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Particles can be represented as wave packet. So how do particles get scattered? Waves superimpose on one another, they don't bounce off of on one another.

It can be seen from picture there is a difference between two.

enter image description here

This illustration shows waves remain unchanged after superposition.

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  • $\begingroup$ your diagram A is not correct, even for water waves. When two waves meet interference happens , they do not go thr ough unscathed unless there is no interaction , as with electromagnetic waves in vacuum. see the water: youtube.com/watch?v=fjaPGkOX-wo $\endgroup$ – anna v Oct 31 '14 at 14:31
  • $\begingroup$ Interference happens when the two meet each other. Afterwards they are on their way as before. $\endgroup$ – Self-Made Man Oct 31 '14 at 15:14
  • $\begingroup$ I disagree, he video shows a different wave front than would have existed if there were no "scattering" all the way out $\endgroup$ – anna v Oct 31 '14 at 16:20
  • $\begingroup$ acs.psu.edu/drussell/Demos/superposition/pulses.gif @annav $\endgroup$ – Self-Made Man Oct 31 '14 at 16:22
  • $\begingroup$ this is a computer simulation, one dimensional at that, and I do not know the functions. The link is real waves in water. $\endgroup$ – anna v Oct 31 '14 at 19:34
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The word "particle" is being used in two different senses here. First let's talk about that, and then we'll talk about how we can think of scattering via wave packets.

When you find a way to represent a particle by a wave packet, you are finding a way to represent the particle as a localized entity. The formalism works well for some things: we can find, for example, the group velocity and interpret that as the speed of the particle. But we always know that what we are finding is a wave function, which we interpret as the probability amplitude for an interaction to occur at a particular place. Taken in that sense, the wave packet does not really represent the particle itself.

But as you point out, when we say "particle" we think of collisions. We like to think of objects bouncing off of one another. This seems to have little to do with waves. But the concepts of "wave packet" and "bouncing particle" can be reconciled. There are two ways to think about it.

The first is to start with a wave packet, and watch what happens to it when it encounters another object, often represented by a fixed static potential. The initially localized, coherent wave packet becomes distorted as it passes by the potential. In some cases, the distortion is such that the wave that leaves the vicinity of the object is again a wave packet, whose group velocity is the same as the classically-predicted particle velocity. The wave packet has scattered. In some cases, especially with complex potentials like that which would represent a crystal lattice, the outgoing wave would have components going out in many different directions. The particle still scatters, but we can't identify one particular direction that any given particle will go.

Here's a second way of thinking. Quantum mechanics provides another way of looking at collisions. In this picture (called "second quantization"), a collision goes as follows: The original particle is destroyed, and a new particle is created. The new particle can have a different momentum and energy than the original if the scattering object takes up (or gives up) the difference, assuring energy and momentum are conserved. In this picture, the collision really looks like a bounce.

The connection between the two pictures is that it is the wave function, and its interaction with the scattering potential, that determines where and how the bounces can occur. So you can think of a collision as a particle represented by a wave packet getting distorted by the potential, or you can think of a collision as a particle, whose probable position is represented by a wave packet, bouncing (destroy - create - conserve $E$ and $p$) off of the potential. Hope that helps.

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I'm going to try to give a very concise answer. Your illustration and presumably your intuitive picture are predicated on the superposition principle. You're thinking that if each wave packet is separately a solution of the wave equation, then their sum will be as well.

That is not the case. The superposition principle is a property of linear wave equations, such as those describing a free theory (i.e., with no interactions and therefore no scattering). Interactions are typically modeled in field theory as terms of higher than quadratic order in the Lagrangian which then show up as nonlinearities in the wave equation. E.g. a typical equation of motion for a phi fourth classical field is

$$ \square \phi +m^2 \phi + \lambda \phi^3 = 0$$

which, as you may readily verify, does not satisfy the superposition principle.

In other words, you're correct in one sense: if the superposition principle holds there can be no nontrivial interaction. If you want scattering, you must violate the superposition principle, and that is typically done by adding nonlinear terms to the Lagrangian.

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A single particle, say in one dimension, has a wavefunction $\psi(x_1)$. You can take for it's initial condition a wavepacket if you like. Or two wavepackets heading toward each other. But it is still only one particle.

Now if you want to have a two particle collision, you have to have two particles! So you need two coordinates and a wavefunction to describe both coordinates ($\psi(x_1,x_2)$) at the same time. This is different then superimposing two wavepackets for a single particle.

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