## Thermodynamic properties of non-interacting bosons

Consider a system of non-interacting bosons that can occupy microscopic quantum states with energies $\epsilon_i$. Since the bosons do not interact, the energy of one quantum state does not depend on the occupation of any of the other quantum states. A macrostate $q$ of this system consists of $N_q$ bosons. The total energy of this macrostate is $E_q$.

$$$\large N_q=\sum \limits_i n_{qi}\hspace{1.5cm}\large E_q=\sum \limits_i n_{qi} \epsilon_i$$$

Here $n_{qi}$ are occupation numbers that specify how many bosons occupy microstate $i$ in macrostate $q$. An arbitrary number of bosons can occupy each microscopic quantum state.

$$$\large n_{qi} \in 0,1,...,\infty$$$

To calculate the thermodynamic properties, the grand canonical partition function $Z_{gr}$ is constructed.

$$$\large Z_{gr} = \sum \limits_q \exp \left( \frac{\mu}{k_BT}\right)^{N_q}\exp \left( \frac{-E_q}{k_BT}\right)=\sum \limits_q \exp \left( - \frac{E_q-\mu N_q}{k_BT}\right)$$$

Here $\mu$ is the chemical potential, $k_B$ is Boltzmann's constant, and $T$ is the absolute temperature. The energy $E_q$ and particle number $N_q$ of the macrostates can be expressed in terms of the microscopic states.

$$$\large Z_{gr} = \sum \limits_q \exp \left( -\frac{\sum \limits_i n_{qi}(\epsilon_i-\mu)}{k_BT}\right)=\sum \limits_q \prod \limits_i \exp \left( \frac{-n_{qi}(\epsilon_i-\mu)}{k_BT}\right)$$$

The index $q$ runs over all macrostates. The lowest energy macrostate has zero bosons in all of the microstates. There are macrostates with just one boson in one microstate and there are macrostates with many bosons in each of several microstates. The sum over all possible macrostates can be written as the sum over all possible microstates.

$$$\large Z_{gr} = \sum \limits_{n_{1}=0}^\infty \sum \limits_{n_{2}=0}^\infty \cdots \sum \limits_{n_{max}=0}^\infty \prod \limits_i \exp \left( \frac{-n_{i}(\epsilon_i-\mu)}{k_BT}\right)$$$

Here $n_{i}$ are the number of bosons in energy state $\epsilon_i$. The $n_i$ are independent of each other and the factors can be pulled through the sums,

$$$\large Z_{gr}= \left[\sum_{n_{1}=0}^{\infty}\exp\left(-\frac{n_{1}(\epsilon_1-\mu)}{k_BT}\right)\right]\left[\sum_{n_{2}=0}^{\infty}\exp\left(-\frac{n_{2}(\epsilon_2-\mu)}{k_BT}\right)\right]\cdots \left[\sum_{n_{max}=0}^{\infty}\exp\left(-\frac{n_{max}(\epsilon_{max}-\mu)}{k_BT}\right)\right]$$$

All of the factors in square brackets are the same. This can be rewritten as a product where the indices are not necessary anymore,

$$$\large Z_{gr} = \prod_i \left[ \sum_{n=0}^{\infty}\exp \left( -\frac{n(\epsilon_i-\mu)}{k_BT} \right) \right].$$$

At low particle densities and high temperatures, the average number of bosons in all of the microscopic states will be less than one. Thus the chemical potential will be less than the lowest microscopic energy level, $\mu<\epsilon$. In this case, the sum over $n$ is a geometric series "$1+x+x^2+\cdots=\frac{1}{1-x}$" where $x=\exp\left(- \frac{\epsilon_i-\mu}{k_BT} \right) < 1$. Summing this series yields,

$$$\large Z_{gr} = \prod_i \left[ 1-\exp\left( -\frac{\epsilon_i-\mu}{k_B T} \right) \right]^{-1}.$$$

As the temperature decreases or the particle density increases, the chemical potential will increase. As $\mu\rightarrow \epsilon_0$ the occupation of the lowest energy level becomes large and the bosons undergo a Bose-Einstein condensation. Here, we will only consider the case where $\mu<\epsilon_0$.

