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

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Homework Help Overview

The discussion revolves around the measurement probabilities of angular momentum operators Lz and L^2 in quantum mechanics, specifically in the context of a wavefunction proportional to sin^2(theta). Participants are exploring the implications of this wavefunction on the possible measurement outcomes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the possible values of L^2 and Lz based on the given wavefunction, while expressing uncertainty about calculating the probabilities. Some participants question the lack of phi dependence in the wavefunction and discuss the implications of this on the problem. Others suggest using spherical harmonics to express sin^2(theta) and explore linear combinations of these functions.

Discussion Status

Participants are actively engaging with the problem, offering hints and exploring different interpretations of the wavefunction. There is recognition of the need to express the wavefunction in terms of spherical harmonics, and some guidance has been provided regarding the linear combinations necessary for this representation.

Contextual Notes

There is a noted assumption regarding the wavefunction's dependence on the angular variable phi, which is under discussion. The original poster also mentions attempts to use ladder operators and matrix representations, indicating a complex understanding of the problem's mathematical framework.

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.

[tex]-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1[/tex]

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.

[tex]-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1[/tex]

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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