Is there a relation between the eigenvalues of L^2 and Lz^2?

In summary, the equation for Lz^2 is less than or equal to the equation for L^2 when Ly^2 and Lx^2 are both zero. This relationship is enforced by the uncertainty relation.
  • #1
mmwave
647
2
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 ?
 
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  • #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.
 
  • #3
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.
 
  • #4


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
 

1. What are eigenvalues of L^2 and Lz^2?

The eigenvalues of L^2 and Lz^2 are the possible values that can be obtained when measuring the total angular momentum (L^2) and the component of angular momentum along the z-axis (Lz^2) for a quantum mechanical system. They represent the quantized energy levels of the system.

2. How are eigenvalues of L^2 and Lz^2 related?

The eigenvalues of L^2 and Lz^2 are related through the quantum mechanical operator Lz, which represents the component of angular momentum along the z-axis. The eigenvalues of Lz are the possible values that can be obtained when measuring the angular momentum along the z-axis, and they are also the eigenvalues of Lz^2.

3. Why are the eigenvalues of L^2 and Lz^2 important in quantum mechanics?

The eigenvalues of L^2 and Lz^2 are important in quantum mechanics because they provide information about the energy levels and angular momentum of a quantum mechanical system. They also help in determining the possible outcomes of measurements and predicting the behavior of particles at the quantum level.

4. How do you calculate the eigenvalues of L^2 and Lz^2?

The eigenvalues of L^2 and Lz^2 can be calculated by solving the corresponding eigenvalue equations, which involve the quantum mechanical operators L^2 and Lz. These equations can be solved using mathematical techniques such as diagonalization or using the ladder operator method.

5. Can the eigenvalues of L^2 and Lz^2 have any value?

No, the eigenvalues of L^2 and Lz^2 are quantized, meaning they can only take on certain discrete values. These values are determined by the quantum numbers associated with the system, such as the principal quantum number and the magnetic quantum number. This is a fundamental aspect of quantum mechanics and helps explain the discrete energy levels observed in atoms and other quantum systems.

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