- #1

mmwave

- 211

- 2

Given f is an eigenfunction of both L^2 and Lz

L^2f = [lamb] f Lz f = [mu] f and L^2 = Lx^2 + Ly^2 + Lz^2 then

<L^2> = < Lx^2 + Ly^2 + Lz^2> = <Lx^2> + <Ly^2> + <Lz^2>

<L^2> = [inte]f* L^2 f dr = [inte]f* [lamb] f dr

= [lamb] [inte]f*fdr = [lamb] <f> = [lamb]

<Lz> = [inte]f* Lz^2 f dr = [inte]f* [mu]^2 f dr = [mu]^2 [inte]f*f dr

= [mu]^2

Now

<lz^2> = <L^2> - <Lx^2> - <Ly^2> substituting gives

[mu]^2 = [lamb] - <Lx^2> - <Ly^2>

or [mu] squared is less than or equal to [lamb] with equality only when <Lx^2> = <Ly^2> = zero

I would like to know if the above is correct and is this just another way of saying that Lz^2 <= to L^2 ?