Editing Bose–Einstein condensate (section)

=== Bose Einstein's non-interacting gas ===
{{Main|Bose gas}}
Consider a collection of ''N'' non-interacting particles, which can each be in one of two [[quantum state]]s, <math>|0\rangle</math> and <math>|1\rangle</math>. If the two states are equal in energy, each different configuration is equally likely.

If we can tell which particle is which, there are <math>2^N</math> different configurations, since each particle can be in <math>|0\rangle</math> or <math>|1\rangle</math> independently. In almost all of the configurations, about half the particles are in <math>|0\rangle</math> and the other half in <math>|1\rangle</math>. The balance is a statistical effect: the number of configurations is largest when the particles are divided equally.

If the particles are indistinguishable, however, there are only <math>N+1</math> different configurations. If there are <math>K</math> particles in state <math>|1\rangle</math>, there are <math>N-K</math> particles in state <math>|0\rangle</math>. Whether any particular particle is in state <math>|0\rangle</math> or in state <math>|1\rangle</math> cannot be determined, so each value of <math>K</math> determines a unique quantum state for the whole system.

Suppose now that the energy of state <math>|1\rangle</math> is slightly greater than the energy of state <math>|0\rangle</math> by an amount <math>E</math>. At temperature <math>T</math>, a particle will have a lesser probability to be in state <math>|1\rangle</math> by <math>e^{-E/kT}</math>. In the distinguishable case, the particle distribution will be biased slightly towards state <math>|0\rangle</math>. But in the indistinguishable case, since there is no statistical pressure toward equal numbers, the most-likely outcome is that most of the particles will collapse into state <math>|0\rangle</math>.

In the distinguishable case, for large ''N'', the fraction in state <math>|0\rangle</math> can be computed. It is the same as flipping a coin with probability proportional to <math>\exp{(-E/T)}</math> to land tails.

In the indistinguishable case, each value of <math>K</math> is a single state, which has its own separate Boltzmann probability. So the probability distribution is exponential:

:<math>\,
P(K)= C e^{-KE/T} = C p^K.
</math>

For large <math>N</math>, the normalization constant <math>C</math> is <math>1-p</math>. The expected total number of particles not in the lowest energy state, in the limit that <math>N\rightarrow \infty</math>, is equal to
: <math>\sum_{n>0} C n p^n=p/(1-p) </math>
It does not grow when ''N'' is large; it just approaches a constant. This will be a negligible fraction of the total number of particles. So a collection of enough Bose particles in thermal equilibrium will mostly be in the ground state, with only a few in any excited state, no matter how small the energy difference.

Consider now a gas of particles, which can be in different momentum states labeled <math>|k\rangle</math>. If the number of particles is less than the number of thermally accessible states, for high temperatures and low densities, the particles will all be in different states. In this limit, the gas is classical. As the density increases or the temperature decreases, the number of accessible states per particle becomes smaller, and at some point, more particles will be forced into a single state than the maximum allowed for that state by statistical weighting. From this point on, any extra particle added will go into the ground state.

To calculate the transition temperature at any density, integrate, over all momentum states, the expression for maximum number of excited particles, <math>p/(1-p)</math>:

:<math>\,
 N = V \int {d^3k \over (2\pi)^3} {p(k)\over 1-p(k)} = V \int {d^3k \over (2\pi)^3} {1 \over e^{k^2\over 2mT}-1} </math>
:<math>\,
p(k)= e^{-k^2\over 2mT}.
</math>

When the integral (also known as [[Bose–Einstein integral]]) is evaluated with factors of <math>k_B</math> and <math>\hbar</math> restored by dimensional analysis, it gives the critical temperature formula of the preceding section. Therefore, this integral defines the critical temperature and particle number corresponding to the conditions of negligible [[chemical potential]] <math>\mu</math>. In [[Bose–Einstein statistics]] distribution, <math>\mu</math> is actually still nonzero for BECs; however, <math>\mu</math> is less than the ground state energy. Except when specifically talking about the ground state, <math>\mu</math> can be approximated for most energy or momentum states as&nbsp;<math>\mu \approx 0</math>.