mhill
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discrete spacetime means discrete momentum ??
the question is using De Broglie's Wavelength [tex]\lambda = h|p|^{-1}[/tex]
then in case space is discrete would mean that there is a minimum possible wavelength in nature , for example [tex]\lambda = k l_{p}[/tex] for Planck's length this would mean that the maximum allowed momentum modulus would be [tex]h(k l_{p})^{-1}[/tex] 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 [tex]\lambda = h|p|^{-1}[/tex]
then in case space is discrete would mean that there is a minimum possible wavelength in nature , for example [tex]\lambda = k l_{p}[/tex] for Planck's length this would mean that the maximum allowed momentum modulus would be [tex]h(k l_{p})^{-1}[/tex] 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