Perturbation Theory with Symmetric Rotator

[Marthius]

Homework Statement

Given the Hamiltonian and perturbation below, what are the energy shifts for the states with l=1
Given [tex]H_{0}=(L^2)/(2I)[/tex]
[tex]H_{1}=E_{1}cos\vartheta[/tex]

Homework Equations

L= r x P

The Attempt at a Solution

in order to find the first order correction to the energy i used:
[tex]<lm|H_{1}|lm>[/tex]
substituting in 1 for l
But now i am stuck because I don't really understand what the cos theta is doing in the Hamiltonian, or how to solve for this equation.

I was thinking I could build the perturbation out of operators i already knew, but i couldn't find a way to build this particular one.

 

[Felicity]

I am struggling with the same problem, I know the spherical harmonics Ylm are involved and I think we can use [tex]\Delta [/tex] E = < l,m|Ecos [tex]\theta[/tex] |l,m> because Ecos[tex]\theta[/tex] is the hamiltonian here but I'm not sure how to put it together.

If you've figured it out please let me know!

-Felicity

 

[turin]

Do you know how to write the expectation value as an integral?

 

[Felicity]

I would write E[tex]\int[/tex](Y_lm)²cos[tex]\theta[/tex] but i know there is more to it than that, a coefficient or something I am missing. Also I'm not sure about the bounds.----wait, just realized that I am looking for polar or spherical coordinates!

 

[turin]

So often, sloppy integrals are the culprit for problems with understanding. I'm not pointing my finger specifically at you, Felicity, and you may have just abreviated the full expression that you know on paper, but just in case:

[tex]
\langle\mathcal{O}(\theta,\phi)\rangle_{lm}=\frac{\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}d\phi{}d\theta{}\sin\theta{}\left|Y_{lm}(\theta,\phi)\right|^2\mathcal{O}(\theta,\phi)}{\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}d\phi{}d\theta{}\sin\theta{}\left|Y_{lm}(\theta,\phi)\right|^2}
[/tex]

The denominator takes care of any normalization issues that you might be worried about. You will also probably find the orthogonality of the spherical harmonics (and trig functions) useful (but, for the trig functions, be careful about the integration limits).

 

[Felicity]

Thank you for your help! You're right, I often make mistakes when quickly jotting down an integral without thinking, and the normalization tecnique will help me alot.

 

[turin]

Whoops! I had better fix this. What I wrote works for your specific case, but it is not true in general. I should have just said:

[tex]
\langle\mathcal{O}\rangle_{lm}=\frac{\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}d\phi{}d\theta{}\sin\theta{}Y_{lm}(\theta,\phi)^*\mathcal{O}Y_{lm}(\theta,\phi)}{\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}d\phi{}d\theta{}\sin\theta{}\left|Y_{lm}(\theta,\phi)\right|^2}
[/tex]

where [itex]\mathcal{O}[/itex] is understood on the R.H.S. to be represented appropriately. Now who's being sloppy?