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I came across a post on this in the community but wasn't satisfied with the answer,so i am making a new post. Why do we need short wavelength to measure the position of a particle accurately?

The above post has the green marked answer similar to the below description.

I saw in a video that to detect subatomic particles like electrons,we will need shorter wavelengths. And the reason they gave goes along the picture i attached. According to them, if the wavelength of the light is long,the wave the i.e red coloured curvy part won't hit the particle and so we can't see them. But if the wavelength is short,the probability of the curvy part hitting the particle will be more,in short they are saying if $E=hf$ is more then interaction will be more. But i don't understand how it works. What i learnt is

Light is the oscillation of electric and magnetic fields where the thing that oscillates is the fields and not any physical entity. And the light just moves along the straight line in $z$ axis,perpendicular to both the electric and magnetic fields. So,i can't understand how the red curvy part termed as wave is a physical thing. And if that red curvy part is not a physical thing,how can it hit the particle? Please shed some light whether the explanation given in the video is wrong or not,if they are right and my understanding is wrong,kindly enlighten me with the correct concept.

madness
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  • Are you talking about Compton scattering. https://www.nuclear-power.com/nuclear-power/reactor-physics/interaction-radiation-matter/interaction-gamma-radiation-matter/compton-scattering/ including links to the question or videos would help understand your question. The oscillating part is the frequency of the wave, higher energy photons have higher frequencies. A a lot of cases a charged particle responds to the fields of the photons. – UVphoton Oct 25 '22 at 14:05
  • Yes but my question doesn't focus on the experiment or the calculations actually. – madness Oct 25 '22 at 14:06
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    The Thomson cross section is independent of frequency (https://en.wikipedia.org/wiki/Thomson_scattering. – John Doty Oct 25 '22 at 16:30
  • Thanks for replying @JohnDoty. If we talk for general cases,how would you describe the above phenomenon for shorter wavelengths being able to measure accurately? – madness Oct 25 '22 at 16:33
  • https://en.wikipedia.org/wiki/Diffraction-limited_system#The_Abbe_diffraction_limit_for_a_microscope – John Doty Oct 25 '22 at 16:38
  • Thanks but please answer in accordance with my post. I needed to know the reason behind the phenomenon and also whether the explanation i gave above is correct or not. – madness Oct 25 '22 at 16:41
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    @madness I can't make sense of your explanation. In quantum theory, use wave models to calculate wave phenomena, particle models to calculate particle phenomena. The connection is that the squared wave amplitude is the probability of a particle showing up. – John Doty Oct 25 '22 at 16:51
  • @JohnDoty Actually the explanation says that if a light of small wavelength is passed,then the chances of the wave hitting the electron or particle will be higher(And by this it means the wave like curved line will hit the electron). But how is that even possible? The wavelike curved lines surely don't exist physically for them to hit or interact with the electron,the wave like curved line merely means the oscillation of electric field,so how can it interact with the electron? – madness Oct 25 '22 at 16:55
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    @madness The Thomson cross section is independent of wavelength, so the chance of the photon "hitting" the electron do not depend on wavelength (for gamma rays, the story is more complicated). – John Doty Oct 25 '22 at 17:07
  • @JohnDoty Thanks for replying but you are missing my point. In the post and the one i linked with my post,what was hitting the electron was the crest or troughs of the wave(according to them). Is this even possible? Those crests or trouths are merely denoting amplitude of electric field which do not exist physically to "hit" the electron. – madness Oct 25 '22 at 17:12
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    @madness In the wave picture, the wave makes the electron move back and forth, which causes it to radiate a wave. In the particle picture there is no wave. You cannot connect the two pictures together the way you are attempting. The squared amplitude of the wave is the probability that a particle will show up. Welcome to quantum theory: it doesn't work the way you think it should, and so cannot be explained in those terms. – John Doty Oct 25 '22 at 17:51
  • Could anyone please give an answer that corrects the explanation and clear my doubts as mentioned rather than just giving links to specific phenomena? – madness Oct 26 '22 at 06:54
  • The simple answer is that we don't need "short wavelengths". The position uncertainty of a quantum is simply the size of the detector. In general a quantum doesn't even have a wavelength. That's just a common misconception about quantum mechanics. – FlatterMann Oct 26 '22 at 07:07
  • @FlatterMann are the troughs and crests of light even physical to hit a particle? – madness Oct 26 '22 at 08:51
  • Specific phenomena are the foundations of physics: once you understand how the models were constructed to match the phenomena, you'll be able to understand the models. – John Doty Oct 26 '22 at 12:45
  • Classically, a plane electromagnetic are transverse. If the wave is vertically polarized, that means it has an electric field that oscillates up and down. Those are your troughs and crests. Classically, an electron exposed to these moves up and down, and radiates some of the incoming energy in other directions. – John Doty Oct 26 '22 at 12:54
  • Maxima and minima are emergent phenomena. Are they "physical"? Yes. Are they "precise" phenomena? No. – FlatterMann Oct 26 '22 at 18:31

