Question

Two particles of mass mare attached to the ends of a massless rigid rod of length a. The system is free to rotate in three dimensions about the center (but the center point itself is fixed).

(a) Show that the allowed energies of this rigid rotor are

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}\mathbf{n}\left(n+1\right)}{{\mathbf{ma}}^{\mathbf{2}}}$, for n=0,1,2,...

Hint: First express the (classical) energy in terms of the total angular momentum.

(b) What are the normalized Eigen functions for this system? What is the degeneracy of the${\mathit{n}}^{\mathbf{t}\mathbf{h}}$energy level?

Short answer

1. It is shown that $E_{\mathit{n}}\mathit{=}\frac{{\sout{\mathit{h}}}^{\mathit{2}}\mathit{n}\mathit{(}\mathit{n}\mathit{+}\mathit{1}\mathit{)}}{\mathit{m}{\mathit{a}}^{\mathit{2}}}$.
2. The Eigen functions is ${\Psi }_{nm}\left(\theta ,\varphi \right)=Y_n^m\left(\theta ,\varphi \right)$. The degeneracy of the nth energy level is 2n+1 .

Step-by-step solution

Step 1: Definition of Normalized eigen functions

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.

(a) Verification of the given equation

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.

(b) Determination the Normalized eigen functions

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

${\Psi }_{nm}\left(\theta ,\varphi \right)=Y_n^m\left(\theta ,\varphi \right)$

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 ${\Psi }_{nm}\left(\theta ,\varphi \right)=Y_n^m\left(\theta ,\varphi \right)$ and also the degeneracy of the nth energy level is 2n +1 .