# Can particle collisions be represented by the Dirac delta function?

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#### jeffreyH

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##### Can particle collisions be represented by the Dirac delta function?
« on: 06/10/2015 23:56:24 »
When a particle has a definite position you get the spike in the wave showing the position. Basically a temporary wave function collapse. It is a particle, as I understand it. So if we have a collision of two particles does this show two fleeting superimposed spikes? Two Dirac deltas? Please correct me if I am talking rubbish.
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#### PmbPhy

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##### Re: Can particle collisions be represented by the Dirac delta function?
« Reply #1 on: 08/10/2015 15:40:14 »
Quote from: jeffreyH
When a particle has a definite position you get the spike in the wave showing the position. Basically a temporary wave function collapse. It is a particle, as I understand it.
Sure. You can model it that way. However, as I understand it (I think I read this in the Feynman lectures), to be exact, the wavefunction collapses to a very sharp spike. However it's not an perfect Dirac delta function. There will still be some uncertainty in position. But you can approximate it very well using the Dirac delta function. This is done very frequently in practice.

Quote from: jeffreyH
So if we have a collision of two particles does this show two fleeting superimposed spikes? Two Dirac deltas? Please correct me if I am talking rubbish.
You have to keep in mind that it's a Dirac delta function only at the moment when you measure the position. And since the particle is moving it will be a function of time. The mathematical description of measuring the position of a point particle once is as following. Let us first define some terms. We'll work in one dimension for simplicity's sake.

Let:
X = position operator
x = eigenvalue of position = the resulting number which represents the position
|x> = eigenstate of position
$$\delta$$(x - x') = <x|X|x> = $$\Psi$$(x) = Wavefunction of position

In reality the wave function will be $$\Psi$$(x, t).

Let's leave the time part out to explain this more simply because with time as a parameter it becomes more complicated and its been years since I've worked with quantum mechanics with time as a parameter. It would take half the day to review that part of quantum mechanics. However today I have other things on my schedule.

The process of measuring position is described by the expression

X|x> = x|x>

The wave function then becomes

<x|X|x'> = x$$\delta$$(x − x')

If you wish to read this part of the Feynman Lectures then go to http://bookos-z1.org/book/850372/739ddf , download the text and then go to page 20-12. Do you have a DJVU viewer? If not then the PDF version is at
http://bookos-z1.org/book/2220494/2ec0eb
« Last Edit: 08/10/2015 15:44:19 by PmbPhy »

#### jeffreyH

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##### Re: Can particle collisions be represented by the Dirac delta function?
« Reply #2 on: 09/10/2015 05:54:41 »
Thanks Pete.
Fixation on the Einstein papers is a good definition of OCD.