Diffraction with X-Rays is a very, very standard characterization technique to determine the structural parameters of materials. So in that sense, people do the "double-slit" or analogues of it all the time.

I just did a very short google search. People even did it with single-photon x-rays. You probably find more references within:

https://opg.optica.org/oe/fulltext.cfm?uri=oe-32-11-19294&id=550047

X-ray diffraction and using the wavelike nature of x-rays is a very common analytical tool. X-Rays have wavelength in the order of picometers, while typical lattice constants of crystals are on the order of a few hundred to a few thousand picometers. So, they are ideal to measure the distance of atoms, the orientation of crystal, the homogeneity of materials, etc. to very, very high precision.

This is relevant for semiconductors, metallurgy, chemistry, geology, etc.

By the way, diffraction with particles, like electrons, neutrons, etc. is also, very, very common technique. Depending on their energy they also have very short wavelength, but different scattering properties to Xrays. So, they also can provide similar information.

Starting point is probably Wikipedia:

https://en.wikipedia.org/wiki/X-ray_diffraction

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

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

This is very informative Thank you for the respective sources links

Michael Rüsing May I ask one more question to add to this discussion I just had a quick look at the references and realized that the double-slit for x-rays would be the crystal Right? This is the first part of my question The second is : Suppose we use a register screen or a register plate behind a double slit (which I assume is a crystal) Will the picture on the register screen look something like a reflection of a crystal lattice Pls see the attached picture

Thank you

Yes, in the case of x-ray, each atom (more precisely its electron cloud) acts as a "point-scatterer" for the x-rays. So, when using xrays for a typical XRD characterization task, the "slits" are the atom layers in a crystal latters. Keep in mind for x-rays of short wavelength, which are required for lattice investigations, the material is usually rather transparent, i.e. you cannot make as easily a "slit" based on a "shadow" in the right order of magnitude. To observe diffraction the size of the slits and their distance should be somewhat comparable to the wavelength. But even with short wavelength x-rays you could design a kind of double slit by making use of the Babinet principle. Here, the diffraction pattern of a "slit" and the inverse, i.e. a line of material, is the same. So you could, for example, arrange atoms in a long molecule and place two next to each other and you would get a diffraction pattern (almost) identical to a double-slit, but for extremely short wavelength.

I haven't checked the paper I linked and what wavelength of xrays they used and how they realized their "double-slit". But as mentioned above, there are ways.

If I get your questions right, the diffraction pattern from a single crystal will form a cloud of points on the detector, which distribution and pattern depend on the orientation of the crystal planes with respect to the incoming beam, the lattice constants of the crystal, the placement of the detector with respect to those axes and the wavelength of the x-rays.

This method is used to orient crystals before cutting, find their high symmetry planes, etc.

The method you are looking for is called "Laue method" after Max von Laue.

https://physicsopenlab.org/2018/01/18/laue-diffraction/

You get these beautiful point-pictures of crystals.

I just googled a random paper. Check for example here in Fig. 5:

https://www.researchgate.net/publication/341135204/figure/fig2/AS:887594229440512@1588630363656/Laue-photo-showing-an-x-ray-diffraction-pattern-for-CoSb-2-O-6-Typical-exposure-times.png

By the way, mathematically: The diffraction pattern is the Fourier-Transform of the lattice/spatial distribution of scatters.

A place to start:

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

Mathematical part is more or less clear

Michael Rüsing

One more point if I may Will there be time delays for the photons reaching the screen as in Laue Diffraction experiment Let's say if we 'bombard' a crystal with x rays Will the picture on the screen be changing with the time? The density of dots will increase decrease or remain the same? Thanks

Individual photons will reach/appear at the screen at different times. If you turn the intensity down to the single photon level then just single dots will appear at - what will appear as random locations. With time, though, they will form the diffraction pattern. You can imagine the pattern of points as a probability pattern. Therefore, with time probability will still result in the same diffraction pattern. Here, X-rays are not different from electrons or photons with visible wavelength.

If you illuminate the same crystal for long and watch the screen, the picture will stay the same - given the crystal does not change.

With prolonged exposure time the intensity of each peak will get stronger and stronger. Imagine a classical, analogue fotofilm, where over time points will just be overexposed the longer you illuminate. If your detector can manage the contrast, though, the picture will not be different.

