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I was looking at this experiment by Veritasium. https://www.youtube.com/watch?v=a8FTr2qMutA&t=4s

Here he shines a green light through a narrow slit and we see the light smear as the slits becomes narrower. Momentum of a photon is $h/\lambda$. Since $h$ is a constant, shouldn't the uncertainty in momentum manifest as uncertainty in wavelength instead of position on the screen? So instead of the light smearing, shouldn't we see random colors on the screen instead of the smeared, steady, green?

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  • $\begingroup$ Where would the needed energy for the change in energy come from/go to? $\endgroup$ Commented Oct 3, 2022 at 4:57

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The smearing of the light in the intensity distribution behind a slit, is not a quantum effect. It has nothing to do with the Heisenberg uncertainty principle. It is a result of the modulation of the optical field by the transmission function of the slit, which determines the classical diffraction pattern of the light. This diffraction pattern is valid even though the light is monochromatic. In fact, if the light was not monochromatic, there would be a slight scaling in the diffraction pattern for the different wavelengths.

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Momentum of a photon is $h/\lambda$. Since $h$ is a constant, shouldn't the uncertainty in momentum manifest as uncertainty in wavelength instead of position on the screen?

If the experiment is with a monochromatic beam of light, the uncertainty in $\lambda$ is very small by construction.

This means that the spread in momentum of the photons that build up the classical wave is small. For individual photons the Heisenberg Uncertainty holds for measuring both momentum and position at the same time.

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements.

Italics mine.

As the momentum is known within an accuracy of the monochromaticity of the beam, the position can vary within the HUP. This is what is seen in single photon measurements.

you ask:

So instead of the light smearing, shouldn't we see random colors on the screen instead of the smeared, steady, green?

The single photons at a time on your retina are not measured to the accuracy required if you put down the numbers. Your retina is large and the accuracy of color determination in the brain is also non local.

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    $\begingroup$ @FlatterMann the exact solution for the probability distribution of the photons going through slits (see sps.ch/artikel/progresses/… ) has to use the boundary conditions: width of slits, separation of slits, distance of screen from slits etc. The HUP is an envelope fot the exact solution,. I do not understand about "slit changes energy of photon". There will be a width to the energy of the photons dependent on how monochromatic is the source, and some spread due to the exact calculation of the interaction"photon +slits". $\endgroup$
    – anna v
    Commented Oct 3, 2022 at 6:08
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    $\begingroup$ @FlatterMann have you studied quantum mechanics? Can you understand how photons make up light ( classical electromagnetic wave, not field)? $\endgroup$
    – anna v
    Commented Oct 3, 2022 at 13:45
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    $\begingroup$ @FlatterMann The answer is there and appears to be clear, and there is no mention of slit affecting energy/frequency. You wrote "I suggested nicely to change her answer if that was, indeed, what she meant because that would have been wrong." Why would one want to change an answer is there is no mention of it (slit affecting energy etc)? Answers are as written, and any additional assumptions made by the reader of the writer's intent, should be clarified in comments (which was done), and usually not a valid reason for any editing. Thanks. $\endgroup$
    – joseph h
    Commented Oct 3, 2022 at 23:41
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    $\begingroup$ "As the momentum is known within an accuracy of the monochromaticity of the beam" The magnitude of the momentum is known. Momentum is a vector (or the spatial portion of a 4-vector). The uncertainty of the HUP shows up in the uncertainty in the direction. $\endgroup$ Commented Oct 4, 2022 at 1:06
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    $\begingroup$ I don't see anything in this answer that suggests that a slit changes the energy of a photon. I'm not sure why anyone would understand it that way, and I find the suggestion that Anna should revise her answer to be entirely bizarre. $\endgroup$ Commented Oct 4, 2022 at 4:55
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Momentum of a photon is h/λ.

That is the magnitude of the momentum. Momentum is a vector (or, in relativity, the spatial portion of a 4-vector). It have both direction and a magnitude. h and $\lambda$ are both scalar, so they do not fully determine the momentum.

Since h is a constant, shouldn't the uncertainty in momentum manifest as uncertainty in wavelength instead of position on the screen?

Uncertainty in the magnitude of the momentum, and uncertainty in the direction of the momentum, are both consistent with the HUP. However, only the latter is produced by this particular set-up. Light is observed to travel in straight lines because the paths that aren't straight interfere with each other. However, that requires that one path have a path adjacent to it to interfere with it. An obstruction removes the path that would have produced interference, allowing some probability of the light beam bending. The narrower the slit, the more pronounced this phenomenon. There is no corresponding phenomenon that would produce a shift in wavelength.

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There are two different uncertainties at play here. Once is the uncertainty in the magnitude $|\mathbf{k}|$ of the momentum, which is related to the energy of the photons, and the other is the uncertainty in the direction $\mathbf{k}/|\mathbf{k}|$ of the momentum. The latter is more often phrased in terms of the transverse momentum $k_x$, and the related uncertainty principle is $\Delta x \Delta k_x \geq const$.

A plane wave of light has zero transverse momentum, and the uncertainty principle therefore implies that its spread in position is infinite. This is indeed the case. A Gaussian beam, on the other hand, has some finite extent $\Delta x$, and this in turn leads to an uncertainty in the transverse momentum that manifests itself as the divergence of the beam. However, note that the curved wavefront of the Gaussian beam still has a well defined magnitude of its momentum. The beam is simply a weighted superposition of plane waves in different directions, all with the same wavenumber.

The same explanation applies for the narrow slit, which decreases the position uncertainty and therefore increases the uncertainty in the transverse momentum, causing the light to spread out. The magnitude of the momentum vector, though, remains constant.

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