\documentclass[10pt]{article}
\pagestyle{empty}

\parindent=0pt
\parskip=.1in

\newcommand\hs{\hspace{6pt}}

\begin{document}



\centerline{\textbf{The Diatomic Ideal Gas}}
\bigskip


  %% Today we'll be working out properties of a diatomic gas.  To do
  %% this, we'll need to handle situations where there are many
  %% non-interacting particles.  e.g. consider that the Hamiltonian for a
  %% single molecule is $\mathcal{H}^{(1)}$.  If we have many particles, the
  %% total Hamiltonian is:
  %% \begin{align}
  %%   \mathcal{H}_\text{tot} &= \sum_i^N \mathcal{H}^{(1)}(i)
  %% \end{align}
  %% When a Hamiltonian is made up of many separate parts added together
  %% like this, we can use the approach called ``separation of
  %% variables.''  We write the total wavefunction as a product of many
  %% individual wavefunctions:
  %% \begin{align}
  %%   \Psi_\text{tot} &= \psi_{n_1}(1)\psi_{n_2}(2)\cdots \psi_{n_N}(N)
  %%   \\
  %%   \mathcal{H}^{(1)}(i) \psi_{n_i}(i) &= E_{n_i}^{(1)} \psi_{n_i}(i) \\
  %%   \mathcal{H}_\text{tot}\Psi_\text{tot} &= E_\text{tot}\Psi_\text{tot} \\
  %%   &= \sum_i^N \mathcal{H}^{(1)}(i) \Psi_\text{tot} \\
  %%   &= \sum_i^N E_{n_i}^{(1)} \Psi_\text{tot} \\
  %%   E_\text{tot} &= \sum_i^N E_{n_i}^{(1)}
  %% \end{align}
  %% Let's work out what the total energy is for this system.

  %% If I have many \emph{identical} particles that are all mixed
  %% up---unlike our previous discussion where we were looking at
  %% separate systems $A$ and $B$.  If we want to sum over all different
  %% eigenstates, we actually need to avoid overcounting...
  I'm going to quickly review and introduce the energy eigenvalues for
  some simple quantum mechanical problems.

  \emph{For each of the following, I'd like to sketch out the
    potential, then sketch the wavefunctions and the spacing of the
    energy levels.}

  The first problem you handled was a particle in an infinite square
  well potential:
  $$
    \mathcal{H} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} 
  $$
  $$
    E_n = \frac{\hbar^2 \pi^2 n^2}{2m L^2}
  $$
  where $n \ge 1$.  We could solve the same problem in three
  dimensions, and we would have:
  $$
    \mathcal{H} = \frac{-\hbar^2}{2m}\nabla^2 
  $$
  $$
    E_{n_xn_yn_z} = \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 + n_z^2\right)}{L^2}
  $$
  The next moderately simple problem is the rigid rotator.  In this
  case, the only energy in the Hamiltonian is the angular kinetic
  energy:
  $$
    \mathcal{H} = \frac{-\hbar^2}{2I}L^2 
  $$
  $$
    E_{lm} = \frac{\hbar^2 l(l+1)}{2I}
  $$
  Finally, we have a simple problem that hasn't yet come up in the
  paradigms, which is the simple harmonic oscillator.  In this case we
  have both kinetic \emph{and} potential energy:
  $$
    \mathcal{H} &= \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}
     + \frac{m\omega_0^2}{2}x^2  
  $$
  $$
    E_{n} &= \left(n + \frac12\right)\hbar \omega_0
  $$
  Of course, you also studied the hydrogen atom, but its solution is
  less general than those we've listed here.  Any diatomic molecule
  behaves like a rigid rotator and a simple harmonic oscillator, and
  like a particle in a box, too!
  
\newpage

  Now, let's consider a diatomic ideal gas, such as nitrogen.  In this
  case, the energy levels of a single molecule are given by:
  $$
    E_{n_xn_yn_zn_vlm}^{(1)} = \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} + \frac{\hbar^2 l(l+1)}{2I} +
    \left(n + \frac{1}{2}\right)\hbar \omega_{0}
  $$
  That's an awful lot of quantum numbers, and that's just one
  molecule, and we're neglecting any possible electronic excited
  states!

  How does this change when we've got $N$ molecules all confined in
  the same box? We've already talked about how energies relate when we
  combine systems:
  $$
    E_\text{tot} = \sum_i^N E_i^{(1)}
  $$
  where I've left out all the quantum numbers, since there are so very
  many.

