Turn an occupation probability into a count using the state measure

Define the state and evidence

The Fermi occupation probability for one state is f(E)=1/[1+exp((E−mu)/(kBT))]. It lies between zero and one and equals one half at E=mu. Occupation probability alone is not a particle count. To count occupied states in an energy interval, multiply by the density of states D(E) and integrate over energy.

Derive the quantitative relation

This original bounded packet supplies a constant DOS of 2 states/eV on a finite band from 0 to 4 eV, including spin. Thus only eight states are available in the continuous DOS approximation. It is deliberately not the square-root DOS of a three-dimensional free-electron metal. With theta=kBT, integration gives N=2theta times (ln[1+exp(mu/theta)]−ln[1+exp((mu−4)/theta)]).

Test a changed model or boundary

Energy needs an additional factor E: U=integral0to4 2E f(E)dE. The workbench evaluates this integral with composite Simpson weights 1,4,2,…,4,1 at 160 and 320 panels and reports their signed difference. A resolution difference is a convergence diagnostic, not a rigorous error bound. A theorem-based bound would require a suitable fourth-derivative bound over the full interval.

Limit the inference

Chemical potential is independently controlled here, so changing temperature can change particle number. A fixed-density material problem would instead solve the count equation for mu at each temperature. At the special symmetric setting mu=2 eV, f(E)+f(4−E)=1, so N=4 at every positive temperature in this finite-band model. This symmetry does not make the total energy temperature independent.

Worked original case

At mu=2 eV and theta=0.5 eV, the eight-state band is half occupied: N=4 exactly by symmetry. Occupation at E=mu is 0.5, whereas occupied state density there is 1 state/eV. Integrating only f would miss the supplied DOS factor and yield the wrong count units.

Investigation

Derive the analytic count integral and check it against numerical integration. Compare 160 and 320 panels for energy, labeling the difference honestly. Change mu and theta separately and distinguish a chemical-potential-controlled calculation from a fixed-particle-number problem.

Changed-case practice

Try the changed case before feedback

For the same constant DOS2 states/eV but symmetric band0 to6 eV and mu=3 eV, find N at positive temperature.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and virtual evidence checked 31 October 2026 against the primary explanation. All questions and numerical packets are original. Supplied theoretical distributions and synthetic records are not actual laboratory measurements. No nuclear, electrical, heating, beam or material-handling procedures are requested. Retain first reasoning, actual criterion critique, revision and a separately assigned unfamiliar task in the course evidence workspace; public keys do not authenticate mastery.

The numerical quadrature rule is independently sourced to OpenStax Calculus II, numerical integration.