4.7 Translational energy of a molecule in an ideal gas

This example is rather more complicated than the preceding ones, but the result is simple and powerful.

The non-interacting atoms of the gas are in a cuboidal box of side lengths $L_x$, $L_y$ and $L_z$, and volume $V\equiv L_x L_y L_z$. The sides of the box are impenetrable, so the wavefunction $\psi$ must vanish there, but inside the box the atom is free and so $\psi$ satisfies the free Schrödinger equation

$$-{\hbar^2\over 2m}\nabla^2 \psi(x,y,z)=E\psi(x,y,z).$$

The equation, and the boundary conditions, are satisfied by

$$\psi(x,y,z)=A \sin\left({n_x \pi x\over L_x}\right)\sin\left... ...y \pi y\over L_y}\right) \sin\left({n_z \pi z\over L_z}\right)$$

with $n_x$, $n_y$ and $n_z$ integers greater than zero. The corresponding energy is

$$\varepsilon(n_x, n_y, n_z)=\left(\left({n_x\pi\over L_x}\rig... ...}\right)^2 \right){\hbar^2\over 2m} \equiv{k^2\hbar^2\over 2m}$$

where $k^2=k_x^2+k_y^2+k_z^2$ and $k_x=\pi n_x/L_x$ etc. So the one-particle partition function is

$$Z_1=\sum_{\{n_x,n_y,n_z\}}\! e^{-\varepsilon(n_x, n_y, n_z)\beta}.$$

In general this cannot be further simplified. However there can be simplifications if $k_{\scriptscriptstyle B}T$ is much greater than the spacing between the energy levels, as we saw in the rotational case. For a volume of 1 litre, that spacing is of order $\hbar^2/2mL\approx 10^{-20}$ eV--truly tiny. Even at the lowest temperatures ever reached, we are in the high-temperature regime! Thus we can replace the sum over levels by an integral. We choose $k_x$, $k_y$ and $k_z$ as the variables, and replace $\sum_{n_x}$ with $(L_x/\pi)\int{\rm d}k_x$, giving

$$\begin{eqnarray*} Z_1\!\!\!&=&\!\!\!{L_x L_y L_z\over \pi^3} \int_0^\infty\!\int... ...(k)\beta}{\rm d}k \qquad\hbox{where $ D(k)\equiv V k^2/2 \pi^2$} \end{eqnarray*}$$

The factor of $1/8$ in the penultimate line comes from the fact that we only integrated over positive values of $k_x$ etc, that is over the positive octant of $k$-space. $D(k)$ is called the ``density of states'' in $k$-space; $D(k){\rm d}k$ is the number of states within range of $k\to k+{\rm d}k$. See here for more on this concept.

This section only depended on the fact that the energy is independent of the direction of $k$. Now we use the actual form of $\varepsilon(k)$ to complete the calculation:

$$\begin{eqnarray*} Z_1\!\!\!&=&\!\!\!{V\over 2\pi^2}\int_0^\infty \!k^2 {\rm d}k\... ...!\!\!V\left({m\over 2 \pi\hbar^2\beta}\right)^{3/2}\equiv V n_Q. \end{eqnarray*}$$

$Z_1$ is a pure number, so ``$n_Q$'' must have dimensions of $1/V$ like a number density; it is called the quantum concentration and is temperature-dependent. From $Z_1$ we can obtain the average single particle energy:

$$\left\langle E_1 \right\rangle =-{\partial \ln Z_1\over\partial\beta}={3\over 2}k_{\scriptscriptstyle B}T$$

as we should have expected.

References

• Mandl 7.1-4, Appendix B
• Bowley and Sánchez 5.9,7.2
• Kittel and Kroemer 3