- #1

amjad-sh

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When we are dealing with a free particle in spherical coordinates,the position eigenfunction of the free particle is [itex]\psi_{klm}(r,\phi,φ)=\langle r\phiφ|klm\rangle=J_{l}(kr)Y_{lm}(\phi,φ)[/itex]. Here appears that the wavefunction describe a free particle of energy Ek of well-defined angular momentum l ,but little information about linear momentum p.

When we are dealing with a free particle in Cartesian coordinates, the position eigenfunction of a free particle is a plane wave∝ [itex]e^{i\vec k\cdot\vec r}[/itex](supposing that V(r) is separable V(r)= V(x) +V(y)+V(z)) which describe a free particle with well-defined linear momentum but gives little information about angular momentum.

But we can write [itex]e^{i\vec k\cdot\vec r} =\sum_{l=0}^\infty\sum_{m=-l}^{l}a_{lm}J_{l}(kr)Y_{lm}(\phi,φ)[/itex]

Can we deduce from this that the linear momentum can't be defined for a free particle moving in a central potential V(r)? I mean can we say that a free particle of linear momentum p is moving in a spherical central potential field?

same question for a free particle moving in a non spherical central potential, can we say that a particle of angular momentum l is moving in a potential V(r)=V(x)+V(y)+V(z)?

Is this related to uncertainty or somthing...

can anybody explain more to me about this? thanks!