Let the Hamiltonian operator of a rigid dumbbell (such as $O_2$) be $\hat{H}=\hat{L}^2/(2I)$, whereas I is the moment of inertia and $L_i$ is the angular momentum of the molecule by rotation around its center of gravity. Determine the energy eigenvalues, the multiplicity of the degeneracy and the eigenvectors.

I'm having trouble finding an approach to this. My first idea was that the energy eigenvalues are the same as those of L, but I don't have an explicit state so I can't really determine the particular eigenvalues. Similarly, I don't know how to get the multiplicity of the degeneracy. Any ideas?

  • $\begingroup$ The eigenvectors will be the same as the eigenvectors of $ \hat{L}^2 $, and their eigenvalues will be the eigenvalues of $ \hat{L}^2 $ divided by $ 2I $. The eigenvalues and eigenvectors of $ \hat{L}^2 $ are a famous part of the study of angular momentum. $\endgroup$ Sep 27, 2016 at 23:41
  • $\begingroup$ They are the $ \left | lm \right \rangle $ states. $\endgroup$ Sep 27, 2016 at 23:42
  • $\begingroup$ The degeneracy will be the number of possible $ m $'s for each $ l $ - namely, $ 2l + 1 $. $\endgroup$ Sep 27, 2016 at 23:43

1 Answer 1


The eigenvectors of the Hamiltonian will be the same as the eigenvectors of $ \hat{L}^2 $, and their eigenvalues will be the eigenvalues of $ \hat{L}^2 $ divided by $ 2I $. The eigenvectors of $ \hat{L}^2 $ are the $ \left | lm \right \rangle $ states (which are also eigenvectors of $ \hat{L}_z $). The eigenvalues will be $ \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) $ and the degeneracy for each value of $ l $ will be the number of possible values of $ m $ - namely, $ 2l + 1 $.

Perhaps check out this wikipedia section. If you want "explicit" states, check out this other wikipedia section and the spherical harmonics.


Let's say $ \left | lm \right \rangle $ is an eigenvector of $ \hat{L}^2 $. We see:

\begin{align} \hat{H} \left | lm \right \rangle & = \frac{1}{2I} \hat{L}^2 \left | lm \right \rangle = \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) \left | lm \right \rangle \end{align}

Therefore $ \left | lm \right \rangle $ is an eigenvector of $ \hat{H} $ with eigenvalue $ \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) $. Furthermore, if $ \left | \psi \right \rangle $ is an eigenvector of $ \hat{H} $:

\begin{align} \hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle & = \frac{1}{2I} \hat{L}^2 \left | \psi \right \rangle \\ \hat{L}^2 \left | \psi \right \rangle & = 2 I E \left | \psi \right \rangle \end{align}

This shows that every eigenvector of the Hamiltonian is also an eigenvector of $ \hat{L}^2 $. Therefore, each eigenvector of the Hamiltonian must be one of the $ \left | lm \right \rangle $ states.

It is in general true that operators proportional to each other share eigenvectors with eigenvalues related by the same proportionality relationship. The above proof works with $ \frac{1}{2I} $ as the general proportionality constant between two general operators.


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