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Stato Solido Capitoli:
Appendici |
QuantumStatisticsThe syllabus includes extensive classical mechanics and statistical mechanics, with glimpses at their quantum counterpart e.g. here. The main problem is that when you are doing statistical mechanics properly you do not really know what quantum means, therefore you may miss the point. A similar embarrassment emerges from the actual historical development of the subject: for instance Bose and Einstein, as well as Max Planck, were guessing what made the classical Maxwell-Boltzmann approach fail in certain conditions (solids, atoms, ratiation), but there was no quantum paradigm to apply yet. Here we assume that you have already taken your QM course, so you know what a boson or fermion is. Namely a particle of integer or semi-interger spin, respectively, that you are willing to treat as non interacting with other particles of the same kind, at least in first approximation. The first definition implies no restriction on the number of particles occupying a given eigenstate, whereas the second implies the Pauli exclusion principle, at most one particle per non degenerate eigenstate. On that basis we want to derive two important tools, the average number of particles, for both bosons and fermions, also known as the Bose-Einstein and Fermi-Dirac distributions. But we want to stress the physical meaning of the tool we are describing, assuming that that part is not yet totally familiar. The easy way is in the grand canonical ensemble. Bose-Einstein distributionBosons eigenvalues are {$E_l=\hbar \omega_l$}. Think of phonons, with distinct eigenvalues {$\omega_l(\mathbf k)$} for each different branch, {$l$}, and each different wavevectors, {$\mathbf k$}, or photons with just one eigenvalue {$\hbar c |\mathbf k|$} per wavevector per polarization. We assume that each eigenvalue is represented by a grand canonical ensemble, i.e. with variable number of particles {$n_l$}, called occupation number, fluctuating as dictated by a common chemical potential {$\mu$}. We want to compute the average number of bosons at a given temperature {$T=1/k_B\beta$}as {$$\sum_l \langle n_l\rangle = \frac 1 \beta \frac {\partial}{\partial \mu} \log Z$$} where {$Z$} is the full grand partition function. Since occupation numbers are independent of each other, the full grand partition function is the product of the partition function for each eigenvalue, {$Z_l= \sum_{n_l}e^{-\beta n_l (E_l-\mu)}$} {$$Z=\prod_l\sum_{n_l}e^{-\beta n_l (E_l-\mu)}$$} Assuming that {$E_l-\mu>0$} for every eigenvalue {$E_l$}, and that {$n_l$} ranges from 0 to {$\infty$} the sum of the geometrical series is {$ \sum_{n_l}(e^{-\beta (E_l-\mu)})^{n_l} =(1-e^{-\beta(E_l-\mu)})^{-1}$}, hence {$$Z = \prod_l \frac 1 {e^{-\beta(E_l-\mu)}}$$} Hence, from the definition {$$\sum_l \langle n_l\rangle = \frac 1 \beta \frac {\partial}{\partial \mu} \sum_l \log \frac 1 {1-e^{-\beta(E_l-\mu)}}$$} The calculation gives the Bose Einstein distribution for each eigenstate {$$\langle n_l \rangle = \frac 1 {e^{\beta(E_l-\mu)}-1}$$} In the case of phonons, we pose {$\mu=0$}, {$E_l = \hbar \omega_l(\mathbf k)(n+1/2)$}. The reason why we set {$\mu=0$} is because phonons (or photons) are considered in thermal equilibrium with matter: energy is conserved only in conjuction with matter and the number of phonons (photons) is not conserved. An own chemical potential for the phonons (photons) would mean that there is a specific cost in energy for creating or destroying them, in conclusion a phonon (photon) temperature, and a related entropy, while matter could have a different temperature (cfr. existence of a spin temperature) |