What Are the Measurement Probabilities for Lz and L^2 in Quantum Mechanics?

rabbit44
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Homework Statement


A system's wavefunction is proportional to sin^2p. What are the possible results of measurements of Lz and L^2? Give the probabilities of each possible outcome.

I'm using p for theta and q for phi.

Homework Equations


The Attempt at a Solution


So I believe that the value of L^2 is 6 (as l is 2), and the possible values of Lz are -2 and +2. But I can't find the probabilities. I tried:

|psi> = a|2,2> + b|2,-2>

where the kets are |l,m> and the spherical harmonice are Y(l,m)

<2,2|psi> = int dpdq <2,2|p,q><p,q|psi>
<2,2|psi> = int dpdq Y(2,2)* [aY(2,2) + bY(2,-2)]

And then inserted the spherical harmonic expressions in, but I just get a=a, so I guess this isn't the way.

I also thought about using the ladder operators, applying L+|psi> = 2a|2, -1> but I don't think this is helpful.

Any help wouldbe much appreciated!

EDIT: Another thing I tried was saying Lz|l,m>=m|l,m>. So I thought if I had a matrix representation of Lz, I could maybe have

Lz (a*, b*) = 2(a*, -b*)

Where ( , ) is a column matrix. But I don't know the matrix representation of Lz or what basis it should be in. Please help!
 
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the wavefunction has NO phi dependence?
 
malawi_glenn said:
the wavefunction has NO phi dependence?

Yeah that was my first reaction. But I don't think it is possible to have no phi dependence and be proportional to sin^2, so I assumed that it meant it was proportional to sin^2 as well as some function of phi. I may be wrong though?
 
The spherical harmonics are basis functions for angular functions, so one should be able to find a linear combination of them to build up sin^2 theta.

Now if I give you this hint, you should be able to do it.

-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1

Also recall that one of the S. harm's does not depend on any angle at all.
 
malawi_glenn said:
The spherical harmonics are basis functions for angular functions, so one should be able to find a linear combination of them to build up sin^2 theta.

Now if I give you this hint, you should be able to do it.

-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1

Also recall that one of the S. harm's does not depend on any angle at all.

Ahhhh I forgot about that one.

I see now, I just need Y(2,0) and then need to subtract an amount of Y(0,0) to get rid of the constant, and normalise.

Thank you!
 
rabbit44 said:
Ahhhh I forgot about that one.

I see now, I just need Y(2,0) and then need to subtract an amount of Y(0,0) to get rid of the constant, and normalise.

Thank you!

correct observation :-)

Good luck
 
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