Bose–Einstein and Fermi–Dirac statistics

Two particles and three available states

Imagine two identical, noninteracting particles in a small system with three complete one-particle states. Call them a, b and c, with energies 0, Δ and 2Δ, where Δ is positive. Can both particles occupy a? For bosons, yes. For identical fermions in this same internal channel, no. That difference changes the equilibrium occupations even though we have introduced no interparticle force.

A complete state specifies every relevant one-particle quantum number, including spin or another internal label. We deliberately supplied three nondegenerate complete states, not three spatial orbitals with hidden spin choices. The comparison is an assigned teaching model, not a measurement or a three-level approximation claimed valid for every real gas.

Write an occupation list as (n_a, n_b, n_c), meaning how many particles occupy each state. The list (2,0,0) is allowed for our bosons and excluded for our fermions. Exchanging imaginary labels ‘particle 1’ and ‘particle 2’ does not create another physical occupation state. Our question is how this counting rule becomes a prediction for a gas.

Choose a first reading and return path

On a first reading, follow the allowed configurations, the mean occupation of one state, the dilute limit, the filled Fermi sea and the condensate fraction. A state is an available one-particle alternative; its occupation counts particles in it. An average occupation need not be an integer even when every individual reading is.

The advanced return path derives grand partition functions, compares fixed-number and fluctuating-number descriptions, counts continuum states, and connects low-temperature expansions and dimensionality to the limits of the ideal-gas model. The published probability, ensembles and partition-functions lessons supply useful preparation. Necessary meanings are introduced here before their formulas.

This lesson concerns equilibrium statistics of identical particles and their ideal-gas consequences. Detailed quantum dynamics, interactions, superfluid response, photon spectra and critical scaling have separate permanent topics. Connections to them are useful; replacing this lesson with their entire theory would hide the counting question we started with.

What exchange changes, and what it does not

For ordinary identical particles in three spatial dimensions, bosonic many-particle states are symmetric under exchanging two particles, while fermionic states are antisymmetric. Antisymmetry makes a state with two identical fermions in the same complete one-particle state vanish. This is the Pauli exclusion principle. The spin–statistics connection is a deeper result; the gas calculations here use it rather than prove it.

The restriction concerns a complete state, not a location drawn on a diagram. Two electrons with opposite spin labels can occupy the same spatial orbital because their complete states differ. Conversely, merely being at different positions at a particular instant does not turn identical particles into different species.

Bose enhancement and fermionic exclusion change state counting and correlations without adding a classical attraction or repulsion to our noninteracting Hamiltonian. Real particles may also interact; those interactions are extra physics. Composite particles behave as bosons or fermions only within the regime where their internal structure is not resolved in the relevant process. Exotic quasiparticles in two-dimensional systems require further care.

Weight the finite system at fixed number

Let T be a positive bath temperature, k_B Boltzmann’s constant and β=1/(k_B T). In the canonical ensemble the particle number remains exactly two. For each allowed quantum occupation configuration with total energy E, its probability is exp(−βE)/Z₂, where Z₂ sums that weight over all its allowed configurations. No chemical potential is needed for this finite fixed-number calculation.

Choose βΔ=ln2. Raising the energy by Δ halves its Boltzmann weight. For bosons, the six unnormalized weights in the opening table are 1, 1/2, 1/4, 1/4, 1/8 and 1/16. Their sum is Z₂,B=35/16. For fermions, retain only the three allowed configurations; their weights sum to Z₂,F=7/8.

The quantum configurations (1,0,1) and (0,2,0) have the same energy and the same bosonic weight. We do not multiply the first by 2 because we can imagine exchanging named particles. A classical two-particle comparator does use different configuration weights, proportional to exp(−βE)/(n_a!n_b!n_c!). Its normalization is Z₁²/2!, with Z₁=1+exp(−βΔ)+exp(−2βΔ). That comparator is not a third exact quantum statistics.

Allow particle exchange to simplify the counting

A large gas is often easier to calculate with the grand canonical ensemble, where a reservoir fixes T and chemical potential μ while particle number can fluctuate. Chemical potential is the thermodynamic energy scale conjugate to adding particles; it is not an extra energy that must be assigned to every particle’s mechanical motion.

