Since the frequencies (or inversely, wavelengths) of photons are part of a continuous realm, doesn't this mean that no photon has exactly the same frequency?

Two photons might have the same apparent wavelength due to our measurement limitations, but that doesn't mean they'd be exactly the same. In a continuous realm, how can anything be exactly the same? Also, there seems to be no uncertainty mechanism that limits the accuracy to which the frequency can be measured, so no help there either.

  • $\begingroup$ Just to be clear, you are asking if the fact that our measurements can't probe the wavelength to infinite precision and the fact that wavelength forms a continuous spectrum means conclusively that no two photons ever have the same wavelength? $\endgroup$ – Jim Mar 9 '15 at 16:17
  • $\begingroup$ @JimdalftheGrey Yes, if not conclusively, then at least it is something we should not assume otherwise. $\endgroup$ – Jiminion Mar 9 '15 at 16:23
  • $\begingroup$ "In a continuous realm, how can anything be exactly the same?" Why would the relation of equality depend on the size of the set it is defined on? I don't understand the question. Things are equal when they are equal. $\endgroup$ – ACuriousMind Mar 9 '15 at 16:24
  • $\begingroup$ @Jiminion There are circumstances where only photons of a particular energy are produced. Since energy=frequency, it stands to reason in those instances that the photons would have equal wavelengths. But apart from those instances, this is like saying that no two snowflakes are the same. Maybe most are just slightly different from each other, but I live in Canada and I know just how many snowflakes there are. Just because it's possible for them to be different doesn't mean every single one is different. I'd bet my life that at least 2 snowflakes have been identical, same with photons $\endgroup$ – Jim Mar 9 '15 at 16:28
  • $\begingroup$ @ACuriousMind, because if the values are continuous, that means they have to be equal to infinite precision. That's a pretty tall order. $\endgroup$ – Jiminion Mar 9 '15 at 16:45

The uncertainty principle limits our ability to determine the wavelength of a particle with infinite precision. At the same time, there is no fundamental reason why any two photons (even if generated by exactly the same process) should produce exactly the same wavelength; however, you can be sure that there will be plenty that are the same within the limits of nature's ability to measure them (note - not our limits, but those of nature).

Whether that is sufficient to say they are the same - or just "not observably different" is an area of philosophy that I don't feel qualified to enter.

  • 1
    $\begingroup$ Sure, you give up just because it's fundamentally impossible to find out any further. Quitter! $\endgroup$ – Jim Mar 9 '15 at 17:13
  • $\begingroup$ I didn't know there were any limits to nature's ability to measure their wavelengths. I don't see Heisenberg uncertainty applying in this case. $\endgroup$ – Jiminion Mar 9 '15 at 17:16
  • $\begingroup$ @Jiminion when are you doing the measurement? How long are you willing to take? Oh dear you have finite energy resolution... Even if you started at the birth of the universe. $\endgroup$ – Floris Mar 9 '15 at 17:18
  • $\begingroup$ Also, if that was the case (nature's limitation) then why couldn't the argument be made that their wavelengths ARE discrete at some point? $\endgroup$ – Jiminion Mar 9 '15 at 17:18
  • $\begingroup$ There is no finite energy resolution. Energy is just related to the frequency. $\endgroup$ – Jiminion Mar 9 '15 at 17:19

Surely you'd agree that all electrons and protons are $exactly$ the same (indistinguishable).

Now consider a regular Hydrogen-1 isotope: A proton and an electron bound together. There's definitely more than just one of these atoms in the observable universe.

Well then consider an electron floating around the $n=2$ shell of a Hydrogen-1 isotope, then falling down to the $n=1$ shell (this event will emit a photon) - I think you'd agree this has probably happened more than just once in the history of the universe.

Every time this has happened, a photon was emitted, with a particular wavelength (determined by some equation). Different experiments might yield different measurements of that photon's wavelength (due to limitations of measurements, like you mentioned). However, according to Quantum Mechanics all the photons emitted through the aforementioned process had exactly the same wavelength.

Maybe there's some (unbeknownst to me) QFT corrections which will alter the wavelengths for different settings in the above scenario, but I think you get what I'm getting at. There are certain processes which occur in the universe that are perfectly reproducible, which will emit photons of the exact same wavelength.

EDIT: As many have pointed out, I completely overlooked broadening of the line widths due to the uncertainty principle, so my example doesn't work at all.

  • $\begingroup$ All I know is that there are EFT higher-order corrections for everything. Think 1+1=2? It does plus some $O(\Lambda)$ correction terms that can be safely ignored $\endgroup$ – Jim Mar 9 '15 at 16:55
  • 1
    $\begingroup$ This is not really correct. The probability of emission decreases over time, as the $n=2$ population decreases, which means that the photon wavepacket has a sharp leading edge and then an exponential decay. It therefore does not have a well-defined wavelength, but rather a Lorentzian spectrum. $\endgroup$ – Emilio Pisanty Mar 9 '15 at 16:57
  • 1
    $\begingroup$ I agree that the std. model thinks electrons and protons are the same. They are at least very similar to a very great degree. And I agree that Hydrogens emit very similar photons. But identical to infinite precision? (as in an infinite number of digits past the decimal point) I dunno. That's a pretty tall order. $\endgroup$ – Jiminion Mar 9 '15 at 17:24
  • 1
    $\begingroup$ No, I am sorry, that's not true. The emission lines from atoms have some widths, and not due to measurement limitations, but because the resonant levels have widths. Only the ground level of an atom is of a fixed (no width) energy. $\endgroup$ – Sofia Mar 9 '15 at 21:09
  • $\begingroup$ @Jiminion just for you to know, emission lines have widths because these are resonant states. Otherwise, an excited state would never decay. In some of them the width is so small that for all practical purposes it may be neglected, but in in other cases the width is big, s.t. they are quasi-stable levels, with have big half-life. $\endgroup$ – Sofia Mar 9 '15 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.