What can happen, is that weak peaks that during short measurement times are lost in noise will eventually outgrow the noise level. But still, the overall picture will also be the same. No matter, if measuring for 1s, 1 min or hundreds of hours.

Assuming that the exposure is infinitely long Will it be just a monotonous black spot (dark spot) then or there will be white (contrast) areas within the spot provided that the conditions are as mentioned earlier?

Assuming that your detector or film cannot be overexposed, then the diffraction pattern of spots will stay the same.

Please note, you find both ddark films with bright spots indicating high intensity or bright film/detector, which turn dark in the exposed areas.

Ok Let me paraphrase the question If we move the detection screen at regular intervals of time (equal) for relatively long period and get hundreds or thousands of photos of patterns Will these photos differ in any shape or density etc. or will they remain identical provided that the intensity of emitted photons remains unchanged?

I will try to explain the reason behind my questions If we assume that the photon is a particle that has some energy and it bombards the screen it means that this energy should spread on the or inside the screen so the effects of hitting the screen will appear not only at the place of impact but elsewhere (for example round the spot or to the left and right of it and if the structure of the screen's surface or inner structure of the parts of the moving detector screen differs so will differ the pattern on it

That is the neat thing. In that moment, that the "wave" is measured, it acts as a particle.

If I shoot a single photon (X-ray, visible light, ...), an electron, a neutron or any other quantum mechanical particle at a double slit, it will diffract. As long as it is unobserved (undisturbed) it will act as a wave and "propagate" through both slits and form an interference pattern at the other side. But - this is important - the moment the single particle is measured, i.e. it hits the screen, it is a particle, because it cannot spread. Here, we will only see one dot appearing somewhere. The probability, where it hit, is given by the diffraction pattern, which is nothing else than its probability distribution. However, a diffraction pattern with peaks and dips etc. will only appear if you repeat this experiment many, many times more. But during each experimental run with just a single particle, just a single dot will appear at the screen each time.

it is nicely shown here:

https://en.wikipedia.org/wiki/Double-slit_experiment

See the chapter "Interference from individual particles" and the gif.

Or check this gif:

https://upload.wikimedia.org/wikipedia/commons/7/7d/Wave-particle_duality.gif

One the screen it will act "particle-like", because the energy is quantized. In the experimental setup and "in-flight" it will act wavelike. Quantum mechanics does not predict the result of one experiment, but just an ensemble/statistics of many experiments.

Let's just consider a particle as a spinning object hitting the screen In that instance there will be an interraction with the screen right It will impact the inner structure of it perhaps changing it Another point here whether the next photon will have to 'adjust' to a new environment impacting the so called 'hit' screen which will affect the pattern after the second shot For the reference if we compare a detector hit by two photons and two detectors hit by the photons time each and then compare the pattern by placing patterns on the screen one against another will they coinside or not? Thank you

For me as non-physicist it's hard two accept any duality of light I'd rather lean to spinning particle theory or even a medium in which light travels Agai in many cases indicating the double wave particle characteristics the commentors very rarely state what wave they have in mind transversal or longitudinal If the experiments were set in the space vacuum would the result be the same as the result for the light travelling in atmospheric gases etc. etc.?

I don't know, if I get you right:

"Let's just consider a particle as a spinning object hitting the screen In that instance there will be an interraction with the screen right It will impact the inner structure of it perhaps changing it"

Yes. The whole point of "detecting" means that something at the "detector is changing", e.g. an electron is excited in some CCD, a bond is broken or created in the chemicals of your photoplate or photofilm, an electron is excited in flourescent material will be excited and recombine, etc. How strong the reaction is and what exactly will happen, depends on the type of detector/screen, its internal energy levels (inner or outer photoeffect), as well as the energy of the hitting particle, as well as the conversion efficiency/quantum efficiency of the process. Here, an x-ray photon might create several charges because it carries a lot of energy in it, while the photoexcitation energy in the photoplate or CCD is only low.

Yes, depending on the design of the detector and all the given parameters, the spot that was hit before can be "burnt out/overexposed", i.e. it cannot detect a second hit at the same spot anymore. This is like overexposing a film on a camera. Everything will look white, but that doesn't mean there is not light anymore on the overexposed areas. A detector can only take so much, before it stops counting in that spot. But this is a matter of correct experimental design - not a quantum mechanical effect.

