Benzene ground state

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)?

• the values of the quantum numbers I have in the final expression: What is your final expression? – Thomas Fritsch Mar 8 '20 at 14:08
• @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 $2E_n$). – JorgeOvi Mar 8 '20 at 14:17
• Why should $n=0$ not be a valid state? – Thomas Fritsch Mar 8 '20 at 14:37
• @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. – JorgeOvi Mar 8 '20 at 14:47

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.