An interactive demo created by Claude Opus 5.5, based on the lecture notes by Sang Hoon Lee (Chapter 7, Section 7.6, Bose–Einstein Condensation)
Cool a gas of identical bosons and, below a critical temperature, a macroscopic fraction of the atoms abruptly piles into the single lowest-energy state. No need to reach absolute zero. The reason lies in how identical particles are counted.
Photons and phonons can be created freely, so their \(\mu = 0\). Now consider more ordinary bosons, such as atoms with integer spin, whose number is fixed from the outset. Then \(\mu\) is a nontrivial function of density and temperature. At \(T = 0\), trivially, every atom is in the ground state. In a box of volume \(V = L^3\) that state has energy
and at any temperature the average number of atoms in it is
For \(N_0\) to be large, \(\mu\) must sit a tiny bit below \(\varepsilon_0\); at \(T = 0\), \(\mu = \varepsilon_0\). The question is how low the temperature must be for that to happen.
The condition that fixes \(\mu\) is that the occupancies of all states add up to \(N\). Converting the sum to an integral with the density of states (half that of the electrons in Section 7.3, since these bosons have spin zero),
This can’t be done analytically, so try \(\mu = 0\), which should work at very low temperature:
This is absurd as it stands, since \(N\) can’t grow as \(T^{3/2}\). It is correct for exactly one temperature, the condensation temperature:
Above \(T_c\), \(\mu\) must be negative: a negative \(\mu\) makes the denominator larger, compensating for the rising \(T\). Below \(T_c\), the integral fails. Near \(\varepsilon = 0\), \(\sqrt{\varepsilon}/(e^{\varepsilon/k_{\mathrm{B}}T}-1) \approx k_{\mathrm{B}}T/\sqrt{\varepsilon} \to \infty\), a spike that doesn’t represent the sum over the actual, discretely spaced states. The integral still correctly counts atoms in the vast majority of states, with \(\varepsilon \gg \varepsilon_0\), so it gives the number of atoms in excited states, and the rest must be in the ground state:
Energies in units of \(k_{\mathrm{B}}T_c\). The shaded area is \(N_{\text{excited}}/N\); the bar at \(\varepsilon = 0\) is the condensate, \(N_0/N\), which no smooth curve can represent.
of the atoms are in the ground state
Above \(T_c\), \(\mu\) is found numerically, by guessing values until the integral equals \(N\).
Above \(T_c\), \(\mu < 0\) and essentially all of the atoms are in excited states. Below \(T_c\), \(\mu \approx 0\) and atoms accumulate abruptly in the ground state. \(N_0(T)\) is continuous but not differentiable at \(T_c\): a phase transition, Bose–Einstein condensation (BEC). These charts follow the slider above.
\(\mu/k_{\mathrm{B}}T_c\): zero below \(T_c\), falling away above it.
With the quantum volume \(v_Q = (h^2/2\pi mk_{\mathrm{B}}T)^{3/2}\), the condition at \(T_c\) reads \(N/V = 2.612/v_Q\): condensation begins just as the atoms’ wavefunctions start to overlap significantly.
The sharp kink belongs to the limit of infinitely many atoms. The lecture notes point out that for a finite number, there is a range of temperatures just below \(T_c\) where the simple picture isn’t accurate (Problem 7.66), and it need not be narrow. Here the sum over the actual box states, \(\varepsilon = (h^2/8mL^2)(n_x^2+n_y^2+n_z^2)\), is done exactly, solving for \(\mu\) at each temperature.
The box walls remove low-lying states compared with the smooth \(g(\varepsilon)\), so the excited states fill up sooner and condensation sets in above \(T_c\). The curve sharpens only slowly as \(N\) grows.
Numerically, \(T_c\) is very small in all realistic experiments, but not as small as you might guess. A single particle in a box is reasonably likely to be in the ground state only when \(k_{\mathrm{B}}T\) is of order \(\varepsilon_0\). With many bosons, most are in the ground state at temperatures only somewhat below \(T_c\), and \(k_{\mathrm{B}}T_c\) exceeds \(\varepsilon_0\) by a factor of order \(N^{2/3}\). The hierarchy is \((\varepsilon_0 - \mu) \ll \varepsilon_0 \ll k_{\mathrm{B}}T_c\).
A logarithmic energy axis, measured from the bottom of the box, so the three scales fit on one line. Tick marks are the actual single-particle levels of the box; \(\mu\) is shown at \(T = T_c/2\).
condensation temperature \(T_c\)
In the experiments of Figure 7.35 (C. E. Wieman, 1996), rubidium-87 atoms held in a magnetic trap were imaged after the trap was switched off, so the pictures map their velocities. Above \(T_c\), at 200 nK, the cloud is broad and isotropic, as Maxwell–Boltzmann predicts; at 100 nK a dense, elongated condensate appears; near \(T = 0\) essentially all atoms are in it. Real traps are harmonic rather than boxes, so this calculator gives only an order-of-magnitude estimate for them. For liquid helium-4, the ideal-gas formula gives about 3.1 K, close to the 2.17 K at which helium actually becomes superfluid, even though the atoms in a liquid interact strongly.
Consider first \(N\) distinguishable, noninteracting particles in a box. Each, separately, has a decent chance of occupying any of the roughly \(Z_1\) single-particle states with energy up to about \(k_{\mathrm{B}}T\), so its chance of being in the ground state is only about \(1/Z_1\). Only a tiny fraction end up there: no condensation.
Look at the whole system at once, though, and there seems to be a paradox. The system state with every particle in the ground state has Boltzmann factor \(e^{0} = 1\), while a typical state with \(U \approx Nk_{\mathrm{B}}T\) has \(e^{-N} \ll 1\). The resolution is that a state is not an energy level. Any particular state with energy of order \(Nk_{\mathrm{B}}T\) is improbable, but there are so many of them that together they win, like the odds of an individual winning the lottery versus the odds that someone does. For distinguishable particles the number of arrangements is \(Z_1^N\), easily beating \(e^{-N}\).
For identical bosons, only the occupation numbers matter, and the count is much smaller:
When \(Z_1 \gg N\), excited states still dominate, as for distinguishable particles. When \(Z_1 \ll N\), the count can’t compensate for \(e^{-N}\): states with all bosons excited become exponentially improbable. The combinatorics gets about one boson into each available excited state, and the rest condense.
One ground state at energy 0, and \(Z_1\) excited states, all at energy \(\varepsilon\). For distinguishable particles, \(P(k\ \text{excited}) \propto \binom{N}{k} Z_1^k e^{-k\varepsilon/k_{\mathrm{B}}T}\). For identical bosons, \(P(k) \propto \binom{k + Z_1 - 1}{k} e^{-k\varepsilon/k_{\mathrm{B}}T}\). The chart shows the resulting probability of each ground-state population, and the boxes show one system state drawn at random from each.
The arrangement counts are for all \(N\) particles excited.
The explanation of BEC lies in the combinatorics of counting arrangements of identical particles. How do we know bosons of a given species really are identical, and that every distinct state of the system deserves the same statistical weight? Good theoretical answers need more quantum mechanics, and even they aren’t completely airtight: some undiscovered interaction might yet tell the atoms apart and make a condensate spontaneously evaporate. So far, no such interaction seems to exist. Invoking Occam’s razor, Schroeder concludes, tentatively, that bosons of a given species are truly indistinguishable, quoting David Griffiths:
“even God cannot tell them apart.”