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 $\tau \rightarrow 0$ .

Answer

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.

Posted in Math and Physics Learning. | Tagged: diatomic molecule, energy, ideal gas, integral approximation, Partition function, PHY452H1S, quantized rotation, specific heat, statistical mechanics | Leave a Comment »

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An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Rotation of diatomic molecules

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

a

b

c

d

e