# Angular Momentum and Expectation Values

1. Aug 2, 2011

### hc91

Can anyone explain to me why the only time that the expectation of L^2 operator and the expectation value of L_3^2 are equal only when there is no angular dependence? And what does this mean? Does this have something to do with being restricted to the z-axis which is what L_3 is associated with? Thanks

2. Aug 2, 2011

### LostConjugate

Angular momentum without angular dependence is just momentum right? So both are equal ... to zero. Just a quick perspective on my part, sorry if that is the wrong answer.

3. Aug 2, 2011

### Bill_K

Because <L2> = <Lx2> + <Ly2> + <Lz2>, and <Lx2 + Ly2> is non zero.

Back in the early days of quantum mechanics, when people were stuck on the idea that subatomic particles should be thought of as little balls running around in definite orbits, they tried to come up with various mechanistic models to "explain" this fact. In the vector model of the atom, L is a vector of length √l(l+1) inclined at whatever angle is necessary to make its projection on the z-axis come out m, and L then precesses around the z-axis like a wobbling top.

Or since Lx, Ly and Lz do not commute, maybe it had something to do with the Heisenberg uncertainty principle. Or, when group theory came into prominence, it was because L2 was the Casimir operator of the three-dimensional rotation group.

Actually it's just because <L2> = <Lx2> + <Ly2> + <Lz2>, and <Lx2 + Ly2> is non zero!

4. Aug 2, 2011

### LostConjugate

I just had a thought. In order to have 100% of your angular momentum in the z direction the particle must have an exact position on the z axis. However it is free to be wherever on the X-Y plane and all the particle's momentum is also on the X-Y plane. Since P_x and P_y both commute with z this scenario seems logical.

5. Aug 22, 2011

### KaL EL

I have a question of my own, regarding angular momentum. Suppose that L=3 (that is the quantum number). How can I show that <L$_{x}$>+<L$_{y}$>+<L$_{z}$> $\leq$ 3$\sqrt{3}$h(bar) ?The mean value is taken with an arbitrary $\psi$, not necessarily with an eigenstate of L$_{z}$,L$^{2}$, i.e. |l,m>...

Last edited: Aug 22, 2011