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

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Izbitzer
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Hello,

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

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

I don't know quite how to start, I guess I have to decompose the wave function in eigenfunctions for [tex]L^2[/tex], 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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Thanks for the reply!

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

The eigenfunctions of [tex]\^{L}^2[/tex] are spherical harmonics, but since the wave function is independent of [tex]\phi[/tex], m = 0, i.e [tex]\Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0[/tex].
The eigenvalues are [tex]l(l+1)\hbar[/tex]. So I just see which value of [tex]l[/tex] corresponds with the values I'm supposed to find the probability of. The probability is calculated with: [tex]|c_{l}|^2 = |<\Psi|Y_l^0>|^2[/tex]. 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 [tex]\^{L}^2[/tex] are spherical harmonics, but since the wave function is independent of [tex]\phi[/tex], m = 0, i.e [tex]\Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0[/tex].
The eigenvalues are [tex]l(l+1)\hbar[/tex]. So I just see which value of [tex]l[/tex] corresponds with the values I'm supposed to find the probability of. The probability is calculated with: [tex]|c_{l}|^2 = |<\Psi|Y_l^0>|^2[/tex]. 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 the[itex]Y^l_0[/itex], but aren't they simple to write in terms of [itex](cos \theta)^n[/itex] ? Then you simply have to Taylor expand [itex]e^{cos \theta)}[/itex]. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the [itex]Y^l_0 (cos (\theta)[/itex], 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 the[itex]Y^l_0[/itex], but aren't they simple to write in terms of [itex](cos \theta)^n[/itex] ? Then you simply have to Taylor expand [itex]e^{cos \theta)}[/itex]. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the [itex]Y^l_0 (cos (\theta)[/itex], 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!