# 6.2: Translational Partition Function

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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

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