How to compute the levy path integral with zero potential?

Question

In quantum mechanics, if we have the quantum particle moving in the potential $V$ then the quantum-mechanical amplitude $K(x_b,t_b| x_a,t_a)$ can be written as

$$K(x_b,t_b|x_a,t_a)=\int_{x_{t_a}=x_a,x_{t_b}=x_b}Dx(t)\exp\left\{-\frac{i}{h}\int_{t_a}^{t_b}dtV(x(t))\right\}$$

where h is the Planck constant. It is known that in Feynman functional measure (generated by the process of the Brownian motion) and with zero potential ($V(x)=0$), the amplitude can be computed exactly, but what is the case with the non-Gaussian case? In the paper of Prof.Nikolai Laskin it said it can be computed with the measure generated by the $\alpha$ stable Levy motion ($1<\alpha<2$), but in this case the probability density function is so different from the Brownian case, so how to compute the amplitude?

Answer 1

The detailed calculations of a free particle propagator for the Levy case are presented: in the Sec 3.1 “Free particle” (pages 405-409) in the book: Fractional Dynamics Recent Advances https://doi.org/10.1142/8087 | October 2011, Pages: 532 Edited By: Joseph Klafter (Tel Aviv University, Israel), S C Lim (Multimedia University, Malaysia) and Ralf Metzler (Technische Universität Munchen, Germany) https://www.worldscientific.com/worldscibooks/10.1142/8087

or

in the Chapter 7 “A Free Particle Quantum Kernel” (pages 101 – 117) in the book: Fractional Quantum Mechanics https://doi.org/10.1142/10541 | July 2018, Pages: 360 By (author): Nick Laskin (TopQuark Inc., Canada) https://www.worldscientific.com/worldscibooks/10.1142/10541

Answer 2

It turns out that for the Levy path integral, the calculation of the amplitude of a quantum particle uses the Fourier translation of the probability density function, since this representation is an integral of an exponential function which is easy to compute.