2

I am reading "A brief history of time" by Stephen Hawking. It is explaining the uncertainty principle:

... However, one will not be able to determine the position of the particle more accurately than the distance between the wave crest of light, so one needs to use light of a short wavelength in order to measure the position of the particle precisely.

Why is that? Why can't we determine the position of a particle accurately if the wavelength is long?

Hooman Bahreini
  • 127
  • 1
    To find out about this, look up 'diffraction' and the theory of the resolving power of a microscope. The basic idea is that when one uses light to form an image so as to precisely locate the source of the light, the image is never perfectly sharp, and it gets harder and harder to get a sharp image when the wavelength is long. – Andrew Steane Oct 30 '19 at 10:35

4 Answers4

2

It is analogous to the separation between the physical marks on a measuring tape or ruler. If you have a tape that is only marked to the nearest inch, you will not be able to make as precise a measurement as you might with a tape marked at 1/16th of an inch intervals.

Marco Ocram
  • 26,161
1

Let's say you are trying to measure the position of a electron with a photon. This is Compton scattering, and is inelastic scattering.

Compton scattering, discovered by Arthur Holly Compton, is the scattering of a photon by a charged particle, usually an electron. It results in a decrease in energy (increase in wavelength) of the photon (which may be an X-ray or gamma ray photon), called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron. Inverse Compton scattering occurs when a charged particle transfers part of its energy to a photon. The effect is significant because it demonstrates that light cannot be explained purely as a wave phenomenon. [3]

https://en.wikipedia.org/wiki/Compton_scattering

The point is, the photon needs to interact with the electron. This means that the photon needs to scatter off of an electron, and that needs an interaction between the particles.

Now I actually asked a question about what really oscillates as the photon propagates, and the answers say, that it is the wavefunction that oscillates as a Gaussian wave. The photon is the excitation of the photon field, and that excitation's wavefunction oscillates as a Gaussian wave.

The strange focus on "oscillation" in the question is also beside the point. The wavefunction that models a freely travelling particle is usually a Gaußian wavepacket. This moves, but it does not "oscillate".

Do photons oscillate or not?

enter image description here

Now whenever the photon and the electron interact, that means that the crest or the trough (in your case the quote says crest but it means the troughs too), the excitation itself of the field (which is the photon itself) oscillate so that they will interact with the electron.

Now if the wavelength is too long, the probabilities of the electron interacting with this excitation (the photon in the photon field) in the field is little, and the position of the electron becomes less certain.

Now you cannot tell the position of an electron by just shooting one photon at it, you need to repeat the experiment, and shoot photons and as per QM, some of the photons will scatter off the electron, thus giving some measurement of the position.

A crest is the point on a wave with the maximum value of upward displacement within a cycle. A crest is a point on a surface wave where the displacement of the medium is at a maximum. A trough is the opposite of a crest, so the minimum or lowest point in a cycle.

https://en.wikipedia.org/wiki/Crest_and_trough

The crest and trough are the maximum level (absolute deviation from the midpoint) of oscillation of the excitation in the photon field, thus giving the maximum probability for interaction, scattering.

As you lower the wavelength, the probability of the photons interacting (the excitation in the photon field to actually scatter off the electron) with the electron becomes higher, and the accuracy of the measurement (of position) becomes higher.

Árpád Szendrei
  • 28,452
  • Thanks a lot for your answer... so it means that it is more likely that a photon hits an electron at it's crest point? and the shorter the wavelength, the more frequent are the crests => better chance of collision with the electron? – Hooman Bahreini Nov 05 '19 at 08:59
  • 1
    @correct, those are the maximum (absolute values) of the excitation (the photon itself), which means a more precise position measurement (if those are more frequent, which means shorter wavelength). – Árpád Szendrei Nov 05 '19 at 15:41
0

The position accuracy (standard deviation of pulse time of arrival) of a radar range measurement is related to its rise (or fall) time of the measuring pulse. Note that $rise \, time \approx \frac{1}{B}$, where $B$ is the bandwidth of the pulse. In general, one has $\sigma_{\tau} \propto \frac{k_{\tau}}{B\sqrt{SNR}}$ where $k_{\tau}=O(1)$ and $SNR$ is the operational signal to noise ratio of the time of arrival measurement, $\sqrt{SNR}\approx \frac{pulse\, amplitude}{std(noise)}$. Higher carrier frequency allows easier wider bandwidth measurements, in practice, sharper and shorter pulses.

hyportnex
  • 18,748
0

In fact, it is possible to use an interferometer to measure position of a macroscopic or mesoscopic object much more accurately than one wavelength under the right circumstances, such as in atomic force microscopy (see the Wikipedia article). It is when the object being measured is extremely small & low mass (like an electron) that accuracy becomes limited by the uncertainty princiole.

S. McGrew
  • 24,774