Past exam of the physics B course of the University of Cambridge / 2021 / Paper 2 / Section C / C9

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

The internal energy is the canonical mean of the energy:Holding the external mechanical parameters fixed, so that the spectrum and degeneracies do not change when differentiating with respect to $\beta$, givesComparison yieldsAn explicitly temperature-dependent effective Hamiltonian would require care: differentiating its partition sum would also differentiate the assigned energy levels. The stated result uses the fixed physical spectrum of the canonical system.

For the supplied linear rigid rotor spectrum, sum over every nonnegative integer $J$ and include all $2J+1$ magnetic substates:It is convenient to introduce the rotational temperatureso that the first terms are $Z_{\mathrm{rot}}=1+3e^{-2{\theta }_r/T}+5e^{-6{\theta }_r/T}+\cdots$. Here $I$ is the moment of inertia and $\hslash$ is the reduced Planck constant, as confirmed in the original PDF. The stipulated degeneracies are used as given, without adding nuclear-spin weights or omitting selected $J$ values.

At low temperature, $T\ll {\theta }_r$, retain the ground level and the first excited level. Their energies are $0$ and ${\hslash }^2/I$, with degeneracies $1$ and $3$. Thereforeso in the notation of the question,$A$ is dimensionless and $B$ has units of kelvins. The first omitted level has three times the first excitation energy, explaining the exponent in the error term.

At high temperature, $T\gg {\theta }_r$, many rotational levels are occupied and the sum can be replaced to leading order by an integral:Set $u=J(J+1)$, so that $du=(2J+1)dJ$. ThenThusThe high-temperature approximation here concerns the rotational spectrum. For a real molecule it should still be used below temperatures where neglected vibrational excitation or dissociation invalidates the rigid-rotor model.

Using $Z_{\mathrm{rot}}\simeq CT=C/(k_B\beta )$ givesThe internal-energy identity from (b), followed by differentiation with respect to temperature at fixed moment of inertia, yieldsFor one mole of independent rotors, the rotational energy is $RT$ and the rotational molar heat capacity is $R\simeq 8.31\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

This agrees with the equipartition theorem. The given spectrum describes a linear rotor with two independent rotational degrees of freedom, each contributing $k_{B}T/2$ to the high-temperature mean energy. Rotation about the molecular axis is not an additional degree of freedom of this ideal linear-rotor model, so the result is $k_B$ rather than $3k_B/2$ per rotor. These are rotational contributions alone; translational and vibrational heat capacities have not been included. At low temperature the discrete excitation gap freezes rotational excitation out, so the rotational heat capacity tends to zero instead of remaining at its classical value.