6.2: Translational Partition Function

First, we derive the density of states that we had already used in computing the distribution functions for quantum gases. We consider a quantum particle in a three-dimensional cubic box with edge length $$a$$. The energy is quantized with integer quantum numbers $$n_x$$, $$n_y$$, and $$n_z$$ corresponding to the three pairwise orthogonal directions that span the cube,

\begin{align} \epsilon_\mathrm{trs} & = \frac{h^2}{8 m a^2} \left( n_x^2 + n_y^2 + n_z^2 \right) \label{eq:trans_quant} \\ & = \frac{1}{2m} \left( p_x^2 + p_y^2 + p_z^2 \right) \label{eq:mom_quant} \ .\end{align}

It follows that momentum is also quantized with $$|p_i| = (h/2a)n_i$$ ($$i=x,y,z$$). It is convenient to consider momentum in a Cartesian frame where $$h/2a$$ is the unit along the $$x$$, $$y$$, and $$z$$ axes. Each state characterized by a unique set of translational quantum numbers $$(n_x,n_y,n_z)$$ ’owns’ a small cube with volume $$h^3/8a^3$$ in the octant with $$x\ge0$$, $$y\ge0$$, and $$z\ge0$$. Since momentum can also be negative, we need to consider all eight octants, so that each state owns a cell in momentum space with volume $$h^3/a^3$$. In order to go to phase space, we need to add the spatial coordinates. The particle can move throughout the whole cube with volume $$a^3$$. Hence, each state owns a phase space volume of $$h^3$$.

By rearranging Equation \ref{eq:trans_quant} we can obtain an equation that must be fulfilled by the quantum numbers,

$\frac{n_x^2}{\left(\frac{a}{h} \sqrt{8 m \epsilon} \right)^2} + \frac{n_y^2}{\left(\frac{a}{h} \sqrt{8 m \epsilon} \right)^2} + \frac{n_z^2}{\left(\frac{a}{h} \sqrt{8 m \epsilon} \right)^2} = 1 \label{eq:trans_sphere}$

and by using Equation \ref{eq:mom_quant} we can convert it to an equation that must be fulfilled by the components of the momentum vector,

$\frac{p_x^2}{\left(\frac{1}{2} \sqrt{8 m \epsilon} \right)^2} + \frac{p_y^2}{\left(\frac{1}{2} \sqrt{8 m \epsilon} \right)^2} + \frac{p_z^2}{\left(\frac{1}{2} \sqrt{8 m \epsilon} \right)^2} = 1 \ . \label{eq:mom_sphere}$

All states with quantum numbers that make the expression on the left-hand side of Equation \ref{eq:trans_sphere} or Equation \ref{eq:mom_sphere} smaller than 1 correspond to energies that are smaller than $$\epsilon$$. The momentum associated with these states lies in the sphere defined by Equation \ref{eq:mom_sphere}) with radius $$\frac{1}{2}\sqrt{8 m \epsilon}$$ and volume $$\frac{\pi}{6}(8 m \epsilon)^{3/2}$$. With cell size $$h^3/a^3$$ in momentum space the number of cells with energies smaller than $$\epsilon$$ is

$\mathcal{N}(\epsilon) = \frac{8 \sqrt{2}}{3} \pi \frac{V}{h^3} \left( m \epsilon \right)^{3/2} \ ,$

where we have substituted $$a^3$$ by volume $$V$$ of the box. The number of states in an energy interval between $$\epsilon$$ and $$\epsilon + \mathrm{d} \epsilon$$ is the first derivative of $$\mathcal{N}(\epsilon)$$ with respect to $$\epsilon$$ and is the sought density of states,

$D(\epsilon) = 4 \sqrt{2} \pi \frac{V}{h^3} m^{3/2} \epsilon^{1/2} \ . \label{eq:density_of_states_derived}$

Partition Function and Accessible States

This density of states is very high, so that we can replace the sum over the quantum numbers $$n_i$$ in the partition function of the canonical ensemble by an integral ,

\begin{align} Z_{\mathrm{trs},i} & = \int_o^\infty e^{-\beta n_i^2 h^2/8 m a^2} \mathrm{d} n_i \ \left( i = x, y, z \right) \\ &= \sqrt{\frac{2 \pi m}{\beta}} \frac{a}{h} \ .\end{align}

