# $\sum_{m=-l, …,l; l=0,1,2,..} e^{\frac{-i E_l (t_f-t_i)}{\hbar}} Y_{lm}(\phi_f,\theta_f)Y_{lm}(\phi_i, \theta_i)$

I have encountered this series while trying to calculate the path integral of a free particle on a sphere. The sum is $$K=\sum e^{\frac{-i E_l (t_f-t_i)}{\hbar}} Y_{lm}(\phi_f,\theta_f)Y_{lm}(\phi_i, \theta_i)$$ $E_l=\frac{\hbar^2}{2I}l(l+1)$, $\frac{\hbar^2}{2I}$ is a constant.

The sum over m is from $-m$ to $m$, and that over $l$ is over the natural numbers $1,2,3...$. $Y's$ are the spherical harmonics on two points of a sphere. The m sum can be done using the addition theorem for spherical harmonics

After doing this, the answer is $K(\theta, t_f-t_i) = \sum_{l=0}^{\infty}\frac{2l+1}{4 \pi} e^{\frac{-\hbar^2}{2I} l(l+1)(t_f-t_i)}P_l(cos(\theta))$.

$\theta$ is the angle between the two position vectors i.e. $\cos{\theta}=\hat{n_f}\cdot \hat{n_i}$, where $n_f, n_1$ are the vectors to the two points.

How do I do the sum over $l$? I am stumped and no idea how to proceed.

Note: I had asked this question on Physics SE here, and got a response, but for the above sum the answer gives reference to a review, which is much beyond my understanding. I am wondering if there is a cute and quick method to do it, or a simplified summary of the method used in the review.

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Is $\theta=\theta_f-\theta_i$? –  Ron Gordon May 28 '13 at 11:09
@RonGordon: No. It is actually the angle between the two position vectors: i.e. if $(\theta_f, \phi_f)$ and $(\theta_i, \phi_i)$ correspond to $\hat{n_f}$ and $\hat{n_i}$, then $\cos{\theta}=\hat{n_f} \cdot \hat{n_i}$ –  ramanujan_dirac May 28 '13 at 11:21
In Math.SE, you have to make that explicit. –  Ron Gordon May 28 '13 at 11:23
@ramanujan_dirac Which sum do you want to compute? Over $m$ or over $\ell$? –  O.L. May 28 '13 at 11:45
I want to compute the $l$ sum, the $m$ sum has been done using the addition theorem for spherical harmonics. –  ramanujan_dirac May 28 '13 at 11:46