau \sqrt{g}(g^{00}\partial_0 X^{\mu}\partial_0 X^{\nu}G_{\mu\nu} -m^2) $$

Now, you need sum over all $X$ and $g$ modulo diffemorphism. From this you are going to find the Feynman propagators. And yes, they are non-zero for space-like interval. It is non-zero for simultaneous events too. This means that the states associated with the boundaries of the path integral, that behaves as states localized in space-time $|X^{\mu}\rangle$, are not orthogonal, and don't provide an orthogonal basis.

You can't think that a particle in the state $|x,t\rangle$ is not at the position $(y,t)$. The solution to this picture is interpret this states as prepared by events of creation of a particle or annihilation of an antiparticle, in an indistinguishable way. Immediately you can see that this preserves the causality. There is no particle faster than the speed of light. There is only a pair of particle and antiparticle "occurring" in some simultaneously surfaces of some inertial frame. A correlation, not a causation.

Now, you are no more summing over trajectories of particles, but over possible sequences of absorption and emission of particles and antiparticles. No particle are faster than light anymore, no imaginary momentum and no conceptual problem.

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  • $\begingroup$ But do we take the change in wavelength into account when the momentum changes for different trajectories? You see, in double slit experiment, the fringe width is hugely determined by the wavelength of the interfering particles. But according to path integral, the momentum of every particle will be different, and so will be its de broglie wavelength. So, the fringewidth will be determined by a complicated formula for the wavelength. Does this happens? Or do we instead consider each trajectory to have the same wavelength corresponding to the energy of the input particles? $\endgroup$ – Prem kumar Oct 24 '16 at 5:41
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    $\begingroup$ @Raja, there is no wave. What is similar to wave and is there is the wave-function. And you can make some analogy with the huygens principle. You can see more here. $\endgroup$ – Nogueira Oct 24 '16 at 15:31

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