Physics 402
The Eigenvalues of $L^2$ and $L_z$

  1. In studying rotational motion, we take advantage of the center-of-mass system to make life easier. Consider the two-particle system shown in the figure including the center-of-mass vector $\bf R$. For convenience we will place our origin at the center-of-mass of the system ($\bf R = 0$). Show the classical mechanical energy of the two-particle system can be written as

    \begin{displaymath} E = {1 \over 2 } \mu v^2 + V(r) \qquad {\rm where} \qquad \m... ...m_2 \over m_1 + m_2} \qquad {\rm and} \qquad v = {dr \over dt} \end{displaymath}

    and $r$ is the relative coordinate between the two particles as shown in the figure. Notice that $V(r)$ depends only on the relative coordinate.

    \includegraphics{L2Lz1.eps}

  2. Starting from the definition of the components of the angular momentum in Cartesian coordinates show that

    \begin{displaymath}[L_x,L_y]= i\hbar L_z \qquad [L_y,L_z] = i\hbar L_x \qquad [L_z,L_x] = i\hbar L_y \qquad . \end{displaymath}

  3. Show that

    \begin{displaymath} L_z L_- \vert\phi_m \rangle = (m - 1)\hbar ~ L_-\vert\phi_m\rangle \end{displaymath}

    and that successive applications of $L_-$ generate the following sequence of eigenfunctions

    \begin{displaymath} ...,\phi_{m-2}, \phi_{m-1}, \phi_m \qquad . \end{displaymath}

  4. Show that

    \begin{displaymath}[L^2,L_x]= [L^2,L_y] = 0 \qquad . \end{displaymath}

  5. Starting with the definitions of $L_+$ and $L_-$ in terms of $L_x$ and $L_y$ show that

    \begin{displaymath} L^2 = L_x^2 + L_y^2 + L_z^2 = {L_+ L_- \over 2} + {L_- L_+ \over 2} + L_z^2 \end{displaymath}

    and

    \begin{displaymath}[L_+, L_-]= 2 \hbar L_z \qquad . \end{displaymath}

    For the second relationship you might find one of the results above useful.

  6. Finally, using some of the results above show that

    \begin{displaymath} L^2 \vert\phi_{m_{min}} \rangle = \hbar^2 \kappa^2 = \le... ...bar L_z - L_z^2 \right ) \vert\phi_{m_{min}} \rangle \qquad . \end{displaymath}