This page is a shooting range : every kind of question the Bose-Einstein parent note can throw at you, worked from zero. Before the examples, we lay out a matrix of every case class so you can see there are no gaps.
Everything below uses one formula, the Bose-Einstein mean occupancy:
n ˉ ( ε ) = e ( ε − μ ) / k B T − 1 1 .
Let us first name every symbol so nothing is used before it is earned.
Definition The symbols, in plain words
ε energy of one quantum state (think: the price of a seat ). Units: joules (J) or, conveniently, in units of k B T μ chemical potential (the discount ): how much energy the reservoir gives up to hand you one particle. See Chemical potential .T temperature in kelvin (K); k B 1.381 × 1 0 − 23 k B T thermal energy scale , the natural yardstick against which we measure ε − μ We abbreviate the single dimensionless number that runs the whole show:
y ≡ k B T ε − μ . n ˉ = e y − 1 1 Everything below is just: pick y n ˉ
y ε μ T y = ( ε − μ ) / k B T seat price measured in thermal units, after the discount . Small y y
Look at the curve: it dives to infinity as y → 0 + decays like e − y for large y
Every question about n ˉ case classes . The examples that follow each cover one (or more) cells.
Cell
Case class
Defining condition
Covered by
A
Photon gas (no μ μ = 0 y = ε / k B T Ex 1, Ex 6
B
Intermediate y (quantum bunching)y ∼ 0.1 1 Ex 2
C
Classical tail (dilute limit)y ≫ 1 Ex 3
D
Degenerate: y → 0 + (near-condensation)μ → ε m i n − Ex 4
E
Forbidden input (y < 0 μ > ε n ˉ Ex 5
F
Real-world word problem back out T ε
Ex 6
G
Exam twist : ratio / BE-vs-FDcompare two statistics at same y
Ex 7
H
Temperature sweep (limiting behaviour)fixed ε − μ T
Ex 8
Recall The whole matrix in one sentenceQuestion ::: Every BE problem reduces to "what is y = ( ε − μ ) / k B T n ˉ = 1/ ( e y − 1 ) y > 0 y → 0 y → ∞
Worked example Ex 1 · A light mode with energyε = k B T Statement. A cavity holds photons at temperature T ε = k B T
Forecast: guess before computing — more than 1, or less than 1? (Photons are social, but this seat is not cheap…)
Step 1 — Set μ = 0
Why this step? Photons can be created and destroyed by the cavity walls, so their number is not conserved ; the thermodynamic cost of adding one is zero, giving μ = 0 Planck radiation law derivation). Then y = k B T ε − 0 = k B T ε
Step 2 — Plug in ε = k B T
Why this step? We're told the numerator equals the denominator, so y = 1 n ˉ = e 1 − 1 1 = 2.71828 − 1 1 = 1.71828 1 .
Step 3 — Evaluate.
n ˉ ≈ 0.582.
Verify: n ˉ < 1 0.582 × ( e − 1 ) = 0.582 × 1.718 = 1.000 k B T why blackbody intensity falls at high frequency.
Worked example Ex 2 · Bosons pile up nearμ Statement. A boson level has y = ( ε − μ ) / k B T = 0.1 n ˉ
Forecast: the net seat is very cheap (y
Step 1 — Recognise the small-y
Why this step? y = 0.1 ≪ 1 n ˉ
Step 2 — Compute the exponential then subtract one.
Why this step? The whole quantum effect lives in the "− 1 e 0.1 = 1.10517 , e 0.1 − 1 = 0.10517.
Step 3 — Invert.
n ˉ = 0.10517 1 ≈ 9.51.
Verify: For small y e y − 1 ≈ y n ˉ ≈ 1/ y = 1/0.1 = 10 9.51 y n ˉ ≈ k B T / ( ε − μ ) e − 0.1 = 0.905 10× smaller : this factor-of-10 gap is boson bunching , the seed of lasers and Bose-Einstein condensation .
Worked example Ex 3 · High-energy state recovers Maxwell–BoltzmannStatement. A boson level has y = 5 n ˉ BE n ˉ MB = e − y
Forecast: far to the right on s01 — do BE and classical agree here, or clash?
Step 1 — BE value.
Why this step? Direct substitution; we expect the "− 1 e 5 e 5 = 148.413 , n ˉ BE = 148.413 − 1 1 = 147.413 1 = 0.006784.
Step 2 — Classical value.
Why this step? This is the Maxwell-Boltzmann statistics estimate — the limit BE should approach.
n ˉ MB = e − 5 = 0.006738.
Step 3 — Percent difference.
0.006738 0.006784 − 0.006738 × 100% = 0.68%.
Verify: They agree to under 1% ε − μ ≫ k B T − 1 n ˉ
Worked example Ex 4 · Ground state asμ → ε m i n − Statement. The ground state sits at ε m i n n ˉ 0 μ y = 0.01 y = 0.001
Forecast: the discount μ
Step 1 — y = 0.01
Why this step? We probe the approach to the forbidden point y = 0 n ˉ 0 = e 0.01 − 1 1 = 0.0100502 1 ≈ 99.5.
Step 2 — y = 0.001
Why this step? Halving-and-more of y 1/ y n ˉ 0 = e 0.001 − 1 1 = 0.00100050 1 ≈ 999.5.
