1. Distinguishability & The Quantum Regime
In classical statistical mechanics (Maxwell-Boltzmann statistics), particles are treated as distinguishable. We can track individual trajectories of molecules in a gas. However, quantum mechanics dictates that identical subatomic particles are fundamentally indistinguishable. Because of their wave-like nature, when the wavepackets of identical particles overlap, we can no longer tell them apart.
The threshold for quantum behavior is governed by the thermal de Broglie wavelength (\(\lambda_{th}\)): \[\lambda_{th} = \frac{h}{\sqrt{2\pi m k_B T}}\] If the average distance between particles \(d \approx (V/N)^{1/3}\) is comparable to or smaller than \(\lambda_{th}\), the wavefunctions overlap significantly, requiring quantum statistics (BE or FD). If \(\lambda_{th} \ll d\), the particles can be treated classically, and quantum statistics reduce to Maxwell-Boltzmann statistics.
2. Mathematical Comparison of Microstates
The distribution of \(n_i\) particles across energy levels \(\epsilon_i\) with degeneracies \(g_i\) is found by maximizing the thermodynamic probability (number of microstates, \(W\)) subject to constraints of constant particle number (\(N = \sum n_i\)) and constant energy (\(E = \sum n_i \epsilon_i\)).
Each formalism calculates \(W\) differently based on distinguishability and occupancy rules:
• Maxwell-Boltzmann (Classical, Distinguishable, No Occupancy Limits):
\[W_{MB} = N! \prod_{i} \frac{g_i^{n_i}}{n_i!}\]
• Bose-Einstein (Quantum, Indistinguishable Bosons, No Occupancy Limits):
\[W_{BE} = \prod_{i} \frac{(n_i + g_i - 1)!}{n_i! (g_i - 1)!}\]
• Fermi-Dirac (Quantum, Indistinguishable Fermions, Max 1 Particle per State):
\[W_{FD} = \prod_{i} \frac{g_i!}{n_i! (g_i - n_i)!}\]
Applying Stirling's approximation (\(\ln x! \approx x \ln x - x\)) and maximizing \(\ln W\) using Lagrange multipliers yields the unified distribution function:
\[n_i = \frac{g_i}{e^{(\epsilon_i - \mu)/k_B T} + a}\]
where \(\mu\) is the chemical potential (Fermi energy in metals at \(T=0\text{ K}\)), and:
• \(a = 0\) for classical Maxwell-Boltzmann statistics (with appropriate normalization constant).
• \(a = -1\) for Bose-Einstein statistics.
• \(a = +1\) for Fermi-Dirac statistics.
3. Summary Comparison Table
4. Solved CSIR-NET & GATE Questions
Question 1 (CSIR-NET Physical Sciences)
Two identical particles are to be distributed among three energy levels of energies \(0\), \(\epsilon\), and \(2\epsilon\). Calculate the total number of microstates if the particles are: (a) distinguishable classical particles (MB), (b) Bosons (BE), and (c) Fermions (FD), assuming non-degenerate energy states.
Detailed Solution:
- For Distinguishable Particles (Maxwell-Boltzmann):
Each particle has 3 choices. Since there are \(N = 2\) particles and \(g = 3\) states: \[\text{Microstates} = g^N = 3^2 = 9\] - For Indistinguishable Bosons (Bose-Einstein):
Using the BE formula with \(N = 2\) and \(g = 3\): \[W_{BE} = \frac{(N + g - 1)!}{N! (g - 1)!} = \frac{(2 + 3 - 1)!}{2! (3 - 1)!} = \frac{4!}{2! 2!} = 6\] (The 6 states are: both in level 0, both in \(\epsilon\), both in \(2\epsilon\), and three pairs [0, \(\epsilon\)], [0, \(2\epsilon\)], and [\(\epsilon\), \(2\epsilon\)]) - For Indistinguishable Fermions (Fermi-Dirac):
Using the FD formula with \(N = 2\) and \(g = 3\) (maximum 1 particle per state): \[W_{FD} = \frac{g!}{N! (g - N)!} = \frac{3!}{2! (3 - 2)!} = \frac{6}{2 \times 1} = 3\] (The 3 states are the pairs containing at most 1 particle per level: [0, \(\epsilon\)], [0, \(2\epsilon\)], and [\(\epsilon\), \(2\epsilon\)])
Answer: (a) 9 microstates, (b) 6 microstates, (c) 3 microstates.
Question 2 (GATE Physics)
Explain why liquid Helium-4 (\(^4\text{He}\)) undergoes a superfluid transition at low temperatures (\(T_\lambda = 2.17\text{ K}\)), whereas Helium-3 (\(^3\text{He}\)) does not show superfluidity until much lower temperatures (\(1\text{ mK}\)).
Detailed Solution:
- Analyze the spin characteristics of the nuclei:
• Helium-4 (\(^4\text{He}\)): Contains 2 protons, 2 neutrons, and 2 electrons. The total nuclear spin is \(I = 0\) (an integer). Therefore, \(^4\text{He}\) atoms are Bosons and obey Bose-Einstein statistics.
• Helium-3 (\(^3\text{He}\)): Contains 2 protons, 1 neutron, and 2 electrons. The total nuclear spin is \(I = 1/2\) (half-integer). Therefore, \(^3\text{He}\) atoms are Fermions and obey Fermi-Dirac statistics. - BE Condensation:
As Bosons, \(^4\text{He}\) atoms can undergo Bose-Einstein Condensation (BEC) at a macroscopic temperature (\(2.17\text{ K}\)), where a significant fraction of particles collapse into the zero-momentum ground state, flowing without viscosity (superfluidity). - Fermionic pairing:
Since \(^3\text{He}\) atoms are Fermions, they obey the Pauli Exclusion Principle and cannot condense into the ground state individually. Superfluidity in \(^3\text{He}\) is only achieved when atoms pair up to form Cooper-like pairs (which act as composite Bosons) via weak attractive forces, which requires extremely low temperatures (\(\sim 1\text{ mK}\)).
Answer: \(^4\text{He}\) is a Boson and undergoes Bose-Einstein condensation directly, while \(^3\text{He}\) is a Fermion and requires Cooper-like pairing at millikelvin temperatures.
Dr. Saju K John
Senior Physics Faculty | PhD (NIT Calicut) | GATE AIR-52Dr. Saju K John obtained his PhD in Physics from NIT Calicut. He cleared CSIR-NET and secured GATE AIR 52. He manages the Advanced Thermodynamics and Statistical Physics syllabus tracks at Benzil Academy.