Solving Angular Momentum in QM: Find Probability of L^2 Values

AI Thread Summary
The discussion revolves around calculating the probability of measuring specific values of L^2 for a particle on a sphere, represented by the wave function \Psi(\theta, \phi) = Ne^{\cos{\theta}}. Participants suggest decomposing the wave function into angular momentum eigenstates, specifically spherical harmonics, and recognizing that the wave function's independence from \phi indicates m = 0. The eigenvalues are identified as l(l+1)\hbar, and the probability calculation involves the coefficients |c_{l}|^2 = |<\Psi|Y_l^0>|^2. One participant successfully solved the complicated integrals through integration by parts, concluding the problem.
Izbitzer
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Hello,

I'm trying to solve a problem dealing with finding the probability of measuring certain values of L^2 for a particle.

The particle is on a sphere and is in the state \Psi (\theta , \phi) = Ne^{\\cos{\theta }}.

I don't know quite how to start, I guess I have to decompose the wave function in eigenfunctions for L^2, and then find the corresponding eigenvalues, and form that find the probability of measuring that particular eigenvalue, but like I said, I don't really know where to start.

Does anybody have any pointers?

Thanks!
 
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Nope, you have to do you wrote. That is try to write your wavefunction as a linear combination of angular momentum eigenstates.

Daniel.
 
Thanks for the reply!

I've worked some more, and I think I've made some progress :smile:

The eigenfunctions of \^{L}^2 are spherical harmonics, but since the wave function is independent of \phi, m = 0, i.e \Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0.
The eigenvalues are l(l+1)\hbar. So I just see which value of l corresponds with the values I'm supposed to find the probability of. The probability is calculated with: |c_{l}|^2 = |&lt;\Psi|Y_l^0&gt;|^2. This is where I'm stuck at the moment. The integrals are really complicated and I can't find them in any books, is there any easier way to calculate this?

Thanks!
 
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Izbitzer said:
Thanks for the reply!

I've worked some more, and I think I've made some progress :smile:

The eigenfunctions of \^{L}^2 are spherical harmonics, but since the wave function is independent of \phi, m = 0, i.e \Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0.
The eigenvalues are l(l+1)\hbar. So I just see which value of l corresponds with the values I'm supposed to find the probability of. The probability is calculated with: |c_{l}|^2 = |&lt;\Psi|Y_l^0&gt;|^2. This is where I'm stuck at the moment. The integrals are really complicated and I can't find them in any books, is there any easier way to calculate this?

Thanks!
A suggestion: I don't recall the exact form of theY^l_0, but aren't they simple to write in terms of (cos \theta)^n ? Then you simply have to Taylor expand e^{cos \theta)}. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the Y^l_0 (cos (\theta), then you can find the coeeficients c_l without doing a single integral.

Patrick
 
nrqed said:
A suggestion: I don't recall the exact form of theY^l_0, but aren't they simple to write in terms of (cos \theta)^n ? Then you simply have to Taylor expand e^{cos \theta)}. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the Y^l_0 (cos (\theta), then you can find the coeeficients c_l without doing a single integral.

Patrick

That probably would have worked, but I managed to solve the damn integrals before I read your reply :smile: All it took was some (a lot of) integration by parts, and it turned out ok. Now I can finally rest. :smile:

Thanks for taking the time guys!
 
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