Equation calculating the amplitude of a particle

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

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

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

where [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are vectors of particle's position (I think?).
and [tex]r_{12} = r_{2}-r_{1}[/tex] 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?
 
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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.

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

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

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.
 
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Redbelly98 said:
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.

Edit added:
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.

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