91影视

This isaseries expansion in terms of the "full" collection of orthonormal Eigen functions for the Sturm-Lowville operator with periodic boundary conditions across the interval.

Consider two particles of mass are attached to the ends of a massless rigid rod of length . In the absence of potential energy, the system's energy is equal to the kinetic energy of the two particles, i.e.

$$\begin{gathered} E=2K \\ =2\left(\frac{p^2}{2m}\right) \\ =\frac{p^2}{m} \\ ...\left(\mathrm{i}\right) \end{gathered}$$

The particles are only allowed to move in a rotating direction, and the rod's length is fixed, which implies that r is always perpendicular to p, where is the momentum of one of the masses. So,the angular momentum is expressed as follows,

$\left|\mathrm{L}\right|=2\left|\mathrm{r}\right|\left|\mathrm{p}\right|$

Here is the distance from one of the particles to the center of the rod, that is $\left|\mathrm{r}\right|=\mathrm{a}/2$.

Substitute localid="1658207215926" $\frac{a}{2}$for r .

$\left|\mathrm{L}\right|=2\left(\frac{\mathrm{a}}{2}\right)p$

$$\begin{gathered} =ap{L}^2 \\ =a^2{p}^2{p}^2 \\ =\frac{L^2}{a^2} \end{gathered}$$

Substitute the above value in equation (i).

$E=\frac{L^2}{m{a}^2}$

Write the eigenvalues of ${\sout{h}}^{2}n\left(n+1\right)$ , for n= 0,1,2,.... .

$E_n=\frac{{\sout{h}}^{2}n\left(n+1\right)}{m{a}^2}$

Hence, the given equation is proved.

Since E is directly proportional to $L^2$, then the Eigen functions are just the ordinary spherical harmonics, that is:

${唯}_{nm}\left(胃,蠒\right)=Y_n^m\left(胃,蠒\right)$

Here, the degeneracy of the energy level is the number of m values for given n , that is 2n+1 .

Thus, the Eigen functions is determined as ${唯}_{nm}\left(胃,蠒\right)=Y_n^m\left(胃,蠒\right)$ and also the degeneracy of the nth energy level is 2n +1 .