# Diatomic Molecule as Rigid Rotor

## Question

Consider a molecule, such as Carbon Monoxide, which consists of two different atoms, one Carbon and one Oxygen, separated by a distance $$d$$. Such a molecule can exist in quantum states of different orbital angular momentum. Each state has the energy $\epsilon_l={\hbar^2 \over 2I}l(l+1)$ where $$I=\mu d^2$$ is the moment of inertia of the molecule about an axis through its centre of mass and $$\mu$$ is the reduced mass defined by $$\frac 1\mu=\frac{1}{m_1} + \frac{1}{m_2}$$. $$l=0, 1, 2, \ldots$$ is the quantum number associated with the orbital angular momentum. Each energy level of the rotating molecule has the degeneracy $$g_l=2l+1$$.

1. Find the general expression for the canonical partition function $$Z$$.
2. Show that for high temperatures, $$Z$$ can be approximated by an integral and calculate this integral.
3. Evaluate the high temperature mean energy $$E$$ and the heat capacity $$C_V$$.
4. Find the low-temperature approximations to the canonical partition function, the mean energy $$E$$ and the heat capacity $$C_V$$.

1. The generic partition function is given by
\begin{eqnarray}
Z&=&\sum_{j=0}^{\infty} g_j e^{-E_j \beta} \nonumber \\
&=&\sum_{l=0}^{\infty} (2l+1) e^{-l(l+1){\hbar^2 \over 2I}\beta} \nonumber
\end{eqnarray}

2. For high temperatures, the energy spacing between the energy levels is small compared to $$k_B T$$, so the summation can be replaced by the integral
\begin{eqnarray}
Z&=&\int_0^\infty(2l+1)e^{-l(l+1){\hbar^2 \over 2I}\beta}dl \nonumber \\
&=&\int_0^\infty e^{{-\beta \hbar^2 l(l+1) \over 2I}}d(l(l+1)) \nonumber \\
&=&{2I \over \beta \hbar^2} \nonumber
\end{eqnarray}

3. Finding the energy in the high-temperature limit.
\begin{eqnarray}
\lt E \gt &=&-{\partial \over \partial \beta}\ln Z \nonumber \\
&=&-{\partial \over \partial \beta}\ln{2I\over \beta \hbar^2} \nonumber \\
&=&\frac 1\beta = k_B T \nonumber
\end{eqnarray}
And the heat capacity: $C_V={\partial \lt E \gt \over \partial T} = k_B$

4. For the low-temperature approximation, most of the particles will be in the ground state, so we can approximation the partition function by simply the first two terms like so:
\begin{eqnarray}
Z&=&\sum_{l=0}^{\infty} (2l+1) e^{-l(l+1){\hbar^2 \over 2I}\beta} \nonumber \\
&=&1+3e^{-\beta \hbar^2 / I} \nonumber
\end{eqnarray}
So the average energy again is
\begin{eqnarray}
\lt E \gt &=&-{\partial \over \partial \beta}\ln Z \nonumber \\
&=&-{\partial \over \partial \beta}\ln\bigg({1+3e^{-{\hbar^2 \over I} \beta}}\bigg) \nonumber \\
&=&-{3·\frac{-\hbar^2}{I}·e^{-\frac{\hbar^2}{I}\beta} \over 1+3e^{-\frac{\hbar^2}{I}\beta}} \nonumber \\
&=&{3\hbar^2 / I \over e^{\beta \hbar^2 / I}+3} \nonumber
\end{eqnarray}
For the heat capacity,
\begin{eqnarray}
C_V&=&{\partial \beta \over \partial T}{\partial \over \partial \beta}\lt E \gt \nonumber \\
&=&-{1 \over k_B T^2} {\partial \over \partial \beta} \bigg({3\hbar^2 / I \over e^{\beta \hbar^2 / I}+3}\bigg) \nonumber \\
&=&{3 \hbar^4 \over k_B T^2 I^2} {e^{\hbar^2 \beta / I} \over \big(e^{\beta \hbar^2 / I} + 3\big)^2} \nonumber \\
&\approx&3k_B\bigg({\hbar^2 \over I k_B T}\bigg)^2 \exp \bigg(-\hbar^2 / I k_B T\bigg) \nonumber
\end{eqnarray}