Bose-Einstein Distribution

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Photon statistics are relatively painless because the number of photons is not conserved. Other bosons, such as $^{87}$ Rb atoms, are conserved. Their number, $N = \sum_s n_s$ , is the same regardless of the system's macroscopic quantum state . Different states have the same particles distributed in different ways among the single-particle states. That means that if we put particles in state , there are only particles left to be distributed among the other quantum states. There are several ways to handle this constraint. All lead us to introduce the particles' chemical potential, $\mu$ . Reif's Fundamentals of Statistical and Thermal Physics describes a straightforward brute-force attack. Kittel's Thermal Physics follows an elegant line of reasoning due to Gibbs. I'll follow the discussion in Feynman's Statistical Mechanics which lacks a little rigor, but is easy to grasp.

Let's say that the system is in equilibrium with a thermal reservoir which provides both heat and particles. Technically, we should move into the Grand Canonical ensemble at this point, as does Kittel. But we can get away without all of that machinery by positing that moving a particle out of the reservoir and into the system costs an amount of (free) energy $\mu$ . This energy cost is paid by the reservoir and thus is independent of the state into which the particle moves. Once again, $\mu$ is a property of the reservoir, not of the system. We continue to transfer particles from the reservoir into the system until the benefit of moving particles into the system's quantum states is offset by the cost in energy to the reservoir as it gives up particles. At that point we stop; and the number of particles in the system depends on the energy $\mu$ required to transfer each one. Our task is to choose $\mu$ so that the desired number of particles end up in the system.

Here's how to do it. A particle in state now has energy $\epsilon_s - \mu$ . The total energy of the system in macroscopic state is therefore

$\displaystyle E_S = \sum_s n_s (\epsilon_s - \mu).$

(40)

The Boltzmann factor for finding the system in that state is

$\displaystyle e^{-\beta \sum_s n_s (\epsilon_s - \mu)}.$

(41)

The system's partition function then becomes

$\displaystyle Z = \sum_{S = \{n_1,n_2,\ldots\}} e^{-\beta \sum_s n_s (\epsilon_s - \mu)}.$

(42)

In principle, we can now compute $\bar n_s$ and fix $\mu$ so that $N = \sum_s \bar n_s$ . Doing so requires more information about the nature of the eigenstates of the single-particle Hamiltonian. But, in principle, it can be done.

Having introduced the chemical potential to take care of the constraint on the number of particles, we can address the bosons' bosonic nature. Each microscopic state can contain any number of particles, $n_s = 0, 1, 2, \ldots$ . Thus, the rest of the derivation follows that of the Planck distribution in Eq. with the additional term of $\mu$ in the exponent. The mean number of bosons in state at temperature is therefore

$\displaystyle \shadowbox{ $ \displaystyle \bar n_s = \frac{1}{e^{\beta (\epsilon_s - \mu)} - 1}. $ }$

(43)

This is known as the Bose-Einstein Distribution. We'll use it to explain Bose-Einstein condensation of alkali metal gases and superconductivity.

As before, we can drop the subscript to obtain an expression for $n(\epsilon)$ , the distribution function for non-interacting bosons in the single-particle potential .

Notice that the effect of imposing conservation of particle number is to introduce the chemical potential. The chemical potential is set by the normalization condition

$\displaystyle \sum_s \bar n_s = N$

(44)

because the thermally-averaged occupation number $\bar n_s$ for each of the one-particle states depends on $\mu$ . Photons can be created and destroyed freely, without any cost to the thermal reservoir: their chemical potential is zero.

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