Luiza Angheluta
<aside> đź“Ś Classical versus quantum statistics: Counting of microstates
Recall that in classical (Boltzmann ) statistics, we count all accessible microstates of one particle weighted by the Boltzmann factor to find the 1*-particle partition function*
$$ Z_1(T) = \sum_{\epsilon_s} e^{-\beta \epsilon_s} \qquad (1) $$
For non-interacting, identical and distinguishable particles, the partition function of N-particles, $Z_N$ factorises into products of $Z_1$
$$ Z_N (T) = Z_1^N,\qquad (2) $$
This is the case for spins localised on a lattice (review paramagnetic spin model).
However, if the identical particles are indistinguishable, we need to divide the partition function by their number of permutations ($N!$), thus
$$ Z_N (T) = \frac{Z_1^N}{N!}\qquad (3) $$
This is case for classical gasses since they are not localised in space and impractical to track their individual trajectories. Because of this N factorial, the partition function is reduced by the particle permutations, which ensures that the thermodynamic potentials (e.g. free energies, entropy) remain extensive state variables. Otherwise, we are dealing with the *Gibbs paradox.* Review classical ideal gas on your own.
This problem of “classical” indistinguishability is circumvented in quantum statistics by a different way of counting microstates. For each quantum state with energy $\epsilon$, we sum over all number of particles that can occupy that state. In this way, we always treat particles in clumps with no need to "label" each of them. This apparently simple switch in the order in which we do the sum has profound implications.
In many cases, the energy level is degenerate, i.e. there are many quantum states with the same energy. This degeneracy is measured by the density of states per energy level $D(\epsilon)$ which can be computed from the Hamiltonian and depends on $\epsilon$. This is discussed in Lecture 19.
In this lecture we focus on describing the statistics of one quantum state of a given energy $\epsilon$. In this case, the partition function is the sum over all number of particles that can occupy this quantum state. Notice that we are not fixing the number of particles. This means that we have changed to the grand-canonical ensemble by controlling instead the chemical potential, $\mu(T)$. Each configuration of N particles in the same quantum state is weighted by the Gibbs factor $e^{-\beta n_\epsilon(\epsilon-\mu)}$ thus the partition function for one state is
$$ Z_\epsilon(T,\mu) = \sum_{n_\epsilon} e^{-\beta n_\epsilon(\epsilon-\mu)} \qquad (4) $$
This partition function determines the probability of occupying a specific quantum state with energy $\epsilon$ at a given temperature $T$ and chemical potential $\mu$. This determines the average occupation number, which is a crucial quantity for the link to thermodynamic properties.
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<aside> đź’ˇ The basic strategy in solving problems involving bosons and fermions is to make use of the corresponding average occupation number to compute the total number of particles N as function of T and $\mu$, and thus the equation of state $\mu(T,N)$. This is given by the general expression of the number of particles an ideal Bose/Fermi gas
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$$ N(T,\mu) = \int_0^\infty d\epsilon \frac{D(\epsilon)}{e^{\beta(\epsilon-\mu)} \pm 1} \\ = \frac{C_d}{2} \left(\frac{2mL^2}{\hbar^2\pi^2}\right)^{d/2} \int_0^\infty d\epsilon \frac{\epsilon^{(d-2)/2}}{e^{\beta(\epsilon-\mu)} \pm 1}\\ = \frac{C_d}{2} \left(\frac{2m}{\hbar^2\pi^2}\right)^{d/2} V \int_0^\infty d\epsilon \frac{\epsilon^{(d-2)/2}}{e^{\beta(\epsilon-\mu)} \pm 1} $$
<aside> đź’ˇ which means that we can write a general expression for the particle density
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$$ \rho(T,\mu) = \frac{C_d}{2} \left(\frac{2m}{\hbar^2\pi^2}\right)^{d/2} \int_0^\infty d\epsilon \frac{\epsilon^{(d-2)/2}}{e^{\beta(\epsilon-\mu)} \pm 1} $$
<aside> đź’ˇ Evaluating the chemical potential relative to the thermal energy and the ground state, we can find asymptotic behavior in the high/low T limits for problems where the general integral is not readily solvable analytically.
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