The contributions along orthogonal spatial coordinates are also independent of each other and factorize. Hence,

$Z_{\mathrm{trs}} = Z_{\mathrm{trs},x} \cdot Z_{\mathrm{trs},y} \cdot Z_{\mathrm{trs},z} = \left( \frac{2 \pi m k_\mathrm{B} T}{h^2} \right)^{3/2} V \ , \label{eq:Z_trs_ideal}$

where we have again substituted $$a^3$$ by $$V$$ and, as by now usual, also $$\beta$$ by $$1/k_\mathrm{B} T$$. The corresponding molar partition function is

$z_\mathrm{trs} = \frac{1}{N_\mathrm{Av}!} \left[ \left( \frac{2 \pi m k_\mathrm{B} T}{h^2} \right)^{3/2} V \right]^{N_\mathrm{Av}} \ . \label{eq:z_trs_ideal}$

At this point it is useful to introduce another concept:

The molecular canonical partition function $$Z$$ is a measure for the number of states that are accessible to the molecule at a given temperature. [concept:accessible_states]

This can be easily seen when considering

$P_i = \frac{N_i}{N} = \frac{g_i e^{-\epsilon_i/k_\mathrm{B}T}}{Z}$

and $$\sum_i P_i = 1$$. If we consider a mole of $$\ce{^{4}He}$$ (bosons) at 4.2 K, where it liquifies, we find that $$z_{\mathrm{trs}}/N_\mathrm{Av} \approx 7.5$$, which is not a large number . This indicates that we are close to breakdown of the regime where Bose-Einstein statistics can be approximated by Boltzmann statistics.

For $$T \rightarrow 0$$ only the $$g_0$$ lowest energy states are populated. In the absence of ground-state degeneracy, $$g_0=1$$, we find $$Z=1$$ and with an energy scale where $$U(T=0)=0$$ we have $$S(0)=0$$ in agreement with Nernst’s theorem.

An expression for the translational contribution to the entropy of an ideal gas can be derived from Equation \ref{eq:Z_trs_ideal}), Equation \ref{eq:z_indist}), and Equation \ref{eq:s_from_z}). We know that $$u = 3 N k_\mathrm{B} T/2$$, so that we only need to compute $$\ln z_\mathrm{trs}$$,

\begin{align} \ln z_\mathrm{trs} & = & \ln \frac{1}{N!} Z_\mathrm{trs}^N \\ & = & -\ln N! + N \ln Z_\mathrm{trs} \\ & = & -N \ln N + N + N \ln Z_\mathrm{trs} \\ & = & N \left(1 + \ln \frac{Z_\mathrm{trs}}{N} \right) \ ,\end{align}

where we have used Stirling’s formula to resolve the factorial. Thus we find

\begin{align} s & = & \frac{u}{T} + k_\mathrm{B} \ln z \\ & = & \frac{3}{2} N k_\mathrm{B} + k_\mathrm{B} N \left( 1 + \ln \frac{Z_\mathrm{trs}}{N} \right) \\ & = & N k_\mathrm{B} \left( \frac{5}{2} + \ln \frac{Z_\mathrm{trs}}{N} \right)\end{align}

By using Equation \ref{eq:Z_trs_ideal}) we finally obtain the Sackur-Tetrode equation

$s = N k_\mathrm{B} \left\{ \frac{5}{2} + \ln\left[ \left( \frac{2 \pi m k_\mathrm{B} T}{h^2} \right)^\frac{3}{2} \frac{V}{N}\right] \right\} \ .$

To obtain the molar entropy $$S_\mathrm{m}$$, $$N$$ has to be replaced by $$N_\mathrm{Av}$$. Volume can be substituted by pressure and temperature, by noting that the molar volume is given by $$V_\mathrm{m} = R T/p = N_\mathrm{Av} V/N$$. With $$N_\mathrm{Av} k_\mathrm{B} = R$$ and the molar mass $$M = N_\mathrm{Av} m$$ we obtain

$S_\mathrm{m} = R \left\{ \frac{5}{2} + \ln\left[ \left( \frac{2 \pi M k_\mathrm{B} T}{N_\mathrm{Av} h^2} \right)^\frac{3}{2} \frac{R T}{N_\mathrm{Av} p}\right] \right\}$