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Systems composed of indistinguishable particles

1. The setup: many identical particles

We look at a system of \(N\) identical, non-interacting particles (a gas, for example).

  • Each particle by itself has some Hamiltonian \(\hat{H}_i\) and eigenstates \(u_i(q)\) with energies \(\varepsilon_i\).
  • For non-interacting particles, the total Hamiltonian is just a sum of the single-particle Hamiltonians.

Because they don't interact, a many-particle energy eigenstate is just a product of single-particle eigenstates, and the total energy is just a sum of the single-particle energies.

So a many-particle state is basically specified by how many particles sit in each single-particle level:

\[ \{n_i\} \quad \text{with} \quad \sum_i n_i = N, \quad \sum_i n_i \varepsilon_i = E. \]

Here, \(n_i\) = number of particles in the single-particle state with energy \(\varepsilon_i\).

2. Classical point of view: particles are distinguishable

In classical physics (or "Boltzmann" counting), we pretend we can label the particles: 1st, 2nd, 3rd, …, \(N\)-th particle.

If you swap the 5th and 7th particle, you say you got a new microstate. So:

  • For a given distribution \(\{n_i\}\), the number of microstates is
\[ \frac{N!}{n_1! n_2! \dots} \]

(That's equation (10).)

So in classical statistics, this factor becomes a "statistical weight" for the distribution \(\{n_i\}\).

Then they mention the Gibbs correction: dividing by \(N!\) to fix overcounting when particles are identical. After that correction, the weight is just

\[ W_c\{n_i\} = \frac{1}{n_1! n_2! \dots} \quad \text{(equation (11))}. \]

But the textbook is pointing out: the only way to justify these corrections properly is to really take indistinguishability seriously using quantum mechanics.

3. Real quantum indistinguishable particles

Quantum mechanically, truly identical particles are indistinguishable at a very deep level:

  • Exchanging the coordinates of two identical particles must not give a new physical state.
  • That is, if \(P\) is a permutation of particle labels, then the probability density must not change:
\[ |P\psi|^2 = |\psi|^2 \quad \text{(equation (13))}. \]

This leads to a key result:

The wavefunction of a system of identical particles can only change at most by a sign under any permutation.

So either

\[ P\psi = \psi \quad \text{for all permutations } P \]

or

\[ P\psi = -\psi \quad \text{for all odd permutations } P \]

and \(+\psi\) for even ones. (Equations (14) and (15).)

  • First case: symmetric wavefunction.
  • Second case: antisymmetric wavefunction.

These two cases correspond to bosons and fermions, respectively.

4. From "Boltzmann" wavefunction to symmetric/antisymmetric ones

They call a naïve product wavefunction (the one that treats particles as if labeled) \(\psi_\text{Boltz}(q)\). This is not suitable for indistinguishable particles, because swapping arguments genuinely changes the wavefunction in a way that shouldn't matter physically.

To fix this, they build from \(\psi_\text{Boltz}\) a new wavefunction that is:

  • Symmetric:
\[ \psi_S(q) \propto \sum_P P\psi_\text{Boltz}(q) \quad \text{(equation (16))} \]

(just add up all permutations with a plus sign)

  • Antisymmetric:
\[ \psi_A(q) \propto \sum_P \delta_P P\psi_\text{Boltz}(q) \quad \text{(equation (17))} \]

where \(\delta_P = +1\) for even permutations, \(-1\) for odd permutations (add all permutations but with plus or minus depending on permutation parity).

So you take the unsymmetrized product wavefunction and "symmetrize" or "antisymmetrize" it.

5. Antisymmetric case → Slater determinant → Pauli principle

The antisymmetric wavefunction \(\psi_A\) can be written as a Slater determinant (equation (18)). Structurally, it looks like:

\[ \psi_A(q) \propto \begin{vmatrix} u_1(1) & u_2(1) & \dots & u_N(1) \\ u_1(2) & u_2(2) & \dots & u_N(2) \\ \vdots & \vdots & \ddots & \vdots \\ u_1(N) & u_2(N) & \dots & u_N(N) \end{vmatrix} \]

Key property of a determinant: if two columns are identical, the determinant is zero.

  • Two equal columns here mean two particles in exactly the same single-particle state.
  • So if two particles try to occupy the same state, the entire wavefunction vanishes → that configuration is forbidden.

This is exactly Pauli's exclusion principle:

Fermions (particles with antisymmetric wavefunctions) cannot share the same single-particle state.

Consequences:

  • For fermions, each \(n_i\) can be only 0 or 1.
  • The statistical weight is
\[ W_\text{F.D.}\{n_i\} = \begin{cases} 1, & \text{if all } n_i = 0 \text{ or } 1 \text{ and } \sum_i n_i = N, \\ 0, & \text{otherwise.} \end{cases} \quad \text{(equation (19))} \]

6. Symmetric case → bosons

For symmetric wavefunctions (bosons):

  • There is no restriction on how many particles can be in the same single-particle state.
  • So
\[ n_i = 0, 1, 2, \dots \]

are all allowed.

  • The statistical weight for a given distribution \(\{n_i\}\) is always
\[ W_\text{B.E.}\{n_i\} = 1 \quad \text{(equation (20))}. \]

These are Bose–Einstein statistics.

7. Who are bosons and fermions physically?

They then remind you of the spin–statistics connection:

  • Particles with integer spin (0, 1, 2, … in units of \(\hbar\)) obey Bose–Einstein statistics → bosons (examples: photons, phonons, gravitons, certain atoms like \(^4\text{He}\), etc.)
  • Particles with half-integer spin \(\left(\frac{1}{2}, \frac{3}{2}, \dots\right)\) obey Fermi–Dirac statistics → fermions (examples: electrons, protons, neutrons, \(^3\text{He}\), many others).

8. Final message

Even though the derivation used a non-interacting gas (because the math is simpler), the conclusion is general:

  • For any system of identical particles, the many-particle wavefunction must be either symmetric (bosons) or antisymmetric (fermions).
  • That requirement completely changes the counting of microstates compared to classical physics, and this is where Bose–Einstein and Fermi–Dirac statistics come from.