Editing Negative temperature (section)

=== Noninteracting two-level particles ===
{{multiple image
   | align     = right
   | direction = vertical
   | footer    = Entropy, thermodynamic beta, and temperature as a function of the energy for a system of {{mvar|N}} noninteracting two-level particles.
   | width1    = 235
   | image1    = Entropy_vs_E_two_state.svg
   | width2    = 235
   | image2    = Beta_vs_E_two_state.svg
   | width3    = 235
   | image3    = Temperature_vs_E_two_state.svg
  }}{{Unreferenced section|date=July 2024}}
The simplest example, albeit a rather nonphysical one, is to consider a system of {{mvar|N}} particles, each of which can take an energy of either {{math|+''ε''}} or {{math|−''ε''}} but are otherwise noninteracting.  This can be understood as a limit of the [[Ising model]] in which the interaction term becomes negligible.  The total energy of the system is

:<math>E = \varepsilon\sum_{i=1}^N \sigma_i = \varepsilon j</math>

where {{mvar|σ<sub>i</sub>}} is the sign of the {{mvar|i}}th particle and {{mvar|j}} is the number of particles with positive energy minus the number of particles with [[negative energy]].  From elementary [[combinatorics]], the total number of [[microstate (statistical mechanics)|microstates]] with this amount of energy is a [[binomial coefficient]]:

:<math>\Omega_E = \binom{N}\frac{N + j}{2} = \frac{N!}{\left(\frac{N + j}{2}\right)! \left(\frac{N - j}{2}\right)!}.</math>

By the [[fundamental assumption of statistical mechanics]], the entropy of this [[microcanonical ensemble]] is

:<math>S = k_\text{B} \ln \Omega_E</math>

We can solve for thermodynamic beta ({{math|1=''β'' = {{sfrac|1|''k''<sub>B</sub>''T''}}}}) by considering it as a [[central difference]] without taking the [[continuum limit]]:

:<math>\begin{align}
  \beta &= \frac{1}{k_\mathrm{B}} \frac{\delta_{2\varepsilon}[S]}{2\varepsilon}\\[3pt]
        &= \frac{1}{2\varepsilon} \left( \ln \Omega_{E+\varepsilon} - \ln \Omega_{E-\varepsilon} \right)\\[3pt]
        &= \frac{1}{2\varepsilon} \ln \left( \frac{\left(\frac{N + j - 1}{2}\right)! \left(\frac{N - j + 1}{2}\right)!}{\left(\frac{N + j + 1}{2}\right)! \left(\frac{N - j - 1}{2}\right)!} \right)\\[3pt]
        &= \frac{1}{2\varepsilon} \ln \left( \frac{N - j + 1}{N + j + 1} \right).
\end{align}</math>

hence the temperature

:<math>T(E) = \frac{2\varepsilon}{k_\text{B}}\left[\ln \left( \frac{(N + 1)\varepsilon - E}{(N + 1)\varepsilon + E} \right)\right]^{-1}.</math>

This entire proof assumes the microcanonical ensemble with energy fixed and temperature being the emergent property.  In the [[canonical ensemble]], the temperature is fixed and energy is the emergent property.  This leads to ({{mvar|ε}} refers to microstates):
:<math>\begin{align}
  Z(T) &= \sum_{i=1}^N e^{-\varepsilon_i\beta}\\[6pt]
  E(T) &= \frac{1}{Z}\sum_{i=1}^N \varepsilon_i e^{-\varepsilon_i\beta}\\[6pt]
  S(T) &= k_\text{B}\ln(Z) + \frac{E}{T}
\end{align}</math>

Following the previous example, we choose a state with two levels and two particles.  This leads to microstates {{math|1=''ε''<sub>1</sub> = 0}}, {{math|1=''ε''<sub>2</sub> = 1}}, {{math|1=''ε''<sub>3</sub>  = 1}}, and {{math|1=''ε''<sub>4</sub> = 2}}.
:<math>\begin{align}
  Z(T) &= e^{-0\beta} + 2e^{-1\beta} + e^{-2\beta}\\[3pt]
       &= 1 + 2e^{-\beta} + e^{-2\beta}\\[6pt]
  E(T) &= \frac{0e^{-0\beta} + 2 \times 1e^{-1\beta} + 2e^{-2\beta}}{Z}\\[3pt]
       &= \frac{2e^{-\beta} + 2e^{-2\beta}}{Z}\\[3pt]
       &= \frac{2e^{-\beta} + 2e^{-2\beta}}{1 + 2e^{-\beta} + e^{-2\beta}}\\[6pt]
  S(T) &= k_\text{B}\ln\left(1 + 2e^{-\beta} + e^{-2\beta}\right) + \frac{2e^{-\beta} + 2e^{-2\beta}}{\left(1 + 2e^{-\beta} + e^{-2\beta}\right)T}
\end{align}</math>

The resulting values for {{mvar|S}}, {{mvar|E}}, and {{mvar|Z}} all increase with {{mvar|T}} and never need to enter a negative temperature regime.