Eigenvalues of L^2 and Lz^2

  • Thread starter mmwave
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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

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 ?
 
211
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.
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,475
20
Originally posted by mmwave
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


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

Everything else looks OK.
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,475
20
Re: further relations between eigenvalues

Originally posted by mmwave
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.
See page 71 of these notes:

http://fafalone.hypermart.net/lectures.pdf
 

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