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Shorter wavelength of photon in detecting a particle

To start with a single photon does not have a wavelength. It is a quantum mechanical entity, and in mainstream physics described as a point elementary particle of energy $hν$ where $h$ is Planck's constant $ν$ is that frequency a large number of photons of that energy, will have as classical electromagnetic radiation. A photon is not a bundle of classical electromagnetic waves, instead, classical waves develop from the underlying quantum field theory of particles. For quantum entities the wavelength appears in the distribution of many particles , depending on the quantum wavefunction of the particular setup

This can be made clear in the double slit experiment single photon at a time, where the footprint of the photon is seen on the screen, and the accumulation shows the classical interference of light.

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. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

You state

According to them, if the wavelength of the light is long,the wave the i.e red colored curvy part won't hit the particle and so we can't see them. But if the wavelength is short,the probability of the curvy part hitting the particle will be more,in short they are saying if E=hf is more then interaction will be more. But i don't understand how it works.

Translated at the quantum level for individual photon interactions, the probability of a low energy photon ( remember frequency is inversely proportional to wavelength, so a long wavelength means small frequency) to interact with a particle is very small, and will not be measurable, no accumulation as in the plot above. The shorter the wavelength the higher the energy of the photon, and then the plot can happen. The photons interact with the particles building up the walls of the slits.

So you are mixing up in your drawing the quantum ( what you draw as an electron particle is also a quantum mechanical point particle if you look at the table), and the classical electromagnetic radiation. the impinging light is a classical representation and is composed of at least hundred of thousand photons ( as seen the double slit experiment above). If one wants to talk of particles and photons one has to stick to the quantum picture, of photon interacting with electron.

You ask in a comment,

Could you please tell the reason behind,as you said,that low energy photons have a lower chance of interacting?

That is the way the mathematical functions that fit the quantum scattering electron-photon come out, and the theory fits the data very well.

If you know the Heisenberg uncertainty principle, the higher the momentum of two interacting particles the smaller the localization in space. High energy is connected with large momentum, so a high energy photon hitting an electron has a probability of finding it in a smaller volume.

These concepts need the serious study of quantum mechanics.

hup

anna v
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  • Thank you very much for replying.Actually the particle shown is not the quantum rather the electron,sorry for not clearing it.Could you please tell the reason behind,as you said,that low energy photons have a lower chance of interacting? That's the part i don't get.It will be very helpful if you kindly clear it out.Since in the diagram,it was claimed by those people that the electron will collide with the crests or troughs of the light more if shorter wavelength is used which makes no sense to me since those crests and troughs don't exist physically and are mere oscillations of electric field. – madness Oct 26 '22 at 14:09
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    Actually, Thomson scattering is wavelength-independent. However, in the long wavelength limit, a single photon doesn't impart enough energy to an electron to be observed. And, you get no interference of the sort in your figure when the wavelength is long relative to the apparatus. – John Doty Oct 26 '22 at 14:22
  • @JohnDoty but the author says the probability of interacting is low,you can see it the para where he confronts my doubt. Imparting low energy and not interacting,aren't they different? – madness Oct 26 '22 at 14:52
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    @madness The probability of an individual photon interacting with an individual electron is extremely small. The Thomson cross section is $\approx 6.66 10^{-29} m^2. You may think of that as the effective area of a photon to interaction with a point electron, or vice-versa. The Compton cross-section, applicable at pm wavelengths and shorter, is even smaller. – John Doty Oct 26 '22 at 18:03