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The title is pretty straightforward. I was wondering if you can have light whose wavelength is not a rational number, but an irrational one. It seems to me there is nothing preventing this from happening, but I am not sure since I've never been exposed to such an instance! Is it possible for light to have a wavelength of, say 100*pi nm?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented Jan 15, 2017 at 4:48
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    $\begingroup$ I once posted an answer to a related question: physics.stackexchange.com/q/127752 $\endgroup$
    – David Z
    Commented Jan 15, 2017 at 4:49

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I agree with you, there is nothing preventing this from happening, not to mention that if it is rational for a certain unit, it could very well be irrational fro another unit (example new unit = $\pi$ meters). And as the choice of unit is arbitrary...

Addendum: for unit dependent quantities, one can chose units that make a given measure rational or not. But there are other quantities where we have no choice as for example $\pi$ or the proton-to-electron mass ratio (those are dimensionless constants).

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    $\begingroup$ But no matter the units, Plancks constant can't be chosen rational, right? $\endgroup$
    – docscience
    Commented Jan 14, 2017 at 21:44
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    $\begingroup$ Scaling can't or shouldn't be able to get you from rational to irrational. Should rather be decided from structural basis of the physical models $\endgroup$
    – docscience
    Commented Jan 14, 2017 at 21:47
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    $\begingroup$ @docscience : Plancks constant is expressed in J⋅s, what do you think happens if you change those units? For example taking "new Joule" = $6.626070040(81)×10^{−34}$ Joule? then $h=1$ new J.s. $\endgroup$
    – user130529
    Commented Jan 14, 2017 at 21:51
  • $\begingroup$ but what physical model determines the existence of PC? A rational interpretation would imply exact quantized quantities. Irrational, not. $\endgroup$
    – docscience
    Commented Jan 14, 2017 at 22:03
  • $\begingroup$ Sorry, I don't understand what you mean, I don't see the what quantization has to do with the choice of units. $\endgroup$
    – user130529
    Commented Jan 14, 2017 at 22:05
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Not really, because irrational numbers are a mathematical concept without an exact analogue in the physical world.

The wavelength of light is related to the momentum carried by each photon according to the de Broglie relation $p = h/\lambda$. And for all quantum-mechanical objects, the uncertainties in momentum $\sigma_p$ and in position $\sigma_x$ are related by the Heisenberg principle, $$ \sigma_p \sigma_x \geq \hbar/2.$$

A mathematical way to think about the Heisenberg Uncertainty Principle is to recognize that a pure, single-frequency sinusoidal wave has infinite spatial extent. If you want to confine your wave to some finite region of space, like "our galaxy" or "planet Earth" or "this laser cavity," your wave has not a single frequency but a distribution of frequencies which interfere constructively where your wave "is" and destructively where your wave "isn't."

Therefore if you have any information about the spatial extent of your wave, $\sigma_x < \infty$, you must have some corresponding uncertainty in its momentum and wavelength. A measurement of the wavelength will give you an estimate of the central value and an estimate of the uncertainty on that central value. In an ordinary measurement, you report the central value truncated to some rational number, where the method of truncation is based on the uncertainty. However the set of non-truncated numbers which are consistent with that central value, which you could have reported instead of that truncated central value without changing your meaning, is an infinite set which contains infinitely many rational and irrational numbers.

The importance of harmonics in describing wave dynamics means that there's a stronger case for describing waves whose wavelengths occur in rational ratios, but the same basic problem applies there. Rational and irrational ratios aren't really consistent with a world where precision is finite.

When my laboratory students measure that one thing is one-third the size of another thing, and record "$\text{ratio} = 0.\overline{333}$" in their lab notebooks, that's when we have this little discussion. There aren't repeating decimals in laboratory experiments, and there aren't nonterminating, nonrepeating decimals either.

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  • $\begingroup$ Indeed there is an uncertainty on the wavelength, but not on the mean value of its probability density which is 100% deterministic. The question could be reworded in that direction... FWIW $\endgroup$
    – user130529
    Commented Jan 14, 2017 at 20:40
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    $\begingroup$ @claudechuber Those of us in the precision measurements business have separate discussions about the width of a distribution ("uncertainty in $p$") and about the precision with which the central value is known ("uncertainty in $\left<p\right>$"). Whether there's a deterministic value for the parameters of a probability distribution "somewhere down there" is a philosophical discussion too complex for these comments, but I tend to think there isn't. $\endgroup$
    – rob
    Commented Jan 14, 2017 at 20:50
  • $\begingroup$ At first I would rather say theoretical rather than philosophical, but as we have no certitude on the theory itself, it may very well become philosophical, so I concede you the point. $\endgroup$
    – user130529
    Commented Jan 14, 2017 at 20:57

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