- #1
RedX
- 970
- 3
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 [tex] \psi [/tex] around a certain momentum p, and once you calculate transition probabilities <f|i> between definite momentum states, integrate over the initial states i: [tex]\int <f|i><i|\psi> di [/tex]
How far off are you if instead of creating a wavepacket [tex]\psi[/tex], 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.
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 [tex] \psi [/tex] around a certain momentum p, and once you calculate transition probabilities <f|i> between definite momentum states, integrate over the initial states i: [tex]\int <f|i><i|\psi> di [/tex]
How far off are you if instead of creating a wavepacket [tex]\psi[/tex], 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.