Solving Angular Momentum Homework: <L^2>,<(L^2)^2>, etc.

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stunner5000pt
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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!
 
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Meir Achuz said:
For your first answer, try 1*2=2, and 0*1=0.
Also 0*1=0 for the second answer.
Back to arithmetic 101.

OOPS

stupid me