# Angular momentum

1. Apr 9, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
For the non stationary state

$$\Psi = \frac{1}{\sqrt{2}} \left(Psi_{100}+\Psi_{110}\right) = \frac{1}{\sqrt{2}} \left(R_{10}Y_{00}e^{-iE_{10}t/\hbar}+R_{11} Y_{10}e^{-iE_{11}t/\hbar}$$

find $<L^2>,<(L^2)^2>,<L_{z}>,<L_{z}^2>,\Delta L^2, \Delta L_{z}$

2. Relevant equations
$$<L^2>=\hbar^2 l(l+1)$$
$$<(L^2)^2>=(\hbar^2 l(l+1))^2$$
$$<L_{z}>=\hbar m_{l}$$
$$<L_{z}^2>=(\hbar m_{l})^2$$
$$\Delta x = \sqrt{<x^2>-<x>^2}$$

3. The attempt at a solution
$$<L^2>=\frac{\hbar^2}{2} \left(0(0+1) + 1(1+1)\right)= \frac{3}{2} \hbar^2$$
the answer is supposed to be hbar^2 ... what am i doing wrong...

$$<(L^2)^2> = \frac{\hbar^4}{4} \left((0(0+1))^2+(1(1+1))^2\right) =\frac{5}{4} \hbar^2$$

$$<L_{z}> = \frac{\hbar}{2} (0+0) = 0$$

$$<L_{z}^2} = \frac{hbar^2}{4} (0) = 0$$

are the values right... i fear that i am sorely mistaken about how to calculate the expectation values

any help would be greatly appreciated!!

2. Apr 10, 2007

### Meir Achuz

Also 0*1=0 for the second answer.
Back to arithmetic 101.

Last edited: Apr 10, 2007
3. Apr 10, 2007

OOPS

stupid me