# Equation calculating the amplitude of a particle

1. May 11, 2008

### Crazy Tosser

If I am right, it's supposed to be this:

When a particle moves from $$r_{1}$$ to $$r_{2}$$, the amplitude of the move can be written as:

$$\langle r_{2} | r_{1} \rangle = \frac{e^{ipr_{12}/\hbar}}{r_{12}}$$

where $$r_{1}$$ and $$r_{2}$$ are vectors of particle's position (I think?).
and $$r_{12} = r_{2}-r_{1}$$ and is the scalar( I think again?)

So my question is.. how can you substitute vectors to get a numerical value for the equation above?

2. May 12, 2008

### Crazy Tosser

Anyone? =[

3. May 12, 2008

### Redbelly98

Staff Emeritus
I'm not familiar with this equation, but it looks like you'd have to know:
The particle's momentum, p
The distance between the two points, r_12

Then just plug the numbers in.

In the exponent, that looks like the dot product of 2 vectors, so you need not only the distance r_12, you actually need the displacement vector r2-r1. Likewise, you need to know the direction of the momentum.

I think it's a scalar, |r_12|, in the denominator of your expression.

Last edited: May 12, 2008
4. May 12, 2008

### lbrits

I didn't actually recognize the expression up to now. It seems the Green's function for the free Schrodinger equation. See:

http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Lippmann-Schwinger_equation [Broken]

In this case the Green's function is actually isotropic, and so the product in the exponent is the product of a scalar and another scalar.

Last edited by a moderator: May 3, 2017
5. May 12, 2008

### Crazy Tosser

Yes, I just realized that, it was hard to see the bold in that equation (it's very small)