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In this video: https://www.youtube.com/watch?v=xsnTrAEiyHg Prof. Walter Lewin showed when a laser beam passes through a very narrow slit the projection of it becomes wider. He claims it is because of the uncertainty principle in action: as we know the position of the light/photons very precisely the momentum becomes uncertain. That implies the direction of the light is no longer determined. But isn't such sense of direction ($p=m \bar{v}$) taken from classical mechanics?

In quantum mechanics the momentum is defined as $p=h/\lambda$. That implies if the momentum becomes uncertain we should get a spread in the wavelength. So instead of the light spreading out on the screen, should we expect some color change?

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    $\begingroup$ You seem to be misunderstanding the logical relationship between $p=mv$ and $p=h/\lambda$. The relation $p=mv$ is neither quantum-mechanical nor classical. It is simply a low-velocity approximation to a relativistic relation. The relation $p=h/\lambda$ does not replace $p=mv$. For example, both $p=mv$ and $p=h/\lambda$ hold for a nonrelativistic electron. $\endgroup$ – Ben Crowell Aug 26 at 23:47
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As far as I can tell, the type of momentum uncertainty he's talking about purely has to do with the direction of the momentum (it is a vector quantity after all). A change in wavelength (which has a corresponding change in absolute value of momentum) would violate conservation of energy at least in the case of a single photon (because $E = hf = hc/\lambda$; if I'm wrong please correct me).

Edit: This arguments needs more clarification. First, all non-ideal laser beams have an uncertainty in momentum and energy. In this case, the narrow slits increase the momentum uncertainty because the photons interact with the slit boundaries. However, there is nothing here that should increase the energy uncertainty, so there is no reason to invoke that.

Edit 2: There is a way color plays a role in diffraction: if you send white light, the colors will be separated by their wavelengths just like in refraction. This is because different colors are diffracted by different amounts by the Fraunhofer diffraction equation explained below. However, separating colors is not the same as changing them. If you shoot a monochromatic laser of red light, the output will be red just like the input.

When people talk about the uncertainty principle, it can actually refer to at least two different things. One thing has to do with the relationship between a function $f(x)$ and its Fourier transform $\mathcal{F}(f(x))$. The other thing has to do with the application of that relationship to a quantum wavefunction from which you get a physical interpretation (namely that there's an inherent limit to what you can measure simultaneously).

In this case, however, you don't need quantum mechanics per se (we don't need to think about this in terms of single photons). Instead of applying the mathematical relationship to a quantum system, we will apply it in a different, equally valid way.

To compute the full diffraction pattern (assuming the target screen is far away enough) of any opening arrangement, you need to use the Fraunhofer diffraction equation. It says the resulting amplitude graph on the screen will be a Fourier transform of the amplitude graph of the light at the slits.

For example, if there's one slit, the input amplitude is a just a box function. The Fourier transform of that is the sinc function, and squaring it will give you the intensity graph of the single-slit pattern.

The uncertainty principle basically says that when you "squeeze" a function horizontally, the Fourier transform of that function will get wider. Since single-slit pattern is the square of the Fourier transform of the input function (the box function in this case), making the slits narrower will make the output function wider and wider.

Although this can be explained classically, this is still a demonstration of the Heisenberg uncertainty principle, because this is a case where the mathematical relationship applies.

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  • $\begingroup$ Sorry I don't follow here- "For example, if there's one slit, the input intensity is a just a box function. The Fourier transform of that is the sinc function, and squaring it will give you the intensity graph of the single-slit pattern". Input intensity is equal to wave function squared. Therefore Fourier transform of it gives the distribution of the 'wave function squared' in the frequency domain. In other word, it gives the probability of finding a photon at a given frequency. As this is 'sinc function' I would expect original color at theta=0 degree, and a gradual color change with theta? $\endgroup$ – Alam Aug 27 at 20:39
  • $\begingroup$ Even about the conservation of energy the uncertainty is there as well. It can be violated in a time interval $hbar/t$. But the expectation value of the energy, i.e., the statistical average from all the photons must be conserved. Which means h times the integral over the sinc function is equal to <E>, and that is conserved. However that doesn't forbid few photons to acquire higher energy than <E>, leading to color change. Some others will have lower energy than <E>, again with some color change? Please correct me if I am wrong. $\endgroup$ – Alam Aug 27 at 21:07
  • $\begingroup$ @Alam A few things. (1) About the quote: I made a mistake; I meant to say input amplitude rather than intensity. See my revised post with the same quote. This is crucial, and I got the words mixed up. Absolutely sorry! (2) To clarify, I'm only taking a "cross-section" of a classical electromagnetic wave at the slit, plotting the amplitude there against the axis parallel to the slit. This means I'm graphing it to an axis that is perpendicular to the direction of travel. $\endgroup$ – SpiralRain Aug 28 at 2:05
  • $\begingroup$ @Alam (3) I'll try to revise my argument at the start by adding a few remarks. Basically, I don't see how the time uncertainty plays any role here. Of course, you're free to change the "accepted answer" if you want. $\endgroup$ – SpiralRain Aug 28 at 2:05
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I'm not an expert here but having watched the video (very good) I would guess the change in momentum is due to direction change only. Likely the momentum imparted is from the slit itself. So total momentum is conserved but the light has spread. Likely the momentum spread is larger in slit but after the slit it is conserved to the original. We do know wavelength changes changes during refraction as c changes and the frequency remains constant. Being in the slit is similar to refraction in that photons are required to interact with the material.

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  • $\begingroup$ "We do know wavelength changes changes during refraction but as c is constant the frequency also changes slightly." Sorry if I'm pestering too much, but doesn't the speed of light change in a refracting material? According to this link, the refracted speed of light is $c' = c/n$ where $n$ is the refractive index. $\endgroup$ – SpiralRain Aug 28 at 1:42
  • $\begingroup$ You are correct in one thing however: both diffraction and refraction separate different colors of light. That's not the same as changing the colors, however. $\endgroup$ – SpiralRain Aug 28 at 1:44
  • $\begingroup$ @SpiralRain yes you are correct, my mistake I'll edit it. $\endgroup$ – PhysicsDave Aug 28 at 2:10

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