  If we want to know the internal energy, we'll need to sum over every
  possible state, with the probability of that particular state.  To
  do this, we'll need to know the partition function, so let's start
  with that.
  $$
    Z = \sum^\text{all states} e^{-\beta E_{\text{this state}}} 
  $$
  $$
    = \sum_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1},
      n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2},\cdots}
    e^{-\beta \left(E_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1}}^{(1)} + E_{n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2}}^{(2)} + \cdots\right)}
  $$
  \ldots except that this isn't quite right.  We can't distinguish
  between the different molecules... when we swap two of them in the
  sum, we're really talking about the same state!  We can fix this
  double-counting by multiplying by an $N!$, which gives us:
  $$
    Z = \frac{1}{N!}\sum_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1},
      n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2},\cdots}
    e^{-\beta \left(E_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1}}^{(1)} +
      E_{n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2}}^{(2)} + \cdots\right)}
  $$
  $$
    = \frac{1}{N!}\sum_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1},
      n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2},\cdots}
    e^{-\beta E_{n_{x_{1}}n_{y_{1}}n_{z_{1}}n_{v_{1}}l_{1}m_{1}}^{(1)}} e^{-\beta E_{n_{x_{2}}n_{y_{2}}n_{z_{2}}n_{v_{2}}l_{2}m_{2}}^{(2)}} \cdots 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1}e^{-\beta E_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1}^{(1)}}\right)
    \left(\sum_{n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2}e^{-\beta
      E_{n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2}^{(2)}}\right)
    \cdots 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x}n_{y}n_{z}n_{v}lm}e^{-\beta
      E_{n_{x}n_{y}n_{z}n_{v}lm}^{(1)}}\right)^N 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x}n_{y}n_{z}n_{v}lm}e^{-\beta \left( \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} + \frac{\hbar^2 l(l+1)}{2I} +
    \left(n + \frac12\right)\hbar \omega_0 \right)}\right)^N 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x}n_{y}n_{z}n_{v}lm}e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}}
    e^{-\beta \frac{\hbar^2 l(l+1)}{2I}}
    e^{-\beta \left(n + \frac12\right)\hbar \omega_0}\right)^N 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x}n_{y}n_{z}}e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}}
    \sum_{n_{v}}e^{-\beta \frac{\hbar^2 l(l+1)}{2I}}
    \sum_{lm}e^{-\beta \left(n + \frac12\right)\hbar \omega_0}\right)^N 
  $$
  $$
    = \frac{1}{N!}
    \left(\sum_{n_{x}n_{y}n_{z}}e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}}\right)^N
    \left(\sum_{n_{v}}e^{-\beta \frac{\hbar^2 l(l+1)}{2I}}\right)^N
    \left(\sum_{lm}e^{-\beta \left(n + \frac12\right)\hbar
      \omega_0}\right)^N
  $$
  Now, if we were computing the internal energy $U$, we'd be able to
  do something very similar:
  $$
    U = \sum_i P_i E_i 
  $$
  $$
    = \frac{1}{N!}\sum_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1,
      n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2,\cdots} P_{\ldots}
    E_\text{tot} 
  $$
  $$
    = \frac{1}{N!}\sum_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1,
      n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2,\cdots} P_{\ldots}
    \left(E_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1}^{(1)} +
    E_{n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2}^{(2)} + \cdots\right) 
  $$
  $$
    = \frac{N}{N!}\sum_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1,
      n_{x_2}n_{y_2}n_{z_2}n_{v_2}l_2m_2,\cdots} P_{\ldots} E_{n_{x_1}n_{y_1}n_{z_1}n_{v_1}l_1m_1}^{(1)}
  $$
  $$
    = N\sum_{n_{x}n_{y}n_{z}n_{v}lm} P_{n_{x}n_{y}n_{z}n_{v}lm} E_{n_{x}n_{y}n_{z}n_{v}lm}^{(1)} 
  $$
  $$
    = N\sum_{n_{x}n_{y}n_{z}n_{v}lm} P_{n_{x}n_{y}n_{z}n_{v}lm} \left( \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} + \frac{\hbar^2 l(l+1)}{2I} +
    \left(n + \frac12\right)\hbar \omega_0 \right) 
  $$
  $$
    = N\left(
    \sum_{n_{x}n_{y}n_{z}} P_{n_{x}n_{y}n_{z}} \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} +
    \sum_{lm} P_{lm} \frac{\hbar^2 l(l+1)}{2I} +
    \sum_{n_{v}} P_{n_{v}} \left(n + \frac12\right)\hbar \omega_0 \right)
  $$
  So you can see, if we can work out the average translational kinetic
  energy, rotational energy and vibrational energy of a molecule in
  this gas, then we'll easily have the total internal energy of this
  system just by adding everything up and multiplying by $N$.

\newpage

\Large \textbf{Activity} \normalsize

  I'm going to divide you into groups.  Each group will have a
  separate task, so that hopefully when we're done as a class we'll
  have an answer for the total internal energy of a diatomic ideal
  gas.

  These sums are pretty challenging, so I'll ask each group to
  consider one of two distinct limits: the low-temperature limit and
  the high-temperature limit.  The low- and high-temperature limits
  have a different meaning for each term in the energy.

  For the translational kinetic energy, the limits will be defined by
  $$
    \beta \frac{\hbar^2\pi^2}{2mL^2} \gg 1
  $$
  or the reverse.  For the rotational energy, it will be
  $$
    \beta \frac{\hbar^2}{2I} \gg 1
  $$
  and for the vibrational energy, it will be
  $$
    \beta \hbar \omega_0 \gg 1
  $$
  These energy scales have large gaps, so the translational energy may
  be in the high-temperature limit while the other two are in the
  low-temperature limit, for instance.

  You should have five groups, so I'd like each group to do one term in
  the energy, in either the low-temperature limit or the
  high-temperature limit, but let's \emph{not} do the low-temperature
  limit of the translational kinetic energy.  The reason we avoid that
  particular limit is that we've ignored the possibility that two of
  our molecules may be in precisely the same state.  You know the
  Pauli exclusion principle for electrons (and fermions in general),
  which means we could have a problem.  For bosons (all particles that
  aren't fermions) there is another problem, which I also won't go
  into.  As long as our temperature is high enough that the
  probability of any given state is pretty low, we won't run into
  these troubles (which could give us things like a Bose-Einstein
  condensate).
\vfill
\leftline{\textit{by David Roundy}}
\leftline{\copyright DATE David Roundy}
\end{document}