I am trying to understand the heat capacity behaviour for an ideal gas as a function of temperature. The text I am learning from is Schroeder's An Introduction to Thermal Physics. This is an exercise from an old laboratory manual for an introductory statistical mechanics course.

For instance, suppose there is a single atom of argon in a 1 m$^3$ box. Since it is monatomic, it has 3 translational degrees of freedom. The mass of an atom of argon is $6.63 \cdot 10^{-26}$ kg. On page 253 of Schroeder, the term for the translational energy for some quantum state $\epsilon(n_x,n_y,n_z)$ is given as

\begin{equation} \epsilon(n_x,n_y,n_z) = \epsilon_0(n_x^2+n_y^2+n_z^2) \end{equation}

where $\epsilon_0 = \pi^2 \hbar^2 / 2mL^2$. The partition function can be computed using

\begin{equation} Z = \sum\limits_{n_x=0}^{\infty} \sum\limits_{n_y=0}^{\infty} \sum\limits_{n_z=0}^{\infty} e^{-\epsilon_0(n_x^2+n_y^2+n_z^2)/kT}. \end{equation}

I have written some code in MATLAB to compute the partition function, average energy and heat capacity as a function of temperature using:

\begin{equation} \overline{E} = \frac{1}{Z} \sum\limits_{s} E_s \, e^{-E_s/kT} \end{equation}

\begin{equation} C_V = \Big(\frac{\partial E}{\partial T}\Big)_{N,V}. \end{equation}

The output is a graph that looks like this (excuse the bad quality):


The lab manual states asks me to explain the 'bump' in the heat capacity. The trouble is that I do not understand what could be causing it. By the equipartition theorem, I expected the heat capacity to be constant because $U = NfkT/2$ and $C_V = \partial U / \partial T = Nfk/2$. ie. the heat capacity doesn't depend on temperature.

Now, I understand that the heat capacity should go to $0$ for $T=0$ and it should be constant for high $T$. The graph shows both of these properties. But what could be possibly causing this bump?

I know that parahydrogen displays this behavior. However, my understanding is that this is due to the restrictions on the energy levels and the residual rotational energy of the molecule. But here, we are dealing with a monatomic atom. The only degrees of freedom are translational. Why am I seeing analogous behaviour?

So my question is this:

What is the explanation for the local maximum in the heat capacity-temperature graph for a monatomic atom?


If I instead change the graph to summation to

\begin{equation} Z = \sum\limits_{n_x=1}^{\infty} \sum\limits_{n_y=1}^{\infty} \sum\limits_{n_z=1}^{\infty} e^{-\epsilon_0(n_x^2+n_y^2+n_z^2)/kT}. \end{equation}

I get the following result:


ie. I get what I was expecting when I change the lower summation index to 1. So my updated question is:

Is the correct summation from $n_{x,y,z} = 1$ instead of $0$? This kind of makes sense to me because the equation for the partition function was defined using the particle in a box with $n$ nodes, and it doesn't make sense to have an $n$ that is 0.


1 Answer 1


The summation indices on $Z$ must match the summation indices on $\bar{E}$, because $Z$ is the normalization constant for the total probabilities of being in every state. The $n=0$ state has zero wavenumber and doesn't exist.

You're right to discard the analogy with para hydrogen. In that case, the peak is caused by interaction of the nuclear spins of the two hydrogen atoms.

This is mostly equivalent to Kittel, Thermal Physics, Chapter 3, Ideal Gas: A First Look. In that, he converts the sum to an integral which should be true for temperatures much larger than $\epsilon_0$. In that case, the result is $U=\tfrac32 \tau$ and $c_V = \tfrac32$. Schroeder did the same thing on page 253-254.

Is your $T$ axis in kelvin? The states are so finely spaced that way before 1000 K, the integral should be correct and the heat capacity should be $\tfrac32 k$. Also, the high-temperature result should be $\tfrac32 k = 2.07\times 10^{-23}$ in SI units. Your order of magnitude is off.

Your summations should go up fairly far. Does doubling the upper bound change your curve? Perhaps you haven't converged yet.

  • $\begingroup$ I have found this resource that says, "the subscript n represents the set of quantum numbers nx, ny, and nz, each of which can be any nonzero, positive integer" on page 296 (equation 6.4). Do you know why this might be? Everything else you have argued makes perfect sense. Yes, my $T$ axis is in Kelvin. I have changed the limits of the summation to larger values (on the order of $n = 1000$), but that didn't seem to affect the bump. $\endgroup$
    – Wise Owl
    Nov 27, 2015 at 14:44
  • $\begingroup$ You're right. I was remembering the harmonic oscillator in which $n$ is the number of nodes. In the infinite square well, $n$ is the number of anti-nodes. $\endgroup$
    – Spirko
    Nov 27, 2015 at 15:18
  • $\begingroup$ You can speed up your $Z$ summation by making it a single sum and cubing it. This is because separation of variables lets you untangle the three summations. $\endgroup$
    – Spirko
    Nov 27, 2015 at 20:37
  • $\begingroup$ Additionally, you can use $\bar{E} = \tau^2 \partial \ln Z/\partial \tau$ with $\tau = kT$ to find the average energy without doing another summation. $\endgroup$
    – Spirko
    Nov 27, 2015 at 20:40

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