Understanding Angular Momentum Measurements: Equations and Solutions

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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?
 
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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.
 
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.
 
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?
 
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 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.
 
malawi_glenn said:
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]
 
Winzer said:
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 ;-)