mhill
- 180
- 1
discrete spacetime means discrete momentum ??
the question is using De Broglie's Wavelength \lambda = h|p|^{-1}
then in case space is discrete would mean that there is a minimum possible wavelength in nature , for example \lambda = k l_{p} for Planck's length this would mean that the maximum allowed momentum modulus would be h(k l_{p})^{-1} for 'k' a constant , then since momentum is finite and spacetime is also finite there wouldn't be any IR or UV divergences but does this make sense
the question is using De Broglie's Wavelength \lambda = h|p|^{-1}
then in case space is discrete would mean that there is a minimum possible wavelength in nature , for example \lambda = k l_{p} for Planck's length this would mean that the maximum allowed momentum modulus would be h(k l_{p})^{-1} for 'k' a constant , then since momentum is finite and spacetime is also finite there wouldn't be any IR or UV divergences but does this make sense