QuantumInfo.StatMech.ThermoQuantities

Given a microcanonical Hamiltonian $H$ with parameters $d \in D$, the partition function $Z(\beta)$ for an inverse temperature $\beta \in \mathbb{R}$ is defined as the integral over the configuration space: $$Z(\beta) = \int_{c \in H.\text{dim}(d)} e^{-\beta E(c)} \, dc$$ where $E(c)$ is the energy of the configuration $c$. If the energy $E(c)$ of a configuration is infinite ($\top$), the integrand is treated as $0$; otherwise, it is given by the exponential of $-\beta$ times the finite value of the energy $E(c)$.

Given a microcanonical Hamiltonian $H$ and a degeneracy $d$, the partition function $Z(T)$ is defined as a function of the absolute temperature $T$. It is calculated by evaluating the $\beta$-dependent partition function $Z(\beta)$ at the inverse temperature $\beta = \frac{1}{T}$.

For a given Hamiltonian $H$ and measure $d$, the internal energy $U$ is a function of the thermodynamic beta $\beta$ (where $\beta = \frac{1}{k_B T}$). It is defined as the negative derivative of the natural logarithm of the partition function $Z(\beta)$ with respect to $\beta$: \[ U(\beta) = -\frac{d}{d\beta} \ln(Z(\beta)) \] where $Z(\beta)$ is the partition function of the micro-Hamiltonian system.

For a given Hamiltonian $H$ with density of states $d$, the Helmholtz free energy $A$ (also denoted $F$) is defined as a function of the absolute temperature $T \in \mathbb{R}$ by the expression $$A(T) = -T \ln Z(T)$$ where $Z(T)$ is the partition function `PartitionZT` of the system at temperature $T$.

Given a microscopic Hamiltonian $H$ and a dimension $d$, the entropy $S(T)$ is defined as the negative derivative of the Helmholtz free energy $A(T)$ with respect to the absolute temperature $T$. That is, $$S(T) = -\frac{dA}{dT}$$ where both $A$ and $S$ are functions of the temperature $T \in \mathbb{R}$.

For a given Hamiltonian $H$ and measure $d$, the entropy $S$ as a function of the inverse temperature $\beta$ is defined as $$S(\beta) = \ln(Z(\beta)) + \beta U(\beta)$$ where $Z(\beta)$ is the partition function (`PartitionZ`) and $U(\beta)$ is the internal energy (`InternalU`).

Let $H$ be a microcanonical Hamiltonian and $d$ be a parameter. The Hamiltonian is said to be $Z$-integrable at a given inverse temperature $\beta \in \mathbb{R}$ if the following two conditions hold: 1. The function mapping a configuration $x \in H.\text{dim}(d) \to \mathbb{R}$ to the Boltzmann factor $e^{-\beta E(x)}$ is Lebesgue integrable. Here, the energy $E(x)$ is defined by the Hamiltonian $H$. If the energy is infinite ($E = \top$), the value is taken to be $0$; otherwise, it is $e^{-\beta \cdot \text{val}(E)}$. 2. The partition function $Z(\beta)$, defined as the integral of this Boltzmann factor, is non-zero.

Let $H$ be a micro-Hamiltonian and $d$ be a measure. The property `PositiveβIntegrable` holds if for all positive inverse temperatures $\beta > 0$, the partition function $Z$ is integrable (denoted as `ZIntegrable`).

Let $H$ be a micro-Hamiltonian and $d$ be a given density or dimension parameter. For any inverse temperature $\beta \in \mathbb{R}$, if the partition function integral converges (i.e., $H$ is $Z$-integrable at $\beta$), then the partition function $Z(\beta)$ is smooth (of class $C^\infty$) at $\beta$.

Let $H$ be a microcanonical Hamiltonian and $d$ be a system parameter. For any temperature $T \in \mathbb{R}$ and inverse temperature $\beta \in \mathbb{R}$ such that $T \cdot \beta = 1$, if the partition function $Z$ is integrable at $\beta$ (denoted by `ZIntegrable`), then the entropy defined in terms of $T$ ($S(T)$) is equal to the entropy defined in terms of $\beta$ ($S_\beta(\beta)$).

{β : ℝ} (hi : H.ZIntegrable d β) : β = (deriv (H.EntropySβ d) β) / deriv (H.InternalU d) β

For a standard microcanonical system with state $d = (N, V)$, where $N \in \mathbb{N}$ is the number of particles and $V \in \mathbb{R}$ is the volume, the pressure $P(T)$ at temperature $T$ is defined as the negative partial derivative of the Helmholtz free energy $A$ with respect to the volume: $$P(T) = -\left. \frac{\partial A(N, V', T)}{\partial V'} \right|_{V'=V}$$ where $A$ is the Helmholtz free energy associated with the Hamiltonian $H$.