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In other words, suppose one creates a particle hologram -- no photo reaction, just a standing waveform in space. Suppose the particle is something we are skilled at manipulating, like an electron.

Next, one continuously blasts the kind of energy the particle would normally absorb at the hologram.

Will the particle, when it is finally detected at some terminating point (like a phosphorous plate) have the same, more, or less "energy"?

And, as a follow on, would there be any mechanism to do this (interact with the diffracted particle), under any conditions?

In short: you can absorb energy from diffractions to record a hologram. Can you push energy into the system?

Disclaimer: I recently asked a question on this topic, and discovered that, yes, particle holograms exist:

Matter Holograms: What would be involved?

Chris
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  • Can you define what you mean by a particle hologram? A hologram is a virtual image created by a diffraction plate, and it's hard to see what the phrase would mean when applied to an electron. – John Rennie Dec 21 '20 at 18:17
  • https://physics.stackexchange.com/q/601933/157259 – Chris Dec 21 '20 at 18:30
  • @JohnRennie just posted a disclaimer: I have reason to believe particle holograms do in fact exist, and all the way up to carbon-60 buckyballs. I asked that question on here a few days ago. – Chris Dec 21 '20 at 18:32
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    What that describes is not a hologram of a particle, it is a holographic image (presumably of some macroscopic object) made with particles. The electrons (or whatever) passing through the diffraction plate are just electrons, like the photons creating a hologram are just photons. All you are doing is changing the probability of detecting the particles at any particular point in space. – John Rennie Dec 21 '20 at 18:50
  • @JohnRennie yes exactly: can you interact with those probabilities as if they are the particle? For example, as I understand it, the definition of a particle is the high likelihood component of its probability distribution function. And, when you create a hologram, you create a large standing volume of the particle's pdf, just waiting for large wavelength, low energy interaction and manipulation, no? – Chris Dec 21 '20 at 19:35
  • The probability $Ψ^*Ψ$for one electron to exist at a specific (x,y,z,t) is a solution of the quantum mechanical equation given the boundary conditions.The distribution appears when a large number of electrons follow the same boundary conditions. Each electron is on its own particle track . See the double slit one electron at a time to understand that the wave is a probability wave, individual electrons are individual points on the screen: https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles – anna v Dec 21 '20 at 19:55
  • @annav yes, however the entire point of the double slit experiment, and the hologram, is that the electron is able to interfere with itself, and is therefore not on a single electron path in an energy minimizing hamiltonian without violating the a-priori probability distributions a-posteriori (after observing the collision spot). So, it is fine, in my humble opinion, to process the collision event as an instantaneous collapse of the probability distribution, rather than evidence of some squiggly particle path..or evidence that the apriori cumulative path probabilities were wrong. – Chris Dec 21 '20 at 20:00
  • @annav in other words, if you were to emit in-phase electron waves such that, just prior to a collision spot there was a local minimum of likelihood...and just at the collision spot a local maximum, the electron collisions observed really would comply with the probability peaks and troughs as you moved your phosphorous plate. Thus, a hologram is possible. If the electron tunneled to achieve this, it would exceed the speed of light... or have a winding and/or fast and slow path through the intervening space. Further, it could not self-interfere if it flew like a bullet. – Chris Dec 21 '20 at 20:05
  • @annav in my opinion, having a particle that meanders through the probability soup as a self contained particle on a grinding path through space implies a much more complicated particle, and a lot more information the universe has to "know about" for the particle to move through space. This seems energy intensive for the light and breezy electron. But if, on the other hand, the particle really does spread out as a fluid probability wave and interfere with itself, you in fact have much simpler, self evident behavior. The universe has to remember nothing. – Chris Dec 21 '20 at 20:16
  • "electron is able to interfere with itself, " this is wrong. Every electron is a little dot, a footprint of a particle,and the electrons in the experiments have simple tracks,they are not spread all over.. It is the accumulation that has wave characteristics. see the electrons here https://hst-archive.web.cern.ch/archiv/HST2005/bubble_chambers/BCwebsite/index.htm . Particles do not self interfere, sorry. – anna v Dec 21 '20 at 20:20
  • @annav this drives diametrically against everything I know about modern quantum mechanics. I'll have to check with the head of my local physics department on that one -- a former bell labs researcher and theoretical physicist specializing in quantum mechanics and optics. And quite a good soccer player. – Chris Dec 21 '20 at 20:24
  • @annav and yes, prior to diffraction, electrons are self contained -- like bullets. Hence the stupendous puzzle of the double slit experiment. – Chris Dec 21 '20 at 20:25
  • @annav It also doesn't mesh well with Feynman's lectures (that nobel winner), existing accepted questions on here, or regular photon holograms (which require the phenomenon we know as noise cancellation to actually work -- the photon has to penetrate through a medium designed to absorb photons without being absorbed... if the photon is actually a self interfering wave, then this is plenty possible; if it is a particle, it will simply register at the surface of the solid hologram film) – Chris Dec 21 '20 at 20:39
  • There are no popes in physics. Only experimental numbers and mathematical models. Difraction offers boundary conditions and a new wave function for the photon or any particle appears, a new probability distribution. It is not a classical phenomenon at the level of a single photon, probabilities are not .interactive.. Single photons would behave the same way as with the double slit I linked, adding up to the classical diffraction pattern in accumulation. – anna v Dec 22 '20 at 05:40
  • @annav that seems like something there should be an experiment for to rule out. Since, as you claim, there are "no popes in physics" (I beg to disagree: I think physics is filled with popes). – Chris Dec 22 '20 at 14:26

