I am calculating the propagator of the free particle on a sphere : $K(\theta_f \phi_f t_f; \theta_i \phi_i t_i)$. The wavefunctions in this case are the spherical harmonics $Y_{lm}(\theta, \phi)=\frac{e^{im \phi}}{\sqrt{2 \pi}}{\Omega(\theta)}$. So the kernel is $$K=\sum e^{\frac{-i E_n (t_f-t_i)}{\hbar}} \frac{e^{im (\phi_f - \phi_i)}}{{2 \pi}}{\Omega(\theta_f)}\Omega(\theta_i)$$

The sum is over $m=-l, -l+1, ....., l$, and $l=0,1,2.....$

How do I calculate this sum?


The summation over the magnetic quantum number can be achieved by using the spherical Harmonics addition formula

$\sum_{m=-l}^{l}Y_{lm}(\hat{\mathbb{n_i}}) Y_{lm}(\hat{\mathbb{n_f}}) = \frac{2l+1}{4 \pi} P_l(\hat{\mathbb{n_i}}.\hat{\mathbb{n_f}})$

Where $\hat{\mathbb{n_i}}$, $\hat{\mathbb{n_f}}$ are the initial and final unit vectors on the sphere and $P_l$ are the Legendre polynomials. The remaining sum has the form:

$ K(\theta, t_f-t_i) = \sum_{l=0}^{\infty}\frac{2l+1}{4 \pi} e^{i l(l+1)(t_f-t_i)}P_l(cos(\theta))$

Where $\theta = \arccos(\hat{\mathbb{n_i}}.\hat{\mathbb{n_f}})$

(It is assumed that the energy is that of a free particle on the sphere):

$E_n =l(l+1)$.

The "closest' form of the remainig sum is by means of a fractional derivative of the propagator on the circle which can be expressed by means of the Jacobi's theta function. Please see http:Camporesi's review (equations 8.38 and 6.35) for the full proof:

$ K(\theta, t_f-t_i) = e^{i/4 (t_f-t_i)} (\frac{1}{2\pi} \frac{\partial}{\partial (\cos \theta + 1)})^{\frac{1}{2}} \frac{1}{2\pi} \Theta_3(\frac{\theta}{2}, -\frac{t_f-t_i}{\pi})$

  • $\begingroup$ What does the superscript $\frac{1}{2} $ in $\partial^{\frac{1}{2}}_{\cos{\theta}+1}$ mean in the review? The subscript I am guessing stands for the derivative wrt to $\cos{\theta}+1$ $\endgroup$ – user7757 May 28 '13 at 9:50
  • 1
    $\begingroup$ @ramanujan_dirac : It means that it is a fractional derivative. One way to understand it is to think about the Fourier transform. The derivative becomes a multiplication by the dual variable $p$ in the Fourier representation. A half-derivative is the multiplication by $p^{\frac{1}{2}}$. Now, a multiplication in the Fourier (momentum) space is a convolution in the position space. This explains the definition 8.14 of the fractional derivative in Comporesi. $\endgroup$ – David Bar Moshe May 28 '13 at 10:16

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