4
$\begingroup$

I saw some videos where a person points a laser through a slit. As they reduce the width of the slit, the diffracted image spreads out, like this:

enter image description here

Can this pattern be viewed as a consequence of Heisenberg's uncertainty principle, applied to photons?

$\endgroup$
  • $\begingroup$ Electromagnetic fields are (nearly) perfectly linear and they do not have self-interaction in the vacuum, at least not at the amplitudes achievable without the latest generation of particle accelerators and lasers, and even with those theoretically available resource we have not actually built a facility that can explore the regime where the electromagnetic field becomes non-linear. The uncertainty principle is simply a consequence of the wave nature of these fields. These optical experiments, by the way, are purely classical, there is no quantum physics involved, at all. $\endgroup$ – CuriousOne Jul 23 '16 at 21:53
  • 2
    $\begingroup$ This is a perfectly ordinary phenomena of electromagnetic waves and doesn't require any quantum mechanics what-so-ever to describe. Indeed to describe the outcome of this experiment in quantum terms is less clear and more confusing up until you look at the path integral formulations (which is quite parallel to classical optics in how it addresses this experiment). $\endgroup$ – dmckee Jul 23 '16 at 22:03
  • 2
    $\begingroup$ But the fact that closing the slit more makes the point stretch out more is related to the Heisenberg's uncertainty principle, right? $\endgroup$ – Christian Jul 23 '16 at 22:15
  • 2
    $\begingroup$ Frankly the way the uncertainty principle is treated in pop-sci sources and most introductory texts can never be more than hand-wavy rule of thumb stuff. Treating it in terms of non-commutation of operators can give you the real answers, but then you have to do real calculation. The hand-waving stuff has it's place, but it doesn't replace the careful treatment. $\endgroup$ – dmckee Jul 23 '16 at 22:30
  • 1
    $\begingroup$ @dmckee: I agree, but the more far more egregious problem, in my mind, is when classical diffraction experiments are being presented as evidence for quantum mechanical behavior. That is plain false. Diffraction had been observed in detail for well over two centuries before Planck stumbled onto the correct explanation for the black body spectrum. $\endgroup$ – CuriousOne Jul 23 '16 at 22:43
6
$\begingroup$

Yes, light diffraction may be viewed both as a classical phenomenon and as a quantum mechanical consequence of Heisenberg's uncertainty principle. However, since both explanations work equally well, it doesn't provide any direct evidence for quantum mechanics.

Let me explain why the two explanations are equivalent. I'll do the classical uncertainty bound first.

You may have noticed that in far-field diffraction, the product of the width of the aperture and the size of the pattern on the screen is a constant: a slit half as wide makes a pattern twice as big. In fact, plugging in the formulas for any kind of diffraction whatsoever will give something like $$\sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim D \lambda$$ where $D$ is the distance to the screen. This is the "uncertainty principle" for classical diffraction.

Now let's view this on the quantum level. The position uncertainty is simply $$\sigma_x = \sigma_{\text{slit}}.$$ The momentum is given by the de Broglie relation $$p = \frac{h}{\lambda}$$ but we want the uncertainty in the $x$-component of the momentum, $p_x = p \sin \theta \approx p\theta$. Now, the uncertainty in the angle $\theta$ is just $\sigma_{\text{screen}}/D$ by trigonometry, so we have $$\sigma_{p_x} = p \sigma_{\theta} = \frac{h}{\lambda} \frac{\sigma_{\text{screen}}}{D}.$$ Putting this all together, $$\sigma_x \sigma_{p_x} = \frac{h}{\lambda D} \sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim h.$$ Up to a constant, this is the usual Heisenberg uncertainty principle; the two pictures are equivalent.


From a mathematical perspective, both uncertainty principles listed above special cases of a more general fact: the product of the width of a function $f$ and the width of its Fourier transform is bounded.

In the quantum case, the Fourier pair is position and momentum. In the classical far-field diffraction case, the pair is the screen and source, as proven here. This result also has applications in signal processing.

$\endgroup$
  • $\begingroup$ If we lower the intensity of the laser beam to the point that just single photons come out in a while, record where the individual photons reach and wait a long time, the profile of recieved photons resembles exactly the above picture, but in this limit we can't use classical electromagnetism. This limit of the experiment is only explained on the quantum level. Right? $\endgroup$ – seyed Jul 24 '16 at 18:19
  • $\begingroup$ @seyed Yes, but that's because you already put the quantum mechanics in by talking about single photons. If you didn't know photons existed already (say you were in the early 1900's), then diffraction by itself produces no evidence for their existence. $\endgroup$ – knzhou Jul 24 '16 at 18:34
  • $\begingroup$ Related question here. $\endgroup$ – knzhou Mar 27 at 16:31

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.