- #1
stunner5000pt
- 1,461
- 2
Homework Statement
For the non stationary state
[tex] \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}[/tex]
find [itex] <L^2>,<(L^2)^2>,<L_{z}>,<L_{z}^2>,\Delta L^2, \Delta L_{z}[/itex]
Homework Equations
[tex] <L^2>=\hbar^2 l(l+1)[/tex]
[tex] <(L^2)^2>=(\hbar^2 l(l+1))^2[/tex]
[tex] <L_{z}>=\hbar m_{l}[/tex]
[tex] <L_{z}^2>=(\hbar m_{l})^2[/tex]
[tex] \Delta x = \sqrt{<x^2>-<x>^2} [/tex]
The Attempt at a Solution
[tex]<L^2>=\frac{\hbar^2}{2} \left(0(0+1) + 1(1+1)\right)= \frac{3}{2} \hbar^2 [/tex]
the answer is supposed to be hbar^2 ... what am i doing wrong...
[tex] <(L^2)^2> = \frac{\hbar^4}{4} \left((0(0+1))^2+(1(1+1))^2\right) =\frac{5}{4} \hbar^2[/tex]
[tex] <L_{z}> = \frac{\hbar}{2} (0+0) = 0 [/tex]
[tex] <L_{z}^2} = \frac{hbar^2}{4} (0) = 0 [/tex]
are the values right... i fear that i am sorely mistaken about how to calculate the expectation values
any help would be greatly appreciated!