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Eigenvalues of L^2 and Lz^2

  1. Nov 8, 2003 #1
    I'm trying to show the relation between L^2 and Lz where L is total angular momentum and Lz is the z component.

    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


    <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 ?
  2. jcsd
  3. Nov 8, 2003 #2
    further relations between eigenvalues

    By using the ladder operator L+ and L- that raise or lower the angular momentum by hbar and considering that Lz cannot be less than zero nor more than some finite value lhbar you can show that
    L^2f = hbar^2 * l * (l+1) f and Lz f = hbar m f
    where l = 0,1/2, 1,... and m = -l, -l+1, ... l-1, l

    comparing L^2f and Lz f means
    l(l+1) <= m^2 <= l^2
    with equality only for l = m = 0.

    Asked to comment the implications of this I can only say that Lz^2 is always less than L^2 meaning you always have some Lx and Ly if you have any Lz. Is there more to this?

    This result is supposed to be enforced by the uncertainty relation
    [sig]Lx * [sig]Ly >= hbar/2 * |<Lz>|

    I can't see how to relate the two let alone demonstrate it. Suggestions on how to relate the two would be much appreciated.
  4. Nov 10, 2003 #3

    Tom Mattson

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    Not quite: You have to integrate over all of R3, not just the radial coordinate.

    Everything else looks OK.
  5. Nov 10, 2003 #4

    Tom Mattson

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    Re: further relations between eigenvalues

    See page 71 of these notes:

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