"Another point here whether the next photon will have to 'adjust' to a new environment impacting the so called 'hit' screen which will affect the pattern after the second shot"

But, for the diffraction pattern and the interference that doesn't matter. The "next" electron/x-ray/photon will not remember or avoid this "burnt out" spot. It can hit it again. It is just a matter of probability, if it hits the same spot. It might not be detected if it hits the same spot, because the response of the detector might be overexposed, but that doesnt stop the particle from hitting that spot. quantum mechanics with a single particle like this, is like rolling dice. If you role dice two times, the dice won't "know" what you rolled before. Each throw will be determined simply by the probability. You might get 10x a 6 in a row. You might not. You can calculate how probable that is. The same situation is with the double slit experiment - except that the probability where a hit is landing is given by the intensity distribution ("the interference pattern") at the detector. The point where the particle hits each time is purely governed by statistics - not by the previous runs or following runs of the experiment. This is a core point of quantum mechanics. It is statistics. You could run the experiment with 1000 photoplates, which you shoot each time with just one particle. Here, each photoplate would have some random single spot somewhere. If you add the 1000 photoplates together it would look similar (not the same because of statistics) then running the experiment a 1000 times on the same detector (assuming the detector is reacting linear and not in danger of being overexposed in some spots). Ofcourse, the pictures wouldn't 100% look identical, because it is just statistics. But the more often you run the experiment, the more it will approach the ideal statistical result.

Sounds off like some nuclear physics experiment Not that I do not appreciate this profound delving into the specifics The thing that puzzles me here is the statistics That would be explainable if it was a batch of photons/electrons hitting the screen How then a single photon deviates from it's trajectory compared to the previous shot remains a mystery Interaction with the medium Maybe? Change of initial conditions Imperfection of the equipment The closest I can imagine is the position on the detection screen proves various angles between the reference plane (crystal inner structure) and the trajectory of a photon e.g energy eg frequency of the incident ray or photon That means that the initial conditions are not identical for the two separate photons flying towards the reference plane(inner surface of the crystal) There will be two planes if I understand it correctly the register plate and the one that the beam reflects from (ie crystal) but for some reason the angle of the anticident beam will change Strange

Speaking of Satistics What is the probability that the incident ray of photons (or a single photon) is rebound or gets trapped by the obstacle (for example crystal) in the aforementioned experiment?

"...How then a single photon deviates from it's trajectory compared to the previous shot remains a mystery Interaction with the medium Maybe?.."

Imagine it like a dice roll. On your first roll there is a 1/6 change to throw a 1. Then the next roll of the dice will still be completely random. You still have a 1/6 change of throwing a 1. Each throw is completely independent of the previous or following experiment. The dice does not know, what was thrown before. It is random. Yes: In statistics if I repeat the dice-rolling 100x, I would expect roughly 16x each number. But that does not tell me anything about each individual dice roll. There is no "magical balance" in the universe that if you roll 4x a 6 in a row, there is a lesser change on rolling a 6 again. It is still 1/6. However, if we look at 5 consecutive roles, the total probability of getting 5x a 6 in a row is (1/6)^5. So to get a reasonable change to get 5x a 6 in a row you need to perform something like 1/(1/5)^5 experiments.

That is the same with the double-slit experiment. each experimental run is independent of the previous run. There is a finite change of hitting the same spot again, as much there is a finite change of rolling two times the same number again. It is just not probably, because in the next experiment the next electron/photon can hit so many spots again.

"What is the probability that the incident ray of photons (or a single photon) is rebound or gets trapped by the obstacle (for example crystal) in the aforementioned experiment?"

Yes, in a real world experiment you might loose photons in between due to many reasons. But this does not change the total statistics. It just means you need to repeat the experiments more times to compensate for the losses.

Well that concludes my discussion then

I've got all answers I need Thank you

Well Thank you very much for recommending our discussion I will do it as well but in order to contribute to it from the mathematical perspective I'd like to refer you to your own reference at the beginning of our talk:

https://physicsopenlab.org/2018/01/18/laue-diffraction/

Laue Diffraction

Now I attached a modified screenshot (credits to the website) to explain how one can mathematically find the number of 'missing' photons

On the register screen after each interval there will appear n dots equally distant from the center They can be regarded as vertices of a polygon After the first layer is completed the second will form and then the third etc.Finally there will be the last (largest) polygon formed by the photons To find the percentage of missing photons one can take time t1 spent on completion of the first polygon vertices minus tn all polygon formation and find a quotient of (1-nt1/tn)*100% The delay in time completing the last layer compared to nt1 multiplied by 100% is the sought after percentage of the missing photons

Your welcome. I hope I could help.

Sir If you don't mind leave some contact address The reason Im asking is because I need publications for doing my research and PHD and if you like I can write an article on this theme together with you as an author or co-author Let me know if it's acceptable Thank you very much again for that useful conversation

The diffraction of light in the double-slit experiment is more than a classical wave phenomenon — it’s the gateway to understanding the deepest mystery in quantum physics: the wave-to-particle transition.

Traditionally, light’s diffraction through slits is explained by Huygens’ principle: each point on a wavefront acts as a new wave source. In the quantum view, however, single photons fired one at a time still produce an interference pattern — as if each photon interferes with itself.

This leads to a central question:

What is actually diffracting — a probability cloud, or a real physical wave?

Quantum Wavefront Dynamics (QWD): A Deterministic Extension

In my recent work, I propose a theory called Quantum Wavefront Dynamics (QWD), which builds on this exact question.

QWD treats quantum particles (like photons) as real wavefronts that:

• Diffract and interfere as extended waves when undisturbed (just like light in the classical case),

• But evolve into particle-like detections when the wavefront is asymmetrically disturbed by the environment (e.g., detectors, edges, decohering fields).

This offers a deterministic and physical interpretation of diffraction and interference:

• No collapse postulate required,

• No dualistic behavior — only one continuous process: wavefront curvature evolution.

So when you see diffraction in the double slit:

• You’re not watching a photon “decide” where to go.

• You’re witnessing a geometrically evolving wavefront, which only localizes (becomes a particle) when distorted by interaction.

This turns diffraction from a passive phenomenon into an active, measurable expression of quantum geometry.

If you’re curious, I invite you to read the full theory:

Quantum Wavefront Dynamics (QWD): A Deterministic Framework for the Wave-to-Particle Transition

Preprint Quantum Wavefront Dynamics (QWD): A Deterministic Framework ...

Let’s stop treating diffraction as mysterious — and start understanding it as the natural behavior of real quantum waves.

—

Adam Dakhil Nasser Alzerkany

Author of QWD

Adam D. Nasser Thanks a lot fot the comment For me there is no controvercy nor mystery in the word combination 'Wave-particle duality' A particle behaves like a wave being a part of a bigger system that's all there is to it

Michael Rüsing

Correct me if Im wrong but if in Laue Diffraction experiment instead of a crystal we are going to place any object the diffraction picture will still hold provided that the object is 'transparent' for x-rays What is interesting is that this picture will be an 8n projection of vertices time dependent which is 8nt?

Ruslan Pozinkevych

The diffraction pattern originates from the fact, that the object under investigation, i.e. the crystal, is made up of tiny sub-objects that act as scattering centers. Keep in mind there needs to be a physical mechanisms allowing for scattering.

In the case for X-rays with keV energy/ 0.1 nm wavelength range, the electron clouds of each atom, which make up the crystal, act as scattering centers. Here, in a simple picture the xrays do some broadband excitation, similar to a plasmon, of the whole electron cloud, which then reemits x-rays at the same frequency but in all directions. What we percieve as the diffraction pattern in the far field is the overlapping of all the spherical waves from each atom. If you place a different object, the pattern will be different, if it does not have scattering centers arranged in the same way.

If you somehow create an object with the same microscopic make-up, the pattern will be the same. but then the object should be very, very similar to your crystal, i.e. being a similar crystal.

Michael Rüsing great but a very common sense dictates the following Any object exposed to x-ray gives a diffraction pattern e.g x-ray picture if it is an 8 bit (octagon shape cluster) dependent then we can well restore the object using additional parameters in terms of its shape and even density If you like I can send you an idea of what could be a commercial implementation of the fact Thank you for responding so quickly btw

As far as I understand, what you are suggesting is what every XRD machine, neutron diffraction or electron diffraction machine does in their softwares already. There are big companies, which offer that already (Bruker, Rigaku, FisherScientific, AntonPaar, ...).

Such machines take the pattern observed from single crystals or powders and can fit the crystal structure to it.

In the Laue method, there is only a very limited amount of crystal structures (limited by math!), that are possible for any 2D and 3D materials. There is only a finite amount of symmetries and crystal classes, which are possible. Therefore, there is only a limited amount of diffraction pattern symmetries. Here, crystals under investigations are rotated until axis of high symmetry are found. The pattern will immediately indicated of its cubic, hexagonal, monoclinic, triclinic, orthorhombic, etc. Feel free to sent me a PM. And the distance of the peaks will automatically determine the lattice planes.

Check crystal classes:

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

finite amount of point groups:

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

E.g. check Rietveld analysis:

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

PS: And a "octagonal" crystal does not exist. It falls to the cubic materials.

Michael Rüsing What Im suggesting is based on the time reference for example half-decay of the elements on the reference screen present a changing picture of the detected dots changing with the time as a (x,y,t) or (x,y,z,t) vector Present it as a function of time f(t) of changing 8 bits (bites) and compare it with let's say 2 D digitalized picture of an object When the two become similar or identical fix t and find out what f(t) is that will be a time dependent function of an image in (x,y) or (x,y,z) coordinates the so-called 'time dependent restoration of an original image" Combining time related and process (decay of a radioactive material) and rate at which (x.y,z) change we will get a picture from the past if you like

I am utterly lost.

What decay are you refering to?

If you check the overlap with a previous/existing image or simulation, this is literally what a fitting, e.g. in Rietveld analysis does?

Im not supposed to do the detailed analysis but I will give a general idea of what is being said:

reference material:

https://physicsopenlab.org/2018/01/18/laue-diffraction/

Picture Sodium Chloride (NaCl) Single Crystal

Pls check the coordinates All points are on the plane 2 units above 0 along z axes I assume that the plane that has points (4,y,z) is being completed first so scattering is 4 units away from 0 in x and y direction or (6,y,z) plane completes first You can spot that all 8 points are on the same plane This plane is completed by 8 points (8 bits) first so we have a speed of information 8 bits of information per t When time is nt we have n8 points or bits Lets say that the screen that has these points is made of the material that decays After some time T the entire picture 'evaporates' so we have T/nt≡ 8T/t equivalent :|T,t ∈Z; bits of information: T/t (0 0 0 1)- vector presentation of the picture Lets take some digital photo of the object which was bombarded by an-x ray lets say that a digital representation of it it terms of (0 0 0 1) It is k (0 0 0 1) Please pay attention that k(0001) is an ordinary digital image We equalize T/t(0001) and k(0001) and find a common factor r:

T/t(0001)=rk(0001) this factor brings us back to the original picture but it compares a 3 d object which was x-rayed to a 2D picture which was ordinarily photoed Now by the new vector of rk(0001) we present a 3 d object and its inner parts projected on the plane

I am still somewhat lots. The numbers referenced in the pictures are the so called hkl or Miller indices. They refer to specific crystallographic planes, which are associated with the respective diffraction maxima. They do not specify anything on the image plane, the refer to planes insight your diffracting crystal.

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

To generate a constructive interference you need the signal from many parallel planes. In principles, these can be any planes you can construct through the crystal. The higher the indices are, usually the less atoms will be there per plane. Therefore, as a rule of thumb, higher indexed planes have weaker intensities.

The indices have nothing to do with the order in which peaks appear. They all appear at the same time. Going back to your original question of single photon scattering. Yes, stronger intensity peaks have a higher probability that single photons will appear there. But it will still be random.

And what you further say, we already compare XRD-patterns with ideal patterns, e.g. simulated, i.e. we make image comparisons. There is tons of software (proprietery and open-source) for that.

To sum it all up The model I propose is an abstract plane Be it time coordinate and the plane x,y or it can be xyz plane whatsoever The main idea is to find the function of coordinate change For example we have (x,y,t) coordinates for two squares (two squares are on the same plane After some time these coordinates will change to (x1,y1,t1) so we have instead of (6,0,2) a (6,0,3) vector Both will lie on the planes parallel to each other difference is last z-oordinate 2 or 3 There is multitude of options as to how the model is built If one wants to experiment with indices factors coordinates that's fine too Honestly I did not check the last reference you sent cause like I said Im not a physicist Im interested in abstract model only

Michael Rüsing

Dear Professor

Over the last few hours I sent you 3 emails containing my recent research paper The reason why I bothered is that I think my research paper is related to the theme of the discussion and the formulas might be used to describe some processes in physics If you have time to read it and make some comments pls do I will be glad to hear them Im not sending it here not to overload the discussion with mathematical specifics Thank you very much again