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Not trying to project the Planck length to have a deeper meaning, but according to current quantum theory, basically E=hc/lamda is the Planck length the smallest difference possible in wavelength between two photons with different wavelengths? So for example, if one photon has a wavelength of x, a different photons could have a wavelength of x+Lp but not x + .5 Lp?

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marked as duplicate by knzhou, Jon Custer, Dvij Mankad, Kyle Kanos, ZeroTheHero Mar 7 at 14:19

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  • $\begingroup$ I'm guessing not because you could doppler shift one photon by moving one source arbitrarily slowly - but it's only a guess ! $\endgroup$ – Martin Beckett Nov 26 '16 at 5:27
  • $\begingroup$ Isn't there a fundamental question of whether the degree of doppler shift would be continuous or quantized? Relativistically it would be continuous, but isn't quantum theory going to not allow arbitrary amounts of doppler shift? $\endgroup$ – Joseph Hirsch Nov 26 '16 at 5:34
  • $\begingroup$ I was wondering that. Time isn't known to be quantized so I assumed that speed wasn't. But I don't know so left it as a comment $\endgroup$ – Martin Beckett Nov 26 '16 at 5:36
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/9720/2451 , physics.stackexchange.com/q/169209/2451 and links therein. $\endgroup$ – Qmechanic Nov 26 '16 at 5:44
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No, the planck length is a blurring, not a grid

The planck length is approximately the smallest length for which distances can be measured. Higher accuracy measurements require more energy to perform (actually it's the center-of-mass energy that matters). When we reach the Planck length, our measurement attempts pack so much energy into such a tiny space we create a black hole.

What does this mean for the wavelength of a photon? Lack of measurability doesn't magically create a lower limit (or a quantization) on the photon's wavelength; there is no upper limit on it's energy. However, at shorter wavelengths a photon is more and more particle-like: It gets very hard to see wave-like behavior such as interference and diffraction. Directly measuring the wavelength to a precision to differentiate multiples of the Planck length would be extremly hard if not impossible.

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  • $\begingroup$ So would it be correct to say that "resolving" or measuring a photon to a precision smaller than the Plank scale would require enough localized energy to create a black hole? (so we would not be able to get the results of such a measurement back out?) $\endgroup$ – Joseph Hirsch Nov 26 '16 at 16:36
  • $\begingroup$ @Joseph: Yes it would if the center-of-mass frame has a smaller-than-planck-length photon hitting a particle. $\endgroup$ – Kevin Kostlan Nov 26 '16 at 17:27

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