# Collisions in QM

1. Sep 29, 2008

### wofsy

As a beginner in QM I have a question that I could not find answered in my books.

Two particles evolve happily according to the Shroedinger equation. How can they collide? In classical mechanics they collide when their trajectories intersect and their motion is then calulated using the conservation of momentum and energy.

But in QM the wave functions just superpose. There is no collision.

Where have I gone wrong?

2. Sep 29, 2008

Staff Emeritus
You have assumed they can't collide and more or less stopped there. Of course they can. Most QM books even cover scattering.

3. Sep 29, 2008

### malawi_glenn

You also assume that particles like protons for instance must collide and react like tiny balls. But the interaction (force) between particles has a certain range, and you also have the wave function, so there is a certain probability for the two particles involved to be confined within a region where the force can act between them.

And as Vanadium_50 said, QM scattering is what you are looking for.

4. Sep 30, 2008

### wofsy

Thanks for the reply but I am still confused.

I did not assume that collisions can't occur. I just said that I could not see how they could occur from the Schroedinger equation since it is linear and its solutions will always superpose. Superposition does not allow collision.

In scattering I guess you are saying that there is a potential that leads to solutions that appear macroscopically like collisions but aren't really. They are just solutions of the Shroedinger equation.

Is this right?

5. Sep 30, 2008

### malawi_glenn

What do you mean with the concept of "collisions"? Two small balls hitting each other? You must abandon such classical pictures when doing quantum mechanics.

6. Sep 30, 2008

### wofsy

Ok then lets take this case. A batter hits a baseball for a home run. The trajectory of the ball is changed after contact with the bat. How do you explain that using the Shroedinger equation?

7. Sep 30, 2008

### malawi_glenn

8. Sep 30, 2008

### wofsy

Ok thanks

In the case of the baseball, QM must be able to explain ths homerun in terms of a change in the group velocity of the wave function. My question was how this arises from the Schroedinger equation.

I will read the lins you sent. Thanks again.

9. Sep 30, 2008

### malawi_glenn

Try to apply QM to macroscopic stuff if you want, it will not work ;-)

Those links are very good and are suitable as introductory books in QM.

10. Sep 30, 2008

### wofsy

if the shroedinger equation is right then it must apply to the baseball.
I read how classical equations are approximated through the group velocity of the wave function

11. Sep 30, 2008

### malawi_glenn

It is quite meningless to talk of a wave function of macroscopic object such as an baseball, but in principle you could do it. I'll suggest you learn QM scattering of electrons, protons etc so you learn how to do it. Then you can apply it to a particle of mass 100g ;-)

The DeBroigle wavelenght of a baseball is many times smaller the R.M.S radius of a proton...

Notice the word "approximation"...

12. Oct 7, 2008

### reilly

No, this is not right. To see this, take two particles with equal masses and no spin, moving in opposite directions, interacting by means of a 1- D "central" force(depends on the difference in coordinates, |x1 - x2|), say a potential well of finite extent. Then, of course, you can translate this problem into one of a single particle dealing with a potential well. Further, the solutions of this problem are well known, and can be found in almost any QM book ever written. Then translate back to the original problem, and see very clearly that 1. a "collision" does happen, and 2. the solutions are not linear in the coordinates of the particles, and 3. superposition is an essential feature of scattering and interactions.

If you want to be a bit more rigorous, use wave packets, which will help in elucidating the collision that occurs.

For a more general discussion, look up scattering theory which will help you to understand collisions as they are described in QM .

Regards,
Reilly Atkinson

13. Oct 8, 2008

### koolmodee

You can have a wavefunction over two particles $$\Psi(x,y)$$ and write down a potential V(x,y), which describes the interactions of this two particle state.

Note also that you can add two solutions of a Schrödinger equation which gives you also a solution again to the Schrödinger equation you dealing with.

But when you have two wave functions being solutions of two different Schrödinger equations ( one for particle x, one for particle y), then you can not add/ superpose these two in order to give you a solution to any interacting.

14. Oct 13, 2008

### wofsy

right. that makes total sense. thanks. could you give me an example of a two particle wave function? Is it the product of the wave functions of the separate particles?

Why can't the measurement problem be solved with multi-particle solutions to the Shrodinger equation?

15. Oct 13, 2008

### wofsy

oh yeah it can not be the product because the potential must involve both of them

16. Oct 13, 2008

### ice109

why well? don't you mean potential barrier? anyway what are you using for the potential