# Angular momentum and the HUT

1. Apr 23, 2005

### broegger

Hi,

The operators representing the components of angular momentum are incompatible, since

$$[L_x,L_y] = i\hslash L_z$$.

If you apply the uncertainty principle to this you get:

$$\sigma_{L_x}\sigma_{L_y} \geq \tfrac{\hslash}2|\langle L_z \rangle|$$.

But what if $$\langle L_z \rangle = 0$$? Then the HUT does not prevent simultanoeus eigenstates of Lx and Ly or what?

Last edited: Apr 23, 2005
2. Apr 23, 2005

### dextercioby

Yes,it's true.U can,under certain conditions,measure eigenvalues of $\hat{L}_{x}$ and $\hat{L}_{y}$ exactly...

I think you meant

$$\left[\hat{L}_{x},\hat{L}_{y}\right]_{-}=i\hbar\hat{L}_{z}$$

Daniel.

3. Apr 23, 2005

### broegger

Oh, could you give a quick example? My book doesn't mention this: it just states that it is impossible to have simultaneous eigenstates of Lx and Ly because they don't commute. Everytime <Lz>=0 (like the ground state of hydrogen, right?) the HUT allows simultaneous eigenstates of Lx and Ly, but is it very uncommon (since my book doesn't mention it)?

Yes, edited.

4. Apr 23, 2005

### dextercioby

I think you found it yourself.H atom in the fundamental state (1s) has

$$\langle\hat{L}_{z}\rangle_{|1,0,0\rangle}=0$$

This means that they don't analyze the cases of the general uncertainty relations...That's bad... :yuck:

Why is it very uncommon.I'd say the H atom is very common...~75% of solar mass is hydrogen...

Daniel.

5. Apr 23, 2005

### broegger

Yeah, the H-atom ground state has <Lz>=0 but that doesn't necessarily imply that it is a simultaneous eigenstate of Lx and Ly: the HUT is an inequality, $$\sigma_{L_x}\sigma_{L_y} \geq 0$$, so I don't know.

6. Apr 23, 2005

### dextercioby

Compute

$$\hat{L}_{x}|1,0,0\rangle$$

$$\hat{L}_{y}|1,0,0\rangle$$

$$\hat{L}_{z}|1,0,0\rangle$$

And then the uncertainties...And verify the uncertainty relation...

Daniel.