Step 3 — Read the trend.
Why this step? From Step 1 to Step 2, y × 10 n ˉ 0 × ∼ 10 n ˉ 0 → ∞ y → 0 +
Verify: The law n ˉ 0 ≈ 1/ y 100 1000 99.5 999.5 divergence is the mathematical face of Bose-Einstein condensation : at μ = ε m i n 1/ ( 1 − x ) μ ε m i n
Worked example Ex 5 · What ifμ > ε Statement. A student sets μ = 2 k B T ε = k B T y = − 1 n ˉ
Forecast: what sign will n ˉ
Step 1 — Form y
y = k B T ε − μ = k B T k B T − 2 k B T = − 1.
Step 2 — Substitute (blindly, to expose the trap).
Why this step? We plug in to see why negative y e − 1 = 0.3679 , e − 1 − 1 = − 0.6321 , n ˉ = − 0.6321 1 = − 1.582.
Step 3 — Reject on physical grounds.
Why this step? A count of particles can never be negative. The formula returned n ˉ < 0 μ > ε forbidden for that level. The condition μ < ε m i n Grand canonical ensemble geometric-sum convergence x < 1 every level.
Verify: − 1.582 × ( e − 1 − 1 ) = − 1.582 × ( − 0.6321 ) = 1.000 physical , not arithmetic. ✓ Rule locked in: always check y > 0 n ˉ
Common mistake Trusting a negative occupancyWhy it feels right: the formula "gave an answer," so it must mean something. Fix: BE is only valid for y > 0 n ˉ screaming that you violated μ < ε m i n
Worked example Ex 6 · Cosmic Microwave Background photon modeStatement. The CMB fills space at T = 2.725 ν = 160.2 ε = h ν h = 6.626 × 1 0 − 34 n ˉ
Forecast: at the spectral peak , is n ˉ k B T
Step 1 — Photon ⇒ μ = 0
Why this step? Same reason as Ex 1: photon number isn't conserved, μ = 0 y = h ν / k B T
Step 2 — Compute ε = h ν
ε = 6.626 × 1 0 − 34 × 1.602 × 1 0 11 = 1.0615 × 1 0 − 22 J .
Step 3 — Compute k B T
k B T = 1.381 × 1 0 − 23 × 2.725 = 3.763 × 1 0 − 23 J .
Step 4 — Form y
Why this step? y y = 3.763 × 1 0 − 23 1.0615 × 1 0 − 22 = 2.821 , n ˉ = e 2.821 − 1 1 = 16.79 − 1 1 = 15.79 1 ≈ 0.0633.
Verify: y ≈ 2.82 ≈ 2.82 radiance peak of a blackbody — a lovely consistency with the Planck radiation law . ✓ Units cancelled to a pure number. ✓ n ˉ ≈ 0.063
Worked example Ex 7 · Samey Statement. At y = 0.5 n ˉ − 1 + 1 e − y
Forecast: which is biggest — the social boson, the antisocial fermion, or the classical middle?
Step 1 — Common exponential.
Why this step? All three share e 0.5 = 1.64872
Step 2 — BE.
n ˉ BE = 1.64872 − 1 1 = 0.64872 1 = 1.5415. Step 3 — FD (see Fermi-Dirac statistics — fermions ).
Why this step? Only the sign flips — the "+ 1 n ˉ FD = 1.64872 + 1 1 = 2.64872 1 = 0.3775. Step 4 — Classical.
n ˉ MB = e − 0.5 = 0.60653.
Step 5 — Rank. n ˉ BE ( 1.54 ) > n ˉ MB ( 0.61 ) > n ˉ FD ( 0.38 ) .
Verify: Bosons bunch above classical, fermions anti-bunch below it — the classical value always sits between the two, exactly as theory demands. ✓ Note n ˉ FD < 1 n ˉ BE
The bar chart makes the ordering visual: cyan boson tower highest, amber fermion lowest, white classical between.
Worked example Ex 8 · Fixε − μ T Statement. A level has fixed ε − μ = Δ n ˉ k B T = Δ y = 1 k B T = 10Δ y = 0.1 k B T = Δ/5 y = 5 T → 0 T → ∞
Forecast: heating a boson gas — does each level's crowd grow or shrink?
Step 1 — (a) y = 1
Why this step? Baseline, same as Ex 1's arithmetic.
n ˉ = e 1 − 1 1 = 0.5820. Step 2 — (b) hot, y = 0.1
Why this step? Raising T y n ˉ = e 0.1 − 1 1 = 9.508. Step 3 — (c) cold, y = 5
Why this step? Lowering T y n ˉ = e 5 − 1 1 = 0.006784.
Step 4 — Read the limits.
Why this step? Extrapolate the trend to the endpoints.
T → ∞ y → 0 + n ˉ → ∞ T → 0 y → ∞ n ˉ → 0 Bose-Einstein condensation ).
Verify: From (c)→(a)→(b), heating (T n ˉ 0.0068 → 0.582 → 9.51 y = Δ/ k B T
Mnemonic One-number workflowSee y y Compute y = ( ε − μ ) / k B T y > 0 n ˉ = 1/ ( e y − 1 ) y y B oson = B elow-by-one = the − 1
