Angular momentum of a particle in a spherically symmetric potential

[stunner5000pt]

Homework Statement

A particle in a spherically symmetric potential is in a state described by the wavepacked

$$\psi (x,y,z) = C (xy+yz+zx)e^{-alpha r^2}$$

What is the probability that a measurement of the square of the angular mometum yields zero?
What is the probability that it yields [tex]6\hbar^2 [/itex]?<br /> If the value of l is found to be 2. what are the relative probabilities of m=-2,-1,0,1,2<br /> <br /> <b>2. The attempt at a solution</b><br /> <br /> i think the first part is simply aking to calculate $<L^2>$<br /> <br /> but the carteisna coords are throwing me off... Should i convert to spherical polars?? Till now whenever the angular momentum L^2 and Lz were required, they were gotten using<br /> $$\hat{L^2} \psi_{nlm_{l}} = l(l+1) \psi_{nlm_{l}}$$<br /> <br /> really from the spherical harmonics... however conversion to spherical polars doesn't yield any familiar spherical harmonic either.<br /> <br /> can it written in a way that yields familiar spherical harmonics, however??[/tex]

 

[Dick]

Yes. Convert to spherical coordinates. You won't necessarily get a spherical harmonic but you can decompose it into spherical harmonics in the usual way you split a wavefunction relative to an orthonormal basis.