# Peeter Joot's (OLD) Blog.

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## Rotation of diatomic molecules

Posted by peeterjoot on February 28, 2013

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## Question: Rotation of diatomic molecules ([2] problem 3.6)

In our first look at the ideal gas we considered only the translational energy of the particles. But molecules can rotate, with kinetic energy. The rotation motion is quantized; and the energy levels of a diatomic molecule are of the form

\begin{aligned}\epsilon(j) = j(j + 1) \epsilon_0\end{aligned} \hspace{\stretch{1}}(1.0.1)

where $j$ is any positive integer including zero: $j = 0, 1, 2, \cdots$. The multiplicity of each rotation level is $g(j) = 2 j + 1$.

### a

Find the partition function $Z_R(\tau)$ for the rotational states of one molecule. Remember that $Z$ is a sum over all states, not over all levels — this makes a difference.

### b

Evaluate $Z_R(\tau)$ approximately for $\tau \gg \epsilon_0$, by converting the sum to an integral.

### c

Do the same for $\tau \ll \epsilon_0$, by truncating the sum after the second term.

### d

Give expressions for the energy $U$ and the heat capacity $C$, as functions of $\tau$, in both limits. Observe that the rotational contribution to the heat capacity of a diatomic molecule approaches 1 (or, in conventional units, $k_{\mathrm{B}}$) when $\tau \gg \epsilon_0$.

### e

Sketch the behavior of $U(\tau)$ and $C(\tau)$, showing the limiting behaviors for $\tau \rightarrow \infty$ and $\tau \rightarrow 0$.

### a. Partition function $Z_R(\tau)$

To understand the reference to multiplicity recall (section 4.13 [1]) that the rotational Hamiltonian was of the form

\begin{aligned}H = \frac{\mathbf{L}^2}{2 M r^2},\end{aligned} \hspace{\stretch{1}}(1.0.2)

where the $\mathbf{L}^2$ eigenvectors satisfied

\begin{subequations}

\begin{aligned}\mathbf{L}^2 {\left\lvert {l m} \right\rangle} = l (l + 1) \hbar^2 {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}L_z {\left\lvert {l m} \right\rangle} = m \hbar {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\end{subequations}

and $-l \le m \le l$, where $l \ge 0$ is a positive integer. We see that $\epsilon_0$ is of the form

\begin{aligned}\epsilon_0 = \frac{\hbar^2}{2 M R_l(r)},\end{aligned} \hspace{\stretch{1}}(1.0.4)

and our partition function is

\begin{aligned}Z_R(\tau) = \sum_{l = 0}^\infty \sum_{m = -l}^l e^{-l (l + 1)\epsilon_0/\tau}= \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

We have no dependence on $m$ in the sum, and just have to sum terms like fig 1, and are able to sum over $m$ trivially, which is where the multiplicity comes from.

Fig 1: Summation over m

To get a feel for how many terms are significant in these sums, we refer to the plot of fig 2. We plot the partition function itself in, truncation at $l = 30$ terms in fig 3.

Fig 2: Plotting the partition function summand

Fig 3: Z_R(tau) truncated after 30 terms in log plot

### b. Evaluate partition function for large temperatures

If $\tau \gg \epsilon_0$, so that $\epsilon_0/\tau \ll 1$, all our exponentials are close to unity. Employing an integral approximation of the partition function, we can somewhat miraculously integrate this directly

\begin{aligned}Z_R(\tau) &\approx \int_0^\infty dl (2 l + 1) e^{-l(l+1)\epsilon_0/\tau} \\ &= \int_0^\infty dl \frac{d}{dl} \left( -\frac{\tau}{\epsilon_0} e^{-l(l+1)\epsilon_0/\tau} \right) \\ &= \frac{\tau}{\epsilon_0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

### c. Evaluate partition function for small temperatures

When $\tau \ll \epsilon_0$, so that $\epsilon_0/\tau \gg 1$, all our exponentials are increasingly close to zero as $l$ increases. Dropping all the second and higher order terms we have

\begin{aligned}Z_R(\tau) \approx 1 + 3 e^{-2 \epsilon_0/\tau}\end{aligned} \hspace{\stretch{1}}(1.0.7)

### d. Energy and heat capacity

In the large $\epsilon_0/\tau$ domain (small temperatures) we have

\begin{aligned}U &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln \left( 1 + 3 e^{-2 \epsilon_0/\tau} \right) \\ &= \tau^2 \frac{3 (-2\epsilon_0)(-1/\tau^2)}{1 + 3 e^{-2 \epsilon_0/\tau}} \\ &= \frac{6 \epsilon_0}{1 + 3 e^{-2 \epsilon_0/\tau}} \\ &\approx 6 \epsilon_0.\end{aligned} \hspace{\stretch{1}}(1.0.8)

The specific heat in this domain is

\begin{aligned}C_{\mathrm{V}} = \frac{\partial {U}}{\partial {\tau}}=\left( \frac{6 \epsilon_0/\tau}{1 + 3 e^{-2 \epsilon_0/\tau}} \right)^2\approx \left( \frac{6 \epsilon_0}{\tau} \right)^2\end{aligned} \hspace{\stretch{1}}(1.0.9)

For the small $\epsilon_0/\tau$ (large temperatures) case we have

\begin{aligned}U = \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln \frac{\tau}{\epsilon_0}= \tau^2 \frac{1}{{\tau}}= \tau\end{aligned} \hspace{\stretch{1}}(1.0.10)

The heat capacity in this large temperature region is

\begin{aligned}C_{\mathrm{V}} = \frac{\partial {U}}{\partial {\tau}} = 1,\end{aligned} \hspace{\stretch{1}}(1.0.11)

which is unity as described in the problem.

### e. Sketch

The energy and heat capacities are roughly sketched in fig 4.

Fig 4: Energy and heat capacity

It’s somewhat odd seeming that we have a zero point energy at zero temperature. Plotting the energy (truncating the sums to 30 terms) in fig 5, we don’t see such a zero point energy.

Fig 5: Exact plot of the energy for a range of temperatures (30 terms of the sums retained)

That plotted energy is as follows, computed without first dropping any terms of the partition function

\begin{aligned}U &= \tau^2 \frac{\partial}{\partial \tau} \ln\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right) \\ &= \epsilon_0\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)} \\ &= \epsilon_0\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.12)

To avoid the zero point energy, we have to use this and not the truncated partition function to do the integral approximation. Doing that calculation (which isn’t as convenient, so I cheated and used Mathematica). We obtain

\begin{aligned}U \approx \frac{\int_1^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}}{\int_0^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}}=\epsilon_0 e^{2 \epsilon_0/\tau} \left( 2 + \frac{\tau}{\epsilon_0} \right).\end{aligned} \hspace{\stretch{1}}(1.0.13)

This approximation, which has taken the sums to infinity, is plotted in fig 6.

Fig 6: Low temperature approximation of the energy

From eq. 1.0.12, we can take one more derivative to calculate the exact specific heat

\begin{aligned}C_{\mathrm{V}} &= \epsilon_0\frac{\partial {}}{\partial {\tau}}\left(\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}\right) \\ &= \left( \frac{\epsilon_0}{\tau} \right)^2\left(\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}+\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)^2}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)^2}\right) \\ &= \left( \frac{\epsilon_0}{\tau} \right)^2\left(\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}+ \frac{U^2}{\epsilon_0^2}\right) \\ &= \frac{U^2}{\epsilon_0^2}+\left( \frac{\epsilon_0}{\tau} \right)^2\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

This is plotted to 30 terms in fig 7.

Fig 7: Specific heat to 30 terms

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.