# How does quantun mechanics explain motion of an object [closed]

Could anyone explain to me a simple question about motion of an object.A very simple explanation I have been looking for about motion of an object. I have been tried to understand object how move one place to another place through newton motion and it is well understood. But whenever I tried to put together quantum mechanics and Newtonian motion I cant not possibly understand what should be the best explanation would be motion of an object ?

If any object made of the basic building block such as electron, proton, AND neutron. According to the quantum mechanics Newtonian motion can not explain motion of those basic building block then MY question is how do the objects move. I mean to move any object we require to apply force on that object and if I tried to know more deeper way it seems me that we are applying force on those atoms that the object made of.If Newtonian motion of force do not have any effect on atomic level then how things move?

Just what to know what I got wrong to explain myself? Any help will be appreciated. Anyway thanks in advance.

## closed as unclear what you're asking by sammy gerbil, Jon Custer, Emilio Pisanty, hft, heatherJun 28 '18 at 16:53

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• They don't move, so much as they are created and annihilated at different points in space-time en.m.wikipedia.org/wiki/Creation_and_annihilation_operators – user198207 Jun 26 '18 at 10:45
• Since a particle in quantum mechanics does not have a definite position until measurement, there cannot be any exactly defined path the particle takes. – Aaron Stevens Jun 26 '18 at 10:56
• @Countto10, as I read the question, the OP is wondering how the collection of quantum particles that are an object can 'have Newtonian motion' while the individual quantum particles do not. – Alfred Centauri Jun 26 '18 at 11:31
• – sammy gerbil Jun 26 '18 at 13:05
• @Countto10 "They don't move, so much as they are created and annihilated at different points in space-time" "I don't agree, and I think this comment is incorrect and confuses the OP." – my2cts Jun 27 '18 at 6:45

I think what you are looking for are wave packets. You can construct a superposition of waves that add constructively in one location (with some spread) and destructively everywhere else. Then the motion of this wave packet can be thought of as the motion of the particle it represents. Of course, the wavefunction is just an expression of the probability of finding a particle in a particular place when you make a measurement. But if that probability moves with a classical trajectory, then an analogy can be made to classical motion.

One example is that of a particle in an infinite square well. If you solve for the eigenstates of the energy, you’ll find waves which span the entire box and don’t move. But if you construct a wave packet (say, with a Gaussian envelope), which is itself not an eigenstate, you’ll see that it bounces around in the box like a particle would classically (except it interferes with itself near the edges - it is a wave after all).

EDIT This answer is wrong, because of my incorrect assumption regarding the reasonability of extending scattering theory to the classical world, (I think?), but there is a related post: Scattering and Forces, which addresses a similiar question

From that post:

In quantum field theories, because locality, forces are always produced by interactions, i.e. correlations made by virtual particle exchange, or field fluctuations if you prefer. So, a repulsion between particles are actually a probabilistic phenomena. The probability of two electrons be close together decreases as they get close. This is so because they have a probability to exchange virtual photons that affects the probability of each particle, producing this kind of correlation. Everything in quantum mechanics are probabilistic phenomena.

Also from this post: The Escape of Particles from a Black Hole

A particle is an excitation of a field, not the field itself. In QED, if you set up a static central charge, and leave it there a very long time, it sets up a field $E=kqr^2$. No photons. When another charge enters that region, it feels that force. Now, that second charge will scatter and accelerate, and there, you will have a e−−>e−+γ reaction due to that acceleration, (classically, the waves created by having a disturbance in the EM field) but you will not have a photon exchange with the central charge, at least not until it feels the field set up by our first charge, which will happen at some later time.

Obviously, I am trying to support my very naïve and loose answer with quotes from those with greater experience in QFT than I have, in the hope of improving it.

END EDIT

Have a look at this diagram, which shows the interaction (repulsion) between two electrons.

Please don't read the straight lines of the electron as indicating that a definite trajectory is being followed, it's more of a schematic that anything realistic.

Image source: Wikipedia: Møller scattering

On the smallest possible scale, this is how a "force" is transmitted between two particles.

The word scattering refers to the fact that the electrons have an influence on each other, what we would call a push or pull (force) in Newtonian mechanics.

Because its quantum mechanics, we can't use the Newtonian $F=ma$, as this is only applicable to macroscopic objects with definite positions and trajectories.

Instead, we calculate the probabilities that a certain amount of momentum will be transferred between the electrons.

This calculation takes into account the chances of any electron being in a particular place at a particular time and the chances of a photon being emitted and absorbed to, in effect, transfer momentum between the electrons.

But once you see the word momentum, its not difficult to think as linking back to Newtonian forces for a large group of particles.

That's what we do when we return to the classical world, where the probabilities of finding a larger object like a soccer ball at some place (or following some trajectory) other that where Newton laws say it should be are so small that we can ignore them and use $F=ma$.

• This doesn't answer the question at all. Scattering is different from displacing the physical object. – AtmosphericPrisonEscape Jun 26 '18 at 18:24
• @AtmosphericPrisonEscape thanks for your comment, I learn the most from incorrect answers. But when you look at this post below, I'm not sure exactly what my wrong assumption is: (although obviously scattering covers a large amount of phenomena. physics.stackexchange.com/questions/287780/…. I will of course look up scattering theory myself but as a self study person, wrong assumptions are the rule, unfortunately. No worries about this, I can always post a question, thanks. – user198207 Jun 27 '18 at 5:35

The particles in quantum mechanics are described by four vectors. All you have to do is go the the framework of four vectors, $(E,p_x,p_yp_z)$ when talking of elementary particles and their immediate composites. Each particle can be uniquely described by its energy and momentum, and its mass will be invariant, the "length" of the four vector under the rules of special relativity..

An ensemble of elementary particles has the four vector of the addition of all its components, as for example the proton here, has as a four vector the instantaneous addition of the fourvectors of all the constituents. If hit by another proton there will be a momentum transfer and following energy and momentum conservation.

For a macroscopic object the same holds true, but it is ridiculous to add up order of $~10^{23}$ four vectors (avogadro's naumber), particularly as the velocities for macroscopic objects are very much lower than the velocity of light, and conservation of energy and momentum can be done separately, and masses for large objects are invariant and additive .

When you apply a force on a ball, the center of mass of the ball has to obey newtonian momentum and energy conservation, and all the constituent elementary particles take their share and move as the center of mass moves. You can never apply forces that will give relativistic velocity, so as to worry on the effect to the consituents closer or further away from the point of force. All these can be done with classical newtonian mechanics of rotations and translations, and classical elasticity and deformation if necessary.