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I Free particle in a central potential

  1. Apr 30, 2016 #1
    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!
  2. jcsd
  3. Apr 30, 2016 #2


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    That's just another way of specifying the eigenfunction of free space. ##\psi_{klm}## is a common eigenfunction of the set of operators ##\{H,L^2,L_z\}##, in contrast to the plane wave form which is the common eigenfunction of ##\{H,P_x,P_y\}##. It has nothing to do with the linear momentum being not defined when one is working in a non-Cartesian coordinate.
    By definition a free particle is a particle existing in a space of constant potential which lower than the particle' energy. Nevertheless, there indeed exist a situation in which there is a varying local potential along with a plane wave form as a boundary condition for the wavefunction - it's a potential scattering. In such scattering cases, typically one requires that the wavefunction long before the collision with the target takes the form of a plane wave. But as time flies, the particle approaches the target and the wavefunction around this time will deviate from that of a plane wave.
  4. Apr 30, 2016 #3


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    It can, but its eigenstates are messy superpositions of the eigenstates expressed in the klm basis.
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