Why don't we use relativistic action $L=-~mc^2\sqrt{1-\frac{v^{2}}{c^{2}}}$ while calculating path integral for free particles?

Question

We know that while doing path integral for a free particle moving in one dimension, $\textrm{Kernel}\propto \exp\left\{\frac{im\pi x^{2}}{hT} \right\}$ , where $x$ is the change in the position in time $T$, $m$ is the mass of the body and $h$ is the Planck's constant.

$$\implies K\propto \exp\left\{ {\frac{i\pi mvx}{h}}\right\},$$ where $v=\frac{x}{T}$.

Now, according to de Broglie Equation, $\frac{mv}{h}=\frac{1}{\lambda}$

$$\implies K\propto \exp\left\{{\frac{i\pi x}{\lambda}}\right\}$$

Intuitively, this means that Kernel of a particle at a point is related to the phase of the wave associated with the particle at the point as it takes classical path to move around.

But, we know that more accurately, de Broglie Equation is given by, $\frac{mv\lambda}{\sqrt{1-\frac{v^2}{c^2}}}=h$

So, to retain the physical meaning of the Kernel, relativistic action for free particles $$L=-~mc^2\sqrt{1-\frac{v^{2}}{c^{2}}}$$ must be used to evaluate the path integral.

A practical reason why relativistic action is not used is that it will make any problem really complicated.

So, my question is - ideally speaking, is not the use of relativistic action correct and necessary while calculating the Kernel in Path Integral formulation of Quantum Mechanics?

Given that the answer to the question above is yes, I have few points to note. If we use relativistic action, then considering velocities of motion greater than that of light will become unnecessary because the action for such motion will become imaginary. This should somewhat relieve us from the obligation to consider paths with infinite velocities in the path integral.

But objection can be raised against neglecting velocities higher than that of light while doing path integral. How can a single photon interfere with itself in the double slit experiment if we do not consider one of the paths taken by it to have a velocity faster than that of light to catch up with the other shorter path? We can solve this problem by considering that it is the longer path in which the particle travels at speed of light, while it tends to linger around and waste a little time in the shorter path to compensate for the inequality in the lengths of the paths.

The proposal in the last paragraph can be proven true or false by studying the interference pattern of single photon double slit experiment in detail. Sadly, I could not find detailed description of interference pattern in any of such papers I came across.

Answer 1

I want to say that I appreciate that you came to the correct relativistic free-particle Lagrangian via quantum mechanics, but of course you do not need to come to it from that place.

Paraphrasing, I would say your general question is: "Suppose we have a double-slit experiment, an emitter $E$ fires a particle between two slits $S_1$, $S_2$ to some off-center detector $D$ closer to $S_1$. Specifically since it is off-center we want there to be a difference in the distances $|S_1 - D| < |S_2 - D|$. Now suppose that upstream of $S_{1,2}$ we somehow detect the precise time that the particle was emitted from $E$, so that we know its transit time via either path to the detector $D$, and we can perhaps shut off detection some time between (perhaps by simply shutting off the detector randomly and then postprocessing to find cases where the detector actually shut off between the two light cones, when only signals from one slit could possibly have reached it). Then: can't we filter the data to this narrow time window between when particles from slits 1 and 2 are detected? What would we see?"

Well I think that this experiment is intentionally difficult to reproduce. The theory answer is that indeed if you can filter out this very, very short time interval, this data should ideally show no interference pattern: indeed the wavefunctions of electrons are confined within the light cone and the (negative!) energies needed to ensure this confinement are the very reason that Dirac predicted positrons, so there you have at least one example of an experimental confirmation of the consequences of this idea.

Probably the best setup would be to embed a small positron emitter inside a block of normal matter inside a box at $E$, then check for the entangled photon emitted from the other side: the electron-positron annihilation photons must have nearly-opposite momentum in order to satisfy conservation of momentum, and so the directions are opposite and you know that the other photon went towards the detector. Then you also know the distance to $E$ from the emitter on the other side and you know that the photons both travelled at speed $c$, so you have everything you need for a precise time measurement of the emission.

However I think you run into problems with uncertainty here. It is sort of spatially necessary that the wavefunction for the photon cannot be confined within a distance of less than a few wavelengths; this is the position-momentum uncertainty relation (recall that the momentum precisely defines f and lambda, so we're talking about how purely monochromatic the light is with our momentum uncertainty). This presumably needs to factor into a short time-uncertainty to when the photon is likely to be absorbed, and this time-uncertainty would need to be of the same order as the gap between the transit times between $D$ and $S_1$ vs. $S_2$ because that gap is a projection of the distance between the two slits but you usually see this diffraction when the gap is roughly the size of a few wavelengths. As you spread out the slits you presumably make it harder and harder for the photon that goes from $E$ towards $S_{1,2}$ to hit both slits at once.

So I have the feeling that the double-slit interference pattern is only possible precisely because all of the uncertainties are so wide that you cannot normally do this experiment in practice, and any attempt to narrow those uncertainties to eliminate the interference pattern will necessarily have an alternative explanation in terms of the hoops you had to jump through in order to get there. Quantum mechanics is notoriously difficult to "pin down" like that.

Answer 2

The relativistic Lagrangian for a free particle is

$$L =\frac{mc^{2}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}~\approx~ mc^{2} + \frac{1}{2}mv^{2} + O\left(\frac{v^{4}}{c^{4}}\right)$$