The thermodynamic grand potential can be determined from the grand canonical partition function, $\Phi = U-TS-\mu N= -k_BT \ln(Z_{gr})$. Here $U$ is the internal energy, $S$ is the entropy, and $N$ is the average number of bosons in the system.

$$$\large \Phi = -k_B T \ln(Z_{gr}) = k_B T \sum_i \ln \left[ 1-\exp\left( \frac{\mu-\epsilon_i}{k_BT} \right) \right]$$$

This sum can be approximated by an integral over the density of states $D(E)$. The density of states is defined per unit volume $V$ so the grand potential density is,

$$$\large \phi = \frac{\Phi}{V} = k_B T \int \limits_{-\infty}^{\infty} D(E) \ln \left[ 1-\exp \left( \frac{\mu-E}{k_BT} \right) \right] dE$$$

The particle density is minus the derivative of the grand potential density with respect to the chemical potential.

$$$\large n = -\frac{\partial \phi}{\partial \mu} = k_B T \int \limits_{-\infty}^{\infty} D(E) \frac{1}{1-\exp\left( \frac{\mu-E}{k_BT} \right)}\exp\left( \frac{\mu-E}{k_BT} \right)\cdot \frac{1}{k_BT} dE = \int \limits_{-\infty}^{\infty}D(E)\underbrace{ \frac{1}{\exp \left( \frac{E-\mu}{k_BT} \right)-1}}_{F_{BE}(E)}dE$$$

In this expression we recognize the Bose-Einstein function $F_{BE}(E)=\frac{1}{\exp(E-\mu/k_BT)-1}$ which tells us the mean number of bosons in a state with energy $E$. The Bose-Einstein function was derived here for the case of non-interacting bosons. If there are interactions between the bosons then the Bose-Einstein function does not accurately describe the probablility that a state at a given energy is occupied.

Other thermodynamic quantities can then be calculated used the standard formulas of thermodynamics. Some of these quantities are listed below. Each quantity is expressed as an integral over the density of states.

 Helmholz free energy density: $$\large f=\phi+\mu n = \int \limits_{-\infty}^{\infty} D(E) \left\{ k_BT\ln\left[ 1-\exp\left( \frac{\mu-E}{k_BT} \right)\right] + \frac{\mu}{\exp \left( \frac{E-\mu}{k_BT} \right)-1} \right\} dE$$ Entropy density: $$\large s = -\frac{\partial \phi}{\partial T}=\frac{1}{T}\int \limits_{-\infty}^{\infty}D(E)\left\{ -k_BT \ln \left[ 1-\exp\left( \frac{\mu-E}{k_BT} \right) \right]+\frac{E-\mu}{\exp\left( \frac{E-\mu}{k_BT} \right)-1} \right\}dE$$ Internal energy density: $$\large u = f+Ts = \int \limits_{-\infty}^{\infty} \frac{E D(E)}{\exp\left( \frac{E-\mu}{k_BT} \right)-1}dE$$ Specific heat: $$\large c_v=\frac{\partial u}{\partial T}=\int \limits_{-\infty}^{\infty}\frac{ED(E)(E-\mu)\exp\left(\frac{E-\mu}{k_BT}\right)}{k_BT^2\left[\exp\left(\frac{E-\mu}{k_BT}\right)-1\right]^2}dE$$

An important application of these equations is to the case of photons in vacuum. The number of photons is not conserved. This means that in a cannonical ensemble where the volume, the temperature, and the particle number are held constant, a change in the photon number does not change the free energy.

$$\large \left.\frac{\partial F}{\partial N}\right|_{T,V}=0=\mu.$$

This gives us the important result that the chemical potential of a system of photons is zero. For the case that $\mu=0$, the buttons below can be used to calculate the thermodynamic properties from the density of states.

Photons:

Phonons: