It is kind of mystical that a particle goes through a slit and eventually changes its impulse due to Heisenberg uncertainty. Since the slit is an opening, it must not have interacted with it. Does it interact or not? If yes, than how? By virtual photons, or something else?

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1Might help https://en.wikipedia.org/wiki/Wave_interference – Sep 16 '21 at 16:29
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1@Wolphramjonny I went to the pointed by you article and searched for 'interaction'. I found just 1 result. Reading: "Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other, either because they come from the same source or because they have the same or nearly the same frequency." – Mercury Sep 16 '21 at 18:45
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2@Wolphramjonny What this has to do with interaction of the particle with the walls? Of course a particle is attached with a de Broglie wave and the diffraction is result from superposition of waves started at all points of the slit, but this tells nothing about interaction of the particle with walls. By the way the wave is not physical but square root of probability so it can not interact. – Mercury Sep 16 '21 at 18:45
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2IDK what physicists say about the particle interacting, but any time a wave passes by the edge of an obstruction, the wave is diffracted.* We can use a wave function to predict the liklihood of finding a quantum-scale particle at any given place, and according to all of our experiments, the "wave" (whatever it actually is) is diffracted by the edges of any obstruction. – Solomon Slow Sep 16 '21 at 18:57
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* This works for sound waves, and surface waves on water, and every other kind of wave we know about. Diffraction isn't just for "quantum" wave functions. It's deeply embedded in what "wave" means. – Solomon Slow Sep 16 '21 at 19:00
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"A particle goes through a slit..." is a very classical picture, a particle can travel all possible paths and we can only infer it's position once it has interacted with our detector; uncertainty principle talks about standard deviation associated to measurement of canonical observables like momentum & position of particle – KP99 Sep 16 '21 at 19:07
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It would help answers if you included the standard diffraction pattern for a slit in the question. – JEB Sep 16 '21 at 19:27
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@JEB Yes? Where do you find hint for interaction in the diffraction mechanism? – Mercury Sep 16 '21 at 20:06
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@KP99 Since one has a particle in the detector and out of the source there also is a particle in between. As you say it travels all the paths simultaneously (Feynman alternative formulation of QM) but it does not interact but just pass if no walls there. As the impulse is changed there must be a mechanism and covering the issue under carpet by saying that it is too classical doesn't give any answer. – Mercury Sep 16 '21 at 20:22
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@SolomonSlow All you say is true but trivial and answers nothing. – Mercury Sep 16 '21 at 20:25
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@Mercury, That is why I did not post an "answer." Like I said above: I don't know what a physicist would say about the particle interacting with the edges of the slit. I only understand it as a wave phenomenon. – Solomon Slow Sep 16 '21 at 20:27
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Sure, every interaction should be accounted for in the classical action. I am not sure what kind of potential should define interaction b/w particle & wall/slit, maybe try using a periodic delta type potential – KP99 Sep 16 '21 at 20:40
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@Mercury e.g https://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction If you include that in the question, it makes it easier to post answers as we don't need to define coordinates, variables, and maybe images ourselves. – JEB Sep 16 '21 at 21:18
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Yes they interact. Photons diffract around the edges of all masses. Photons also scatter off the edges. – Bill Alsept Sep 18 '21 at 06:07
8 Answers
The conventional explanations of what happens when light or other waves hit a slit or small aperture in a barrier assume that the barrier is a classical object that blocks the transmission of the incident wave. You might consider it possible, therefore, that there is some quantum-level interaction between the light or matter wave and the material of the barrier, given that the barrier is composed of quantum particles after all.
However the difficulty in considering that is that the diffraction that occurs seems to be largely independent of the material of which the barrier is made and the nature of the incident wave. Electrons and neutrons are both diffracted in a similar way, even though neutrons- having no charge- would not be subject to the same electromagnetic interactions that an electron might experience with the material of the barrier. Likewise, the diffraction pattern does not seem to be affected by whether the barrier is a conductor or an insulator, which again suggests that quantum level interactions between the wave and the barrier are not significant.

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1Perhaps rather than saying that interactions between the wave and barrier are not significant, it would be better to say that to leading order they are universal? To me, "quantum level interaction" is somewhat redundant since fundamentally all interactions are quantum (as far as we know). I think I understand what you're saying (the details of the material don't matter), but clearly it is significant that the diffracted particle interacts with the material, and all interactions are ultimately quantum. – Andrew Sep 21 '21 at 21:18
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1The distinction I was trying to make is that the barrier appears to behave as a classical barrier, simply blocking the propagation of the wavefront associated with the particle. It is impossible, as far as I know, to explain exactly how it is doing that at a microscopic level in terms of forces acting between individual particles. – Marco Ocram Sep 21 '21 at 21:48
Yes the photon interacts with the slit, from a momentum perspective the slit (material) is much much heavier (connected to the apparatus, etc) than the photon, any momentum change is equaled by momentum change in the slit. This is similar to you jumping up and down on the earth, momentum must be conserved but the earth is so large we do not need to consider it.
We know the photon is the same color going in and out and therfore its energy is conserved.
If you are really asking "why does the wave spread out" than this transfer of momentum appears to have a random or Gaussian nature, the slit is full of atoms and structure and yes virtual photons are the ones that transmit forces but do not exchange energy.
If you are really asking "is the slit wall interaction the reason for the interference pattern" then I would say no. The pattern is not a result of any properties of the slit (or it's material) only just the dimensions of the slit. The pattern is a property of the light itself and how it behaves in the EM field .... per Feynman light tends to choose a path that is shortest and has a path length integer multiples of its wavelength.
Integer multiples of wavelength is a fundamental property of resonance and energy transfer ..... laser cavities only lase for example when the mirrors are set integer wavelengths apart.
The final pattern we observe is a combination of the random Gaussian nature combined with the need for light to travel integer paths and shorter path lengths.

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1If there is interaction of the particle with atoms of the walls (via their virtual particles and/or fields) than the momentum distribution is due to this interaction. In my opinion this is a logical consequence of having interaction with the walls. Right? But as momentum and coordinates depend on each other (e.g. the value of the momentum specifies (in QM to some extent anyway) the landing of the particle on the screen (detector) then the interference pattern should also be due to these interactions. – Mercury Sep 23 '21 at 19:44
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Which is in conflict with your answer to the question "is the slit wall interaction the reason for the interference pattern". (Of course I know what the experiments show about the material of the walls and the different types of particles.) That’s why I see a contradiction. A) If there is not interaction how would a particle change its behavior at all? B) If there is interaction why does the interference pattern not depend on material of the walls and/or the type of the particles passing through but only on wavelength and the geometry? – Mercury Sep 23 '21 at 19:45
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To be clearer l use the word photon, not particle. My definition is that a photon is energy or a disturbance in the EM field. Also a photon is different from the virtual photon. – PhysicsDave Sep 24 '21 at 22:41
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What is interesting about a photon is that it determines its own path .... this determination can even begin before the photon has left the atom .....the excited electron is already disturbing the EM field via forces (virtual photons).... an atom can be excited for quite a length of time. Thus the geometry of the entire setup is considered before the photon is even emitted ..... the path has considered the effects off all electrons/atoms in the EM field ..... the slit, the screen, and its own location in the source. Of importance is that the path choose is also "resonant". – PhysicsDave Sep 24 '21 at 22:49
Quantum mechanics theories have been developed so as to model what happens at the micro world of particles , because classical mechanics and electrodynamics cannot model the data nor predict new ones.
In quantum mechanics one solves with the boundary conditions and for the voltages the particular system under study, and the solution gives the probability for the interaction to happen.
For this question a particle perpendicular to the surface, scatters off the surface which has a slit, of a given width, and is detected a given distance away. The probability for scattering back or be absorbed by the surface, is part of the solution. The solution in the region of the slit gives the probability of interacting with the fringe electric fields at the edge of the slit, and being deflected. See the answer here for a single slit diffraction pattern.
Does it interact or not? If yes than how? By virtual photons or else?
For calculations in electromagnetic scattering at the quantum level Feynman diagrams are used, and there, virtual photons appear to carry the energy and momentum changes due to the interaction. The Heisenberg uncertainty and the de Broglie wave are an envelope that allows estimates without going into the unnecessary complications of solving hard equations,as it is proven that they arise from the theory of quantum mechanics.
The double slit experiment has been studied, one electron at a time, example, and it is clear how the accumulation of electrons looking random when one by one, shows the probability wave behavior in the accumulation.

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I appreciate that you think that QM is a model. There are almost 95% thinking it is the ultimate theory and there is no deeper content. I think that the Feynman trajectories are more physically sound in the context of the slit. I think that the interaction with walls maybe included in this formulation of QM. E.g. the particle follows all possible trajectories and in some of them near the edges it interacts with atoms from the wall. Then of course all interfere and give the pattern. It sounds to me like a possible explanation what is really happening. – Mercury Sep 17 '21 at 17:41
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Contrary in the orthodox formulation of QM the change of the impulse seems unphysical - just follows from Fourier transformation without any physical interaction. In that context I think that Feynman trajectories are more physically deeper although there both lead to same results. – Mercury Sep 17 '21 at 17:43
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2@Mercury quantum mechanics is not a model, it is a theory that is used to model observations and data that cannot be modeled with classical physics theories. And quantum field theory (Feynman diagrams) is based on quantum mechanics and its postulates, it is one level higher in complexity to allow for modeling many particle states. – anna v Sep 17 '21 at 18:02
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I don't mean Feynman diagrams but Feynman paths (the integral over all possible trajectories). The alternative formulation of QM. It seems to include explicitly interaction with walls, where Copenhagen does not. The Feynman diagrams appear at the points of the interaction on the walls (in fact are applied to make the calculation of the interaction and must be hidden in the action S). – Mercury Sep 17 '21 at 20:12
Your question and the idea behind it is very reasonable, but you cannot get any other answer than the explanation reached and taught. Science works only with an agreement what is right and what is wrong.
The reasons for rejecting contrarian thought are three:
- the scientific world would not be one, but a chaotic discussion club in which there is no agreement about what fundamental truths are
- what one is taught for years must be correct, and all the more so because the teaching is accompanied by the sentence: "It is mysterious and not comprehensible, but the results should be explained as interpretable"
- how will a scientist feel if he advocates something that later turns out to be an error.
What I have learned here is that you have to back up your idea with calculations. And that the specialists in energy-material interactions (plasmons, polaritons, ...) will not support your idea.
I support your idea with one small detail. Destructive interference for light or electrons is not a dissipation of energy, but a shift to areas of doubled intensity. This shift must have an origin and your idea may be the reason.

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@Mercury, sciences are not a religion. They give technical solutions while religions gave mental orientations. For your question, if it will be possible to get a higher resolution for optical devices by understanding how an electric or magnetic manipulation of the edges of slits get used, that will be fine. – HolgerFiedler Sep 20 '21 at 05:33
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just replace scientific with clerical in your 3 items and you get a description of church. I think in books it would be thought very seriously and honestly about the problems of QM and not to lead students into delusion. I never encountered any explanation in books about the slit or discussion which honestly to admit there is a problem. For me this is just a description. They insist that the wave is probabilistic and only the wave is in the slit and use diffraction. The diffracted part does not interact with walls. Moreover it is probabilistic which excludes interaction. – Mercury Sep 20 '21 at 06:58
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https://physics.stackexchange.com/questions/114384/relationship-between-the-material-properties-of-an-edge-and-the-fringes-behind-t – HolgerFiedler Sep 20 '21 at 07:17
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Search for Möllenstedt, Düker, Beobachtungen und Messungen an Biprisma-Interferenzen mit Elektronenwellen, Zeitschrift f. Physik, Band 145, 1956, S. 377 – HolgerFiedler Sep 20 '21 at 09:48
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@Mercury in the link above the images are for electron diffraction, changing the potential of the metallic aperture is the same as narrowing the slit as viewed by electrons. – PhysicsDave Sep 21 '21 at 20:42
The double slit experiment is intended as a demonstration of interference and diffraction. In its idealized form there is no interaction with the walls. It produces a diffraction pattern proportional to the Fourier transform of the transmission of the slit system by selectively blocking oarticles. In practice there are of course deviations from this idealized behavior due to scattering, residual reflection or transmission, light polarisation, surface excitation etc.
If your question however is about the origin of the diffraction, then it is necessary to use Schrödinger's or Maxwell's wave function. Multiply an incoming plane electron or light wave with the slit system transmission's function. In k-space this results in plane waves in all directions with amplitude equal to the Fourier transform of the slit transmission. There is no interaction with the edges other than transmission versus blockage.

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First I talk about single slit. Second the double and single slit experiments are not intended to teach but are real and as such they require explanation. QM has just invented a genius way to describe what is happening. Third - QM description uses Schrodinger equation and is shows exactly what you say (no interaction other than transmission vs blockage). Fourth: there must be interaction because impulse of the particle is changed. Remember the Einstein Bohr dispute (E. said he can measure P of the slit to calculate p of the photon). – Mercury Sep 19 '21 at 21:41
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This comment contains some inaccuracies. First you talk sbout 'a slit' not about a songle dlit. Second, I did not use the term 'teach' but demonstrate, which implies a real experiment. QM did not invent, it was invented by real people. Third: of course. Fourth: the interaction is with the particles that are blocked by the slit. Finally, if your question is about the Einstein Bohr dispute, please mention that, give some detail and make the question specific. You may consider a new question. – my2cts Sep 20 '21 at 07:26
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Bohr Einstein dispute is about single slit. Einstein said he can measure P of the slit installation for a particle passed by. Bohr does not object this. He objected there is P of the slit installation - just the accuracy of that. That is 100%. The interaction of non passing particles is out of question. – Mercury Sep 20 '21 at 18:09
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The incident wave function $\psi(\mathbf{x},t)$ of the particle (described as a localized wave packet) interacts with the walls and propagates through the openings. It is possible to model this process in non-relativistic QM by studying the evolution of the wave packet $\psi(\mathbf{x},t)$ via the time-dependent Schrödinger equation, with a potential that models the walls (the potential goes to zero at the openings).
This QM way of doing things does not model the exact interaction with the particles in the walls (the potential in the Schrödinger equation is a classical, external, one), but this is probably unnecessary (see the answer of @Marco Ocram), especially if we are interested in the late-time dynamics of the diffracted wave packet, far from the openings.
For a simulation of this Schrödinger evolution leading to diffraction and interference, see this video.
Note: in the video we see that part of the wave function is reflected (this is normal since the wall is modelled as an external potential that has no internal degrees of freedom that can be excited or modified, so this is a sort of "elastic" process). In reality, the particle could also be absorbed by the walls, or it could cede part of its total momentum to the internal degrees of the wall: this really depends on the nature of the walls and how they are modelled. However, the part of the wave function that passes through the openings is much less influenced by the details of the walls (note also that part of the wave function can tunnel the walls potential and interfere with the part that propagates through the openings).

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You say: But as the slit is an opening it must not have interacted with it. This is only true for particles. But QM is about waves and waves have extent. If you take this extent into account, then it is not paradoxical anymore that the screen and the slit and everything else has an influence on the outcome of the experiment even if it finally manifests in a particle that seems to hit a sensor at a specific point.
That’s again the famous wave-particle dualism: an object is “simultaneously wave and particle”. If you have problems understanding this you are in good company: no one understands it. But: the mathematical models to describe the quantum mechanical effects are perfectly well understood and the outcome of experiments is predicted by the models to high accuracy. It is only the interpretation of the effects (predicted by the models and reproduced in experiments) that lacks human understanding.
EDIT
After having read through the other answers and comments, I came to think that your question is another formulation of the measurement problem. In this specific case, the state of the incoming particle comprises a superposition of several states/waves, each of which specifies an exact position of the particle in relation to the slit. Here http://mtnmath.com/faq/meas-qm-2.html the measurement problem is stated as: How does a superposition of different possibilities resolve itself into some particular observation? which, in this particular scenario, is the same as your original question.

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1Just want to mention that the wave function does not exist in real world. It is square root of probability. As such it can not interact with real matter. That the mathematical models are well understood is not surprising. There are lot of math theories more complicated than linear algebra but their understanding brings not too much for physics. – Mercury Sep 17 '21 at 17:29
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@Mercury In that sense particles don’t exist either. Whatever, it is the extent of a wave that explains the paradox situation OP is referring to. I may be wrong, but I think in the deBroglie-Bohm interpretation the wave function actually does exist. – Hartmut Braun Sep 17 '21 at 17:59
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@Mercury I edited the answer and reformulated the statement about interacting waves. – Hartmut Braun Sep 17 '21 at 18:16
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@mercury Your comment is wrong for multiple reasons. For example the probability does not contain the phase. – my2cts Sep 18 '21 at 09:21
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I meant this in sense that $\psi .\psi ^{\ast }$= P. What other multiple reasons you do you see? – Mercury Sep 19 '21 at 21:17
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Yes it is true what you say. After the slit the wave and the slit should be in entangled state $sum ^{\infty }{i}\left| p{i}\rangle \right| X_{i})$ where $| X_{i}\rangle$ is the state of the slit and than collapse at the detector. Nevertheless all probabilities with p(i) not equal to p initial must come from interaction, right? Thats my problem, because nowhere an interaction is said to occur – Mercury Sep 20 '21 at 07:15
First, it is necessary to note that two-slit experiment is a thought experiment - its goal is to teach quantum mechanics, the wave properties of electrons, rather than model reality.
However, since the experiments along this line has been done (e g., solid state Aharonov-Bohm interferometers are a bit of a routine nowadays), it makes sense to ask what are the properties of the screen/slits that assure that the experiment is working.
Without going into too many details, let me point out the two principal ones:
- The screen is very heavy, so that it is not affected by the electrons hitting it. In other words, there is not energy exchange - no decoherence.
- Screen is static, it is not vibrating - otherwise it would randomize the electron phase - no dephasing.
Update
On a more mathematical level, one can describe the two-slit experiment by Hamiltonian
$$
H=\frac{\mathbf{p}^2}{2m} + V(\mathbf{x}, Q)+H_Q,
$$
where $V(\mathbf{x}, Q)$ is the potential describing the slits, which can change as a function of the screen variables $Q$ - these can be, e.g., phonon modes of the screen vibrations, or screen movement as a whole (in case of a nanomechanical system, such as an STM tip), or whatever corresponds to actual physical realization of the experiment.
We now can consider there cases:
- Neglecting screen dynamics - which is the way the problem is usually treated in the textbooks. Then the interference is largely described by the interference between two paths between the origin and the osbervation point on the screen: $$ \left|e^{i\mathbf{kL}_1}+e^{i\mathbf{kL}_2}\right|^2 = 2 + 2\cos\left[\mathbf{k}(\mathbf{L}_1-\mathbf{L}_2)\right] $$
- Only elastic scattering. In this case the momentum of electron does not change in magnitude, but the above expression eneds to be averaged over the screen vibrations. $$\langle\left|e^{i\mathbf{k}\mathbf{L}_1(Q)}+e^{i\mathbf{k}\mathbf{L}_2(Q)}\right|^2\rangle_Q = 2 + 2\langle\cos\left[\mathbf{k}(\mathbf{L}_1(Q)-\mathbf{L}_2(Q))\right]\rangle_Q \approx\\ 2 + 2e^{-r}\cos\left[\mathbf{k}(\langle\mathbf{L}_1(Q)\rangle-\langle\mathbf{L}_2(Q)\rangle)\right] $$ Although the details of averaging vary depending on the physical implementations, we generally expect suppression of the oscillations, as described by factor $e^{-r}$.
- Finally, if the inelastic scattering is taken into account, we will also need to average over the distribution of the electron momenta after the scatteirng, which will further suppress the interference.
If we exclude the inelastic pricesses, the interference picture can be
References
- D. Sanchez and K. Kang, Validity and breakdown of the Onsager symmetry in mesoscopic conductors interacting with environments, Phys. Rev. Lett. 100, 036806
- D. Sanchez and M. Buttiker, Magnetic-Field Asymmetry of Nonlinear Mesoscopic Transport, Phys. Rev. Lett. 93, 106802

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How is there no energy exchange when p of particle can change dramatically also by module which follows from HUP. Maybe you mean no change of v because of the huge mass? – Mercury Sep 19 '21 at 21:49
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@Mercury HUP does not decsribe the change of momentum or any other quantity - it imposes the limits on the precision of measurement. The evolution of particle (and its momentum) is described by Schrödinger equation (or von Neuman equation for the density matrix). – Roger V. Sep 20 '21 at 07:09
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Of course HUP is follow up of Schroedinger equation. But it admits the change. And this change happens somewhere. Before the detector the slit and the particle are in superposition and as the detector clicks the particle has a new impulse and HUP says it can be much bigger than the initial p especially when the slit is narrow. I don't believe the detector can supply energy to the particle. Energy must be taken from the walls of the slit, don't you agree.? – Mercury Sep 21 '21 at 09:28