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From a QFT point of view, the difference between bosons and fermions is that their creation/annihilation operators ($a^{\dagger}$, $a$ and $c^{\dagger}$, $c^{\dagger}$ respectively) obey the following relations:

$$ [a_i, a_j^{\dagger}] = \delta_{ij},$$

$$ \{c_i, c_j^{\dagger}\} = \delta_{ij}.$$

How can I derive/relate this microscopic and single-particle view, with the statistical mechanical Bose-Einstein and Fermi-Dirac distributions?

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For a single species of boson/fermion with no interactions, the Hamiltonian is \begin{align} H &=\sum_k \omega_k a_k^\dagger a_k \hskip1cm\text{(boson)} \\ \\ H &=\sum_k \omega_k c_k^\dagger c_k \hskip1cm\text{(fermion)} \tag{1} \end{align} with \begin{align} a_ia_j - a_j a_i = 0 \hskip1cm a_ia_j^\dagger - a_j^\dagger a_i &=\delta_{ij} \hskip1cm\text{(boson)} \\ \\ c_ic_j + c_j c_i = 0 \hskip1cm c_ic_j^\dagger + c_j^\dagger c_i &=\delta_{ij} \hskip1cm\text{(fermion)}. \tag{2} \end{align} The vacuum state $|0\rangle$, with zero particles, satisfies \begin{align} a_k|0\rangle &=0 \hskip1cm\text{(boson)} \\ \\ c_k|0\rangle &=0 \hskip1cm\text{(fermion)} \tag{3} \end{align} for all modes $k$. Each application of $a_k^\dagger$ or $c_k^\dagger$ to the vacuum state creates a particle in mode $k$. The operator \begin{align} N_k &= a_k^\dagger a_k \hskip1cm\text{(boson)} \\ \\ N_k &= c_k^\dagger c_k \hskip1cm\text{(fermion)} \tag{4} \end{align} counts the number of particles in the $k$-th mode, because a state $|\psi\rangle$ that satisfies $$ N_k|\psi\rangle=n_k|\psi\rangle \tag{5} $$ has $n_k$ particles in the $k$-th mode. To see this, use equations (2) to deduce \begin{align} N_k a_j^\dagger &= a_j^\dagger (N_k+\delta_{jk}) \hskip1cm\text{(boson)} \\ \\ N_k c_j^\dagger &= c_j^\dagger (N_k+\delta_{jk}) \hskip1cm\text{(fermion)}. \tag{5b} \end{align} A state that satisfies $$ H|\psi\rangle=E_\psi|\psi\rangle \tag{6} $$ has total energy $E_\psi$.

The adjoint of the lower-left equation in (2) implies $(c_k^\dagger)^2=0$, so $n_k\in\{0,1\}$ for fermions. The boson version of equation (2) does not impose any such restriction, so $n_k\in\{0,1,2,3,...\}$ for bosons.

Here's how this is used in statistical mechanics. If the boson/fermion system is in thermal equilibrium with some other (unmodelled) system, then the expectation value of any observable $X$ associated with the boson/fermion system is $$ \rho(X) = \frac{1}{Z}\sum_\psi e^{-\beta E_\psi} \frac{\langle \psi|X|\psi\rangle}{ \langle \psi|\psi\rangle} \hskip2cm Z\equiv \sum_\psi e^{-\beta E_\psi} \tag{7} $$ where the sum is over states satisfying (6). For photons, the sum is over all states satisfying (6). For a system of matter bosons (or fermions), the sum is typically restricted to states with a given total number of particles.

The Bose-Einsten and Fermi-Dirac distributions are obtained by using (7) to calculate $\rho(N_k)$, the average occupation number in a given mode. This calculation can be done using the operator identity $$ H = \sum_k\omega_k N_k \tag{8} $$ to get $$ E_\psi = \sum_k\omega_k n_k, \tag{9} $$ where $E_\psi$ and $n_k$ are defined by equations (5)-(6). Use this in (7) to get $$ \rho(N_k) = \frac{\sum_\psi e^{-\beta E_\psi}n_k}{ \sum_\psi e^{-\beta E_\psi}} \tag{10} $$ which can also be written $$ \rho(N_k) = -\beta^{-1}\frac{\partial}{\partial\omega_k} \log Z \tag{11} $$ with the partition function $Z$ defined in (7) regarded as a function of the energy-coefficients $\omega_k$.

Derivations of the Bose-Einsten and Fermi-Dirac distributions in typical statistical-mechanics books, such as chapter 9 in Reif's Statistical and Thermal Physics, start with these ingredients:

  • equations (9) and (11), which are equations (9.2.1) and (9.2.5) in Reif, respectively;

  • the fact that $n_k$ is unrestricted for bosons, and restricted to $n_k\in\{0,1\}$ for fermions (because of equation (2), as mentioned above), which are equations (9.2.13) and (9.2.15) in Reif;

  • the constraint (if any) on the total number of particles $\sum_k n_k$, which is equation (9.2.14) and (9.2.16) in Reif. This constraint leads to the "chemical potential," usually denoted $\mu$.

The derivation from this point on is standard, so I won't repeat it here.

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