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An ideal gas in a quantum-mechanical microcanonical ensemble

1. What system are we talking about?

  • We have N identical, non-interacting particles (a simple ideal gas).
  • They are in a box of volume V and together have a fixed total energy E.
  • That's exactly what a microcanonical ensemble is: N, V, E are fixed.

Quantum mechanically, each particle can sit in some energy level. Because there are so many very closely spaced levels, we group them into "cells":

  • Cell i contains gᵢ almost-equal energy levels, all around energy εᵢ.
  • nᵢ is the number of particles in cell i.

So:

  • Total number of particles: \(\sum_i n_i = N\)
  • Total energy: \(\sum_i n_i \varepsilon_i = E\)

2. Distributions vs microstates

Two important ideas:

  • A distribution \(\{n_i\}\) just tells you how many particles are in each cell.
  • A microstate tells you exactly which level each particle occupies.

For a given distribution \(\{n_i\}\), there are many possible microstates. Call that number W[{nᵢ}].

The total number of allowed microstates with the given N, V, E is

\[ \Omega(N,V,E) = \sum_{\{n_i\}}' W[\{n_i\}] \]

(the sum runs over all distributions that satisfy the N and E constraints).

Because different cells are independent, the number of microstates factorizes:

\[ W[\{n_i\}] = \prod_i w(i), \]

where w(i) is the number of microstates inside cell i.

So the whole problem becomes: How many ways can we put nᵢ particles into gᵢ levels in cell i, depending on the statistics?

3. Counting microstates for different statistics

Think "balls into boxes".

(a) Bosons – Bose–Einstein (BE)

  • Particles are indistinguishable.
  • Any number of particles can occupy the same level.

This is "n identical balls into g boxes, no limit per box". The number of ways is

\[ w_{BE}(i) = \frac{(n_i + g_i - 1)!}{n_i!\,(g_i-1)!}. \]

So for the whole system:

\[ W_{BE}[\{n_i\}] = \prod_i \frac{(n_i + g_i - 1)!}{n_i!\,(g_i-1)!}. \]

(b) Fermions – Fermi–Dirac (FD)

  • Particles are indistinguishable.
  • At most one particle per level (Pauli principle).

So in cell i we just choose which nᵢ of the gᵢ levels are occupied:

\[ w_{FD}(i) = \frac{g_i!}{n_i!\,(g_i - n_i)!}. \]

So:

\[ W_{FD}[\{n_i\}] = \prod_i \frac{g_i!}{n_i!\,(g_i - n_i)!}. \]

(c) Classical particles – Maxwell–Boltzmann (MB)

  • Particles are treated as distinguishable (classical).
  • No restriction on how many particles per level.

Naively: each of the nᵢ particles can choose any of the gᵢ levels → \(g_i^{n_i}\) ways. But this overcounts because swapping two identical particles doesn't make a new physical state, so we divide by \(n_i!\) (Gibbs correction):

\[ w_{MB}(i) = \frac{g_i^{n_i}}{n_i!}, \]

so

\[ W_{MB}[\{n_i\}] = \prod_i \frac{g_i^{n_i}}{n_i!}. \]

4. Entropy and "most probable" distribution

The entropy of the gas in the microcanonical ensemble is

\[ S(N,V,E) = k \ln \Omega(N,V,E) = k \ln \Big[\sum_{\{n_i\}} W[\{n_i\}] \Big]. \]

For a macroscopic system, one distribution \(\{n_i^*\}\) overwhelmingly dominates this sum: the most probable distribution (the one with the largest W). Then

\[ S \approx k \ln W[\{n_i^*\}]. \]

So the problem becomes:

Find the set \(\{n_i^*\}\) that maximizes \(W[\{n_i\}]\) subject to \(\sum_i n_i = N\) and \(\sum_i n_i \varepsilon_i = E\).

They do this by:

  1. Taking \(\ln W\) (logs turn products into sums).
  2. Using Stirling's approximation: \(\ln x! \approx x \ln x - x\) (valid for large x).
  3. Introducing Lagrange multipliers α and β to enforce the constraints.

To treat BE, FD, and MB at once, they introduce a parameter a:

  • a = −1 → Bose–Einstein
  • a = +1 → Fermi–Dirac
  • a → 0 → Maxwell–Boltzmann

After the math, maximizing \(\ln W\) gives

\[ \ln\left(\frac{g_i}{n_i^*} - a\right) - \alpha - \beta \varepsilon_i = 0. \]

Solving for \(n_i^*\):

\[ n_i^* = \frac{g_i}{e^{\alpha + \beta \varepsilon_i} + a}. \]

This is the general occupation formula:

  • Bosons (BE): \(a = -1\) ⇒ \(n_i^* = \dfrac{g_i}{e^{\alpha + \beta \varepsilon_i} - 1}\)
  • Fermions (FD): \(a = +1\) ⇒ \(n_i^* = \dfrac{g_i}{e^{\alpha + \beta \varepsilon_i} + 1}\)
  • Classical (MB): \(a \to 0\) ⇒ \(n_i^* = g_i e^{-\alpha - \beta \varepsilon_i}\)

These are the familiar Bose-Einstein, Fermi-Dirac, and Maxwell-Boltzmann distributions.

Physically, \(n_i^*/g_i\) is the most probable number of particles per single-particle level at energy εᵢ.

5. Connecting α and β to thermodynamics

From earlier in the book, they identify

\[ \beta = \frac{1}{kT}, \quad \alpha = -\frac{\mu}{kT}, \]

where T is temperature and μ is chemical potential.

So the distributions become the usual forms:

  • FD: \(n_i^* = \dfrac{g_i}{e^{(\varepsilon_i - \mu)/kT} + 1}\)
  • BE: \(n_i^* = \dfrac{g_i}{e^{(\varepsilon_i - \mu)/kT} - 1}\)
  • MB: \(n_i^* = g_i e^{-(\varepsilon_i - \mu)/kT}\)

6. Entropy and pressure

They plug \(n_i^*\) back into the expression for \(\ln W\), and after simplifying get

\[ \frac{S}{k} \approx \sum_i \Big[ n_i^* \ln\Big( \frac{g_i}{n_i^*} - a \Big) + \frac{g_i}{a} \ln\Big(1 - a \frac{n_i^*}{g_i}\Big) \Big]. \]

This can be rewritten as

\[ \frac{S}{k} = \alpha N + \beta E + \frac{1}{a}\sum_i g_i \ln\Big(1 + a e^{-\alpha - \beta \varepsilon_i}\Big). \]

Using thermodynamics, the combination on the right is related to PV / kT, so they arrive at

\[ PV = \frac{kT}{a} \sum_i g_i \ln\Big(1 + a e^{-\alpha - \beta \varepsilon_i}\Big). \]

This is a general expression for the pressure of the gas in terms of the microscopic levels and statistics.

7. Classical limit → ideal gas law

For the Maxwell–Boltzmann limit (a → 0):

  • Use \(\ln(1 + a x) \approx a x\).
  • Then
\[ PV = kT \sum_i g_i e^{-\alpha - \beta \varepsilon_i} = kT \sum_i n_i^* = NkT. \]

So you recover the ideal gas law

\[ PV = NkT, \]

which is independent of the details of the energy levels.