Angular momentum

  • #1
1,444
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!!
 

Answers and Replies

  • #2
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,533
115
For your first answer, try 1*2=2, and 0*1=0.
Also 0*1=0 for the second answer.
Back to arithmetic 101.
 
Last edited:
  • #3
1,444
2
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
 

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