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I’m currently studying my first course on Quantum Physics. Regarding the topic of Sommerfeld’s quantization rules, I’ve come across a problem where I was asked whether a benzene molecule would require more energy to be excited considering a rotation around its perpendicular axis or around an axis connecting two opposite carbon atoms.

My main question relates to the values of the quantum numbers I have in the final expression. Do those numbers start counting at 0 (which means that benzene’s ground state is static), or at 1 (which means that benzene molecules always have a certain rotation)?

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  • $\begingroup$ the values of the quantum numbers I have in the final expression: What is your final expression? $\endgroup$ – Thomas Fritsch Mar 8 '20 at 14:08
  • $\begingroup$ @ThomasFritsch I got the expression $E_n=\frac{n^2 \hbar^2}{12} \cdot \frac{1}{m_C d_C^2 +m_H (d_C+d_H)^2}$ where $m_C$ is the carbon mass, $m_H$ is the hydrogen mass, $d_C$ is the distance between carbon atoms and $d_H$ is the distance between hydrogen and its respective carbon atom. (That's around the Z axis. For the other axis I got $2E_n$). $\endgroup$ – JorgeOvi Mar 8 '20 at 14:17
  • $\begingroup$ Why should $n=0$ not be a valid state? $\endgroup$ – Thomas Fritsch Mar 8 '20 at 14:37
  • $\begingroup$ @ThomasFritsch since I have used the same quantization condition as for Bohr's atom, where you cannot possibly have a $n=0$ state, I'm not quite sure whether angular quantum numbers run from 1 or it just depends on the system. $\endgroup$ – JorgeOvi Mar 8 '20 at 14:47
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Bohr's H Atom (mentioned in your comments) and the rotating molecule are quite different.

In Bohr's H atom the energy levels were $E_n\propto -\frac{1}{n^2}$. There $n=0$ was invalid because $E_0$ would be infinite.

In your rotating molecule you found the energy levels to be $E_n\propto n^2$. So here $n=0$ is no problem, and therefore is perfectly valid.

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