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Angular Momentum(Q&M)

  • Thread starter Winzer
  • Start date
  • #1
598
0

Homework Statement


What are the possible measurements for Lz



Homework Equations


[tex] \psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\phi) sin(\theta) [/tex]
probability Lz quantum


The Attempt at a Solution


Well I'm sure I can expand [tex] sin(\phi)= \frac{e^{i \phi}-e^{-i \phi}}{2 i}[/tex]
Getting m=1,-1.
[tex] \psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\theta) \frac{e^{i \phi}-e^{-i \phi}}{2 i}[/tex]
Should the the probability be the coefficents mod squared?
 

Answers and Replies

  • #2
180
4
Yes, but you have to keep track of normalization; the coefficients should be those of normalized eigenfunctions of Lz. I suggest writing psi in terms of spherical harmonics.
 
  • #3
1,482
3
you are working with two quantum numbers, If I remember correctly you have add up the normalized coefficients squared by increasing their orbital quantum number l up to m.
 
  • #4
598
0
ok. So:
[tex]\psi=\frac{\sqrt{2}i}{2}\left(Y_{1,1}+Y_{1,-1}\right)[/tex] with everything sorted
I have equal probability of measuring [tex]\pm \hbar[/tex] for [tex]L_z[/tex], right?
 
  • #5
598
0
Wait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term [tex]Y_{1,0}[/tex] and then normalize that. That makes more sense.
 
  • #6
malawi_glenn
Science Advisor
Homework Helper
4,786
22
You don't need a Y(1,0) A priori.

If your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.

You also check the normalizability of your result in post #4 by simply integrating it over [itex]r^2 d\Omega[/itex] and see if you indeed get 1.

Y(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.
 
  • #7
598
0
You don't need a Y(1,0) A priori.

If your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.

You also check the normalizability of your result in post #4 by simply integrating it over [itex]r^2 d\Omega[/itex] and see if you indeed get 1.

Y(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.
ok.
In post four the coefficents work out since[tex]\Sigma |c_k|^2=1[/tex]
 
  • #8
diazona
Homework Helper
2,175
6
Wait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term [tex]Y_{1,0}[/tex] and then normalize that. That makes more sense.
You could consider the coefficient of [tex]Y^1_0[/tex] to be 0 ;-)
 

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