5.2 The thermodynamic equilibrium

is defined by an effective zero-flux for all states, i.e. per time period the same number of electrons will flow into each state as are leaving the state. It is impossible to calculate this ”steady state” just from the matrix elements (5.1) taking into account all possible perturbations of the system. The number of particles and parameters is much too high. But especially this high number of particles serves the perfect solution for the problem, the statistical Thermodynamics:
For the system only macro states exist, which have the most probable microstates. This leads to the fundamental equation that the probability \(W_k\) for a microstate \(k\) is proportional to the Boltzmann factor

 \begin{equation*} W_k \propto \exp\left(- \frac{E_k}{kT} \right) \label{boltzmann_1} \end{equation*}(5.2)

If the energy \(E_{k}\) described as a sum of independent states, it is quite easy to use Eq. (5.2) for the mathematical analysis of the system. In the one electron approximation all electrons are independent. But there exists a coupling of all particles due to the constant number of electrons. To overcome this limitation we define the chemical potential \(\mu\). Eq. (5.2) changes to

 \begin{equation*} W_k \propto \exp \left(- \frac{\sum_i \left(E_i - \mu\right)}{kT} \right) = \prod_i \exp\left(- \frac{\left( E_i - \mu\right)}{kT} \right) \\ = \prod_i W_{k,i} \label{boltzmann_2} \end{equation*}(5.3)

where \(i\) sums up all one electron states of the macrostate. \(E_i\) is the energy of the microstate \(i\).
HINT: If a state is not occupied with an electron it will not add to the complete energy.


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© J. Carstensen (Quantum Mech.)