# Angular momentum

1. Feb 20, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
For the spherical solution of the Schrodinger equation in spherical coordinates given the superposition of spherical harmonic functions

$$\frac{1}{\sqrt{14}} (Y_{1,-1}+ 2Y_{1,0}+3Y_{1,1})$$

calculate $<\hat{L_{z}}>$ and $\Delta L_{z}$

2. The attempt at a solution

now from my textbook (brehm and mullin)
$$<\hat{L_{z}}> = \hbar m_{l}$$
$$<\hat{L_{z}}> = \frac{\hbar}{14} (-1 + 4(0) +9(1)) = \frac{8}{14} \hbar = \frac{4}{7} \hbar$$

while $$<L_{z}^2> = (\hbar m_{l})^2$$

this implies that the uncertainty in the Z component of the angular momentum $\Delta L_{z} =0$

but i was marked wrong in my assignment for this...

am i missing something

is there a difference between $<\hat{L_{z}}>$ and $<L_{z}> [/tex]? thanks in advance for any input 2. Feb 20, 2007 ### Meir Achuz <L_z^2>=(+1+0+9)/14=5/7. Thus, you should get Delta L_z=3/7. 3. Feb 21, 2007 ### dextercioby How did you get that $$\langle L_{z}^2\rangle = (\hbar m_{l})^2$$ ? 4. Feb 21, 2007 ### stunner5000pt my textbook says so.... also $$<L_{z}^2>=\int \Psi^{*}_{nlm}\left(\frac{\hbar}{i}\frac{\partial}{\partial\phi}\right)\left(\frac{\hbar}{i}\frac{\partial}{\partial\phi}\right)\Psi_{nlm} d\tau = (\hbar m)^2$$ Last edited: Feb 21, 2007 5. Feb 22, 2007 ### dextercioby Yes, but in your case the state is no longer [itex] \langle r, \theta, \varphi|n, l, m \rangle$ , but a linear combination of spherical harmonics. So blindly using a fomula in the book is a wrong decision...

6. Feb 22, 2007

### stunner5000pt

... im not sure how to proceed then...

do i 'prove' it???

thanks for the help so far...

but could you look at this thread of mine... its in more of ugent need ...