# Is it acceptable to use plane waves?

1. Dec 2, 2009

### RedX

In classical mechanics, scattering depends on initial momentum and position.

In quantum mechanics, the initial condition that is specified is just the momentum. But if this were strictly true, then definite momentum implies the particle is somewhere out of the lab and therefore doesn't scatter at all.

So the correct way would be to try to create a wavepacket $$\psi$$ around a certain momentum p, and once you calculate transition probabilities <f|i> between definite momentum states, integrate over the initial states i: $$\int <f|i><i|\psi> di$$

How far off are you if instead of creating a wavepacket $$\psi$$, you just use a plane wave with momentum p? Is it acceptable in experiment to do this? It seems very bizarre that you are calculating the particle as if it can be anywhere in the universe when you represent it with a plane-wave.

2. Dec 2, 2009

### Bob_for_short

An accelerator has a certain beam diameter. It is determined with diaphragms. If the target is much smaller than the beam diameter, then the wave packet from the accelerator is well approximated with a plane wave.

3. Dec 2, 2009

### RedX

So wouldn't this imply that if you had no diaphragm at all, you would get the same result if you had a diaphragm that was much larger than the target? Because isn't that what a plane wave is, having no diaphragm at all and therefore no semblance of position? So the beam could be outside the lab - in fact, has an overwhelming probability to be outside of the lab! And yet you get the same results calculating it like that as you do with a wavepacket, when the target is small?

4. Dec 2, 2009

### Bob_for_short

There are in nature accelerators much larger than any lab. As soon as D>>d is fulfilled, there is no sensitivity to D in experiment. In human practice D is always finite.

5. Dec 2, 2009

### RedX

The textbook I have tells you how to make scattering calculations involving various potentials, but the potentials seem to have been pulled out of a hat.

The book doesn't seem to take any potentials from real experiments.

So for example, take gold scattering with alpha particles. You would need to know the width of the beam of alpha particles. I looked at some pictures of the experimental set-up, and some radioactive source is put into a box with a hole cut out of it, but it doesn't say what the size of the hole is. The target is gold foil which is larger than the hole, but I guess it's the size of the atoms that matter and not the size of the sheet. So I guess it's appropriate to use the Coloumb potential of a single atom. But in general, the target is not a single atom but a clump of material - do you have to add the potentials of every single atom to get the real potential?

6. Dec 3, 2009

### alxm

Plane waves are useful approximations in some cases, and not very useful in others (e.g. what I do, molecules).

If it makes sense in the particular case, and you're careful to keep in mind that they're not normalizable etc, it's okay to do it.

7. Dec 3, 2009

### Bob_for_short

A beam of finite size is described with a superposition of plane waves with slightly different wave vectors k. If the beam proper deivergence due to differnt k is much smaller than the scattering angle in question, then we can forget about different k and use just a plane wave with some k0.

Concerning a summary potential. Classically sum of potentials may well result in zero because of mutual cancellation from distant sources.

Quantum mechanically we have to distinguish two cases: if the incident wave-length is larger than the inter-atomic distance, then the wave "feels" many atoms and you are obliged to use all of them in the Schroedinger equation. Kind of a collective effect. If the incident wave-length is much shorter than the inter-atomic distance, then the scattering off many atoms is reduced to scattering off a single atom: there is no collective effect on the scattered flux.

Gold foil is made thin on purpose: to avoid multiple scattering (scattering of scattered projectiles).

Last edited: Dec 3, 2009
8. Dec 4, 2009

### RedX

In the normal direction to the beam, the wavelength would be very large, infinite if you use a plane-wave approximation. So would this mean that if you had a very wide screen, the whole first layer of the screen would have to be taken into account in the Schrodinger equation? Does this rule about the size of wavelength and how many atoms you have to take into account come from quantum mechanics?

Also, once the beam is detected, it is generally destroyed is it not? So if you insert a decay factor into the wave-function, then the frequency spectrum will be affected in addition to the spread from creating a wave-packet. Or is this a classical phenomena because if it were quantum, then unitarity would be violated with a decay factor?

thanks all