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Your question is a bit unclear, but I'll answer the question I think you want to ask: "If an unspecified kind of particle is diffracted by a standing wave pattern formed by a particle beam, how is the particle affected?"

The experiment you describe probably has never been done using electron beams to form a standing wave. However, an inverse of the experiment has been done: Light has been used to form a standing wave pattern, and particles have been diffracted by that pattern. Perhaps more to the point, sound waves have been used to form a standing wave pattern, and light has been diffracted by that acoustic standing wave. See Acousto-Optic Modulator.

When light is diffracted by the acoustic standing wave, it does indeed pick up (or lose) energy. Similarly, it is reasonable to expect that a standing wave produced by interfering particle waves would add (or subtract) energy from light or matter waves diffracted by the standing wave.

S. McGrew
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  • "Perhaps more to the point," – Chris Dec 22 '20 at 03:50
  • And what I am driving at next is: suppose you have a certain type of particle or amalgam (like a nucleus) that can absorb only energy which is packed tight enough to "interact" with the particle or amalgam (ie, you can only make a 1000 nm wavelength, but your target is 10nm, let's say). Then, would it not be possible to add energy to your target particle by interacting with the diffraction or standing wave of that particle (which is much larger). I'd like to ask it as a standalone question... but, right off the bat, do you think it is off base? – Chris Dec 22 '20 at 04:02
  • I hesitate to answer because I don't really understand what you're reaching for. Would you like to continue in Chat? – S. McGrew Dec 22 '20 at 04:05
  • yeah -- that would be awesome – Chris Dec 22 '20 at 04:14
  • If I can figure out this chat feature anyways. – Chris Dec 22 '20 at 04:22
  • https://chat.stackexchange.com/rooms/info/117545/holography?tab=general -- alright, I created this room. And, as far as what I am driving at, well, I acquired some 40 year old paper compendiums that chronicled the rise of lasers and holography from the basics all the way to the 1980s and the laser disc from a retired professor's book dump across the hall a few years back. And I flipped through and read a good chunk of them. Then, I started watching this guy, https://ocw.mit.edu/courses/physics/8-421-atomic-and-optical-physics-i-spring-2014/video-lectures/lecture-1-resonance-i/ – Chris Dec 22 '20 at 04:32
  • To be honest, I am not really sure what I am driving at; more or less I am trying to unpack what I've been learning in a way that makes sense to me, and clear the content that I think is junk out of my brain -- I am not on the physics PhD track, and I am no longer in R&D, so I'm trying to figure things out for myself. – Chris Dec 22 '20 at 04:41