For an ideal gas, the total energy is the sum of occupied one-particle energies. Without a fixed total-N constraint, the grand partition function factorizes into separate contributions for each complete state. For one state of energy ε, introduce the dimensionless number x=β(ε−μ). An occupation n has weight exp(−nx).

This factorization relies on noninteracting modes and the specified ensemble. It is not an assertion that occupations of every real interacting gas are independent, or that a two-particle fixed-N system can be replaced exactly by independent fluctuating occupations. We will keep those two uses separate.

Derive the two mean occupations

For a fermionic complete state, n is either 0 or 1. Its grand partition sum is Ξ_F=1+exp(−x). The probability of occupation 1, and therefore the mean occupation, is ⟨n⟩_F=exp(−x)/(1+exp(−x))=1/(exp(x)+1). At positive temperature it lies strictly between 0 and 1. At ε=μ it equals 1/2.

For a bosonic complete state, n can be any nonnegative integer. The geometric sum Ξ_B=Σₙ exp(−nx)=1/(1−exp(−x)) converges only for x>0. Taking its weighted mean gives ⟨n⟩_B=1/(exp(x)−1). This average may exceed 1 because several bosons can occupy the same state.

For a number-conserving ideal Bose gas with lowest one-particle energy ε₀, finite-system grand canonical convergence requires μ<ε₀. When ε₀ is set to zero, that is μ<0. Condensation is associated with approaching the boundary in an appropriate thermodynamic limit; substituting μ=ε₀ into an isolated finite-system geometric sum does not make a normalizable finite distribution.

Read the curves without changing the quantity

At the same positive x, the bosonic mean is larger than exp(−x), while the fermionic mean is smaller. Both approach the dilute expression when x is large and positive. Close to x=0 from the positive side, the bosonic mean grows without bound; a graph that hides that region by clipping its vertical axis must not be read as a finite maximum occupation.

The vertical quantity here is particles per complete state. It is dimensionless. It is not the probability that a randomly selected particle has energy ε, nor the number of available states near ε. A Bose mean above 1 is therefore not a probability greater than 1.

For example, at x=ln2 the Bose mean is 1, the Fermi mean is 1/3 and the dilute comparator is 1/2. The difference is substantial; calling the comparator accurate merely because the temperature has a familiar numerical value would miss the relevant dimensionless condition.

Find the classical limit

Write z=exp(βμ), called fugacity. Then ⟨n⟩=1/(z⁻¹exp(βε)∓1), with the minus sign for bosons and plus sign for fermions. When z exp(−βε) is much smaller than 1, the extra ±1 is negligible and both means reduce to z exp(−βε). This is the dilute occupation limit.

For a large homogeneous nonrelativistic gas, define the thermal wavelength λ=h/√(2πmk_B T), where h=2πℏ and m is particle mass. If g equally populated internal states have the same one-particle energy, nλ³/g much smaller than 1 is the usual classical degeneracy criterion, with n=N/V. Different internal populations require their own appropriate densities rather than automatic division by g.

High temperature can create this regime in a gas with many thermally accessible states, but our permanently truncated three-state model is a useful warning. Two particles cannot become dilute among an ever-growing number of states if we keep only those same three states at every temperature. Its Bose, Fermi and classical comparator distributions do not become exactly identical just by taking βΔ to zero.

Further: fluctuations and number constraints

A fermionic mode has a Bernoulli occupation distribution: Var(n)=⟨n⟩(1−⟨n⟩). A grand canonical bosonic mode has a geometric distribution with Var(n)=⟨n⟩(1+⟨n⟩). The dilute classical grand canonical comparator is Poisson, with variance equal to its mean. These formulas compare fluctuations of a mode’s occupation, not uncertainty in an estimated mean from a finite experiment.

The formulas follow by differentiating the single-mode grand partition sums, or by evaluating their first two moments. They do not supply a relaxation time or an actual random time trace. For temporal correlations we would need additional dynamics, collisions, coupling to a reservoir and a measurement model.

In the finite canonical opening, occupations are constrained by n_a+n_b+n_c=2 in every configuration. Their covariances therefore sum to zero along each row: ΣⱼCov(n_i,n_j)=Cov(n_i,2)=0. They cannot all be independent fluctuating modes. In particular, one must not use a large grand canonical condensate-number variance as an automatic prediction for an isolated fixed-N condensate.

Count available states before counting particles

To turn occupations into a particle total, count the available one-particle states as well. Let D(ε)dε be the number of states with energies between ε and ε+dε. For a large three-dimensional box with quadratic dispersion ε=p²/(2m), D(ε)=gV(2m/ℏ²)³ᐟ²√ε/(4π²). Its units are inverse energy, so D(ε)dε is a dimensionless state count.

The particle number is N≈∫D(ε)⟨n(ε)⟩dε, and the internal energy is U≈∫εD(ε)⟨n(ε)⟩dε. These are continuum approximations to discrete sums, with their large-box and low-energy-boundary limitations. A normalized energy probability density for a randomly chosen particle is D(ε)⟨n(ε)⟩/N, not ⟨n(ε)⟩ alone.

Changing from energy to frequency or momentum also changes the integration measure. Spin degeneracy belongs in the state count only if it has not already been counted among the complete states. The low-energy ground state of a Bose gas will require separate attention rather than being erased by the continuum approximation.

Fill a Fermi sea

At zero temperature, the ideal Fermi gas fills available complete states in order of energy. In a large homogeneous system its occupied momentum states form a Fermi sea. The Fermi energy E_F is the thermodynamic-limit zero-temperature chemical potential; the associated wave number satisfies k_F=(6π²n/g)¹ᐟ³ and E_F=ℏ²k_F²/(2m). Here g counts degenerate internal components with a common population rule.

Integrating the occupied quadratic-dispersion density of states gives U/N=3E_F/5 and pressure P=2nE_F/5 at T=0. This nonzero degeneracy pressure is not thermal heating, and it does not require a separately inserted repulsive force. Compressing the gas changes the available momentum states and the energy required by the exclusion rule.

The Fermi temperature T_F=E_F/k_B is an energy scale expressed in kelvin, not a claim that a zero-temperature gas is secretly warm. It tells us whether T/T_F is small. Ideal-gas ideas give useful starting points for electrons in metals and some astrophysical systems, but bands, interactions and relativistic dispersion can change the quantitative result.

Further: low-temperature fermions at fixed density

At small positive T, only states in an energy window of order k_B T around the Fermi surface can readily change occupation. Deeply occupied states cannot all gain arbitrarily small independent excitation energies because nearby destinations are already occupied. This is why the electronic ideal-gas heat capacity is much smaller than assigning 3k_B/2 to every particle would suggest.

For the homogeneous three-dimensional quadratic ideal gas at fixed N and V, the leading Sommerfeld terms are μ/E_F≈1−(π²/12)(T/T_F)² and U/(NE_F)≈3/5+(π²/4)(T/T_F)². Thus C_V/(Nk_B)≈(π²/2)(T/T_F). These are asymptotic low-temperature expressions, not exact equations across the full crossover.

More generally, at each positive T the chemical potential must be adjusted to preserve the chosen density. With u=ε/E_F, θ=T/T_F and a=μ/E_F, solve (3/2)∫₀∞√u/[exp((u−a)/θ)+1]du=1. Then U/(NE_F)=(3/2)∫₀∞u³ᐟ²/[exp((u−a)/θ)+1]du. Holding μ fixed instead describes a different constraint. The integrals include their entire tails; a finite plotted interval is not permission to renormalize a truncated integral.

The table uses the full fixed-density integrals, not their leading low-temperature approximation. As temperature rises the chemical potential can become negative without making the particle energies negative or the ideal gas unstable. Each temperature is a calculated sample; a line between samples would only guide the eye.

Separate the Bose ground state

For the ideal uniform three-dimensional Bose gas, lower T at fixed density while setting the lowest one-particle energy to zero. The chemical potential approaches zero from below. Excited states can hold only a limited particle density at a given T when μ approaches that boundary.

For one internal component, the limiting excited density is n_ex=ζ(3/2)/λ³, where ζ is the Riemann zeta function and ζ(3/2)≈2.61237535. The continuum density-of-states integral misses the individual lowest state: the √ε factor cannot represent a macroscopic population concentrated in that one discrete state. We must write N=N₀+N_ex and keep N₀ separately.

Define T_c by nλ(T_c)³=ζ(3/2). Equivalently, k_B T_c=(2πℏ²/m)[n/ζ(3/2)]²ᐟ³ for this ideal single-component homogeneous model. Below it, N₀/N=1−(T/T_c)³ᐟ² in the thermodynamic limit. Condensation means a macroscopically occupied quantum state, not that all particles have literally become one particle or lost conservation of their total number.

Further: energy and fugacity of the Bose gas

The ideal model’s ground state has energy zero, while its excited population carries thermal energy. Put t=T/T_c. For 0<t≤1, U/(Nk_B T_c)=(3/2)[ζ(5/2)/ζ(3/2)]t⁵ᐟ². Dividing by t gives U/(Nk_B T) on that positive-temperature domain. At exactly T=0 the latter expression is a 0/0 ratio and is undefined, even though its right-hand limit is zero and U/(Nk_B T_c)=0 is well defined.

Above T_c there is no macroscopic condensate in this ideal thermodynamic-limit model. The fugacity z is determined by Li₃ᐟ₂(z)=ζ(3/2)t⁻³ᐟ², with 0<z<1. The polylogarithm Li_s(z)=Σₗ₌₁∞zˡ/lˢ provides a compact notation for the continuum integrals. The energy follows U/(Nk_B T)=(3/2)Li₅ᐟ₂(z)/Li₃ᐟ₂(z). These relations require solving for z rather than leaving μ fixed while claiming fixed N.

Differentiating the positive-temperature expression below T_c gives C_V/(Nk_B)=(15/4)[ζ(5/2)/ζ(3/2)]t³ᐟ². For the homogeneous ideal gas the heat capacity is continuous at the transition, with nonanalytic temperature dependence in the thermodynamic limit. Trap geometry and interactions can change the density of states and critical behavior; their formulas cannot simply inherit this exponent or prefactor.

Further: dimension, size and superfluidity

Our two-particle system has a finite sum of smooth positive-temperature weights. It does not have a thermodynamic phase transition. A sharp Bose-condensation transition requires an appropriate many-particle limit; finite systems exhibit a rounded crossover whose details depend on the spectrum and constraints.

For a homogeneous ideal gas with quadratic dispersion in d dimensions, the low-energy density of states is proportional to ε^(d/2−1). As μ approaches zero, the excited-particle integrand behaves as ε^(d/2−2). Its integral at zero energy is finite only for d>2. Thus this uniform infinite ideal-gas argument yields no finite-temperature Bose condensate in one or two dimensions. A confining trap changes the density of states, and interacting two-dimensional systems can have a Berezinskii–Kosterlitz–Thouless transition instead; those are different assumptions, not exceptions produced by ignoring the infrared limit.

Bose condensation and superfluidity ask different questions. Condensation concerns occupation and coherence of a state; superfluidity concerns response and dissipation of motion. The simple Landau excitation test uses v_L=inf_{p>0}[ε(p)/p], where p is momentum magnitude and ε(p) is excitation energy. At flow speed v, creating an excitation can be energetically allowed when vp>ε(p) for a favorable momentum direction. For the homogeneous ideal gas’s zero-temperature spectrum ε(p)=p²/(2m), the ratio ε(p)/p=p/(2m) tends to zero, so this test gives no positive critical velocity. This is an energetic test for the assumed excitations, not a complete account of dissipation or nucleation in a particular flow. Interactions can be crucial for real superfluid behavior. A condensate fraction alone is not a complete transport measurement.

Use the statistics within its physical model

Thermal photons and phonons illustrate bosonic occupation factors, but their equilibrium number is not conserved in the way atom number is. With the appropriate thermalization assumptions their chemical potential is zero, and their dispersion and density of states differ from a massive quadratic gas. An externally driven or specially constrained photon system needs its own treatment.

A Fermi–Dirac occupation function can also describe noninteracting or effective quasiparticle states in a solid, but replacing a band structure by ε=p²/(2m) is an additional approximation. Likewise, paired fermions may form bosonic composite degrees of freedom in an appropriate regime; this does not mean the underlying constituent fermions have stopped obeying exclusion.

Before using either formula, ask: what are the complete states, which quantities are held fixed, is the gas ideal, what is the dispersion, what counts the internal degeneracy, and is the finite or thermodynamic-limit description appropriate? Those questions turn a remembered ±sign into a physical model.

Worked examples

Try it yourself

1. How many allowed Bose and Fermi occupation lists occur in the opening model?

   Show the answer and reasoning

   Six Bose and three Fermi lists.

   Enumerate nonnegative occupations summing to 2; then for fermions remove every list with an occupation 2.

2. Why do the two bosonic configurations at total energy 2Δ have equal weights?

   Show the answer and reasoning

   Each is one quantum occupation state with the same energy.

   An exchange of invented particle labels is not another Fock configuration; do not add a label-counting factor 2.

3. What is the probability that the two bosons both occupy c at the opening temperature?

   Show the answer and reasoning

   1/35.

   Its weight is 1/16 and Z₂,B=35/16, so the normalized ratio is 1/35.

4. Why is the fermionic mean energy higher than the bosonic one in this opening comparison?

   Show the answer and reasoning

   Their allowed configuration sets and weights differ; fermions cannot use the doubly occupied lowest state.

   This particular ideal-model comparison follows state counting, not an extra assigned repulsive force or a universal claim about every interacting material.

5. At positive T and ε=μ, what is the Fermi mean occupation?

   Show the answer and reasoning

   1/2.

   The occupation formula gives 1/(exp(0)+1). This finite-T value should not be used to settle every discrete zero-temperature boundary degeneracy.

6. Is the Bose formula at x=0 a valid finite grand canonical distribution?

   Show the answer and reasoning

   No.

   Its geometric partition sum fails to converge. A limiting condensed thermodynamic description keeps the ground-state population and ensemble conditions explicit.

7. Does ⟨n⟩_B=3 mean a probability greater than 1?

   Show the answer and reasoning

   No; it is an average of three particles per state.

   Occupancy is not a normalized probability density over particle energies.

8. Can the single-mode variances be used as independent fluctuations for all three fixed-N opening occupations?

   Show the answer and reasoning

   No.

   The exact constraint n_a+n_b+n_c=2 couples their fluctuations; covariance row sums must vanish.

9. What condition makes the ±1 denominator correction small?

   Show the answer and reasoning

   z exp(−βε)≪1 for the relevant states.

   Then the mode mean is small and both quantum distributions approach the dilute classical occupation.

10. Does taking T→∞ make the fixed three-state/N=2 model exactly classical?

    Show the answer and reasoning

    No.

    The accessible state count is permanently restricted to three, so dilution among increasingly many thermal states does not occur.

11. What changes if spin degeneracy has already been included by explicitly listing both spin states?

    Show the answer and reasoning

    Do not multiply the state count by that same degeneracy again.

    Exclusion and the occupation formula already apply separately to each complete state.

12. Can a zero-temperature ideal Fermi gas exert pressure?

    Show the answer and reasoning

    Yes.

    Exclusion fills a range of nonzero momenta. Compressing the gas changes its ground-state energy even without thermal excitation.

13. In a fixed-density finite-T Fermi calculation, why must μ generally vary with T?

    Show the answer and reasoning

    To keep the density integral equal to the assigned N/V.

    Changing T at fixed μ generally changes N; it is a different set of controls.

14. What is the condensate fraction at T=T_c/4 in the ideal homogeneous 3D model?

    Show the answer and reasoning

    7/8=0.875.

    (1/4)³ᐟ²=1/8, and the remainder occupies the ground state in the thermodynamic-limit model.

15. What is U/(Nk_B T) at exactly T=0 for the ideal Bose endpoint?

    Show the answer and reasoning

    Undefined.

    Both numerator and denominator vanish. Its right-hand limit 0 and the well-defined U/(Nk_B T_c)=0 are different statements.

16. Does observing a condensate fraction alone establish superfluid transport?

    Show the answer and reasoning

    No.

    Occupation and coherence do not by themselves determine dissipation or critical velocity. Interactions and dynamical response matter.

What to carry forward

• Bose and Fermi statistics begin with allowed occupations of complete quantum states, not a newly added mechanical force.
• The ± sign in a grand canonical mean occupation follows the allowed single-state number sum; it is not an energy probability density.
• Fixed-number finite systems and independent grand canonical modes are different descriptions with different fluctuation constraints.
• The classical limit requires dilute occupation of relevant states; a permanently truncated small spectrum is a useful counterexample to a careless high-temperature slogan.
• Fermi filling produces a ground-state energy and degeneracy pressure; low-temperature excitations occupy a restricted window near the Fermi surface.
• Ideal Bose condensation requires the appropriate density of states, ground-state separation and thermodynamic limit. Finite size, dimension, trapping, interactions and superfluid response